INTRODUCTION TO NUMBER ALGEBRAIC ANALYSIS

M.B.Abdullahi,FUK

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Complex AnalysisLimit is limit be it of real valued functions or complex valued function. To analyze C, its analyticity, which has to do with differentiation and the so called reverse of differentiation, integration I mean, should be known. Limit is the base or the flat-form of the aforementioned calculi. Differentiation as we will see, is the limit of a slope and integration, is the limit of Riemann Analogue Sum.In Pure Mathematical Courses like this one, we check differentiability, existence of limit and down the road using intuition. By intuition, understand algebraic not classical approach. To help you understand what I mean, rather than making the classical substitution of finding limit, we use 0<ε<1 and 0<δ<1. These, δ and ε , are Greek Letters to mean small positive numbers. That is, ε , δ∈(0,1). We will see that (0,1) is not countable much less ¿ z∨≤1, the interior of a unit circle. So, there are infinitesimally many values of ε and δ . Preferably, δ<ε, but not a must. The following definition, if you are able to decode it; you have finished up complex and real analysis.Definition: Let f (z) be a complex values function defined in a region except possibly at the point z=z0. Then the limit l of f (z) is said to exist if given ε>0 ,∃ δ>0(depending on ε) ∋ ¿ f ( z )−l∨¿ε whenever 0<|z−z0|<δ.Explanation: For f (z) to be a complex valued function, it must be single-valued f ( z )=±√ z, as an example; not multiple valued f (±√ z )=z, say. f (z) also has to be defined. If f ( z )=1

z , then at z=0 , f (z) is not defined. Thus, f (z) is not a complex function. The except possibly is from the fact that if z=z0, then z−z0=0 and this is contrary to saying 0<¿ z−z0∨¿. There are notions and expressions we need to underline in the definition. They are, with there explanations, as follows:Limit l of f (z) is said to exist: The definition is not finding the true or classical limit, but is telling you, the reader, that the limit of the function exist. The prior truth is helping ancient Greeks and olden generation mathematicians to guess existence of limit of a function, by just looking at the function. By looking at a function, I mean with no pen kissing a paper. If the function is less than another function that is less than ε , then the limit of the function exists.2

Given ε>0 ,∃ δ>0: Now 0<ε<1 and 0<δ<1. They are small positive numbers. For complex valued function δ , ε∈|z|≤1. We will see using Cantor Diagonalization Method that there are infinitely many values of δ and ε in (0 ,1). But ¿ z∨≤1 contains (0,1). Hence δ , ε in a complex plane have uncountably infinitesimmaly many values. Given ε>0 ,∃ δ>0 also mean ∀ ε>0 ,∃ δ>0.δ depending on ε: δ has a formula (expression) containing ε . But since both δ and ε are in the region |z|≤1. Then δ α ε. Direct proposion does not always mean equality. δ may not equal ε . But if δ=ε, this is perfect.|f (z )−l|<ε whenever 0<|z−z0|<δ: This ‘whenever’ is stronger than ‘when’. We will see that when we start existence of limit of a sequence (which is also a function). |f (z )−l|<ε implies and is being implied by 0<|z−z0|<δ. What is |z−z0|? It is

|x+iy−(x0+ i y0 )|=|(x−x0 )+i( y− y0)|=√(x−x0)2+( y− y0)

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This means 0<√(x−x0)

2+( y− y0)2<δ

That is, 02<(x−x0)2+( y− y0)

2<δ2

But δ 2 is even less than δ . Hence0<(x−x0)

2+( y− y0)2<δ, which is an annulus centred at (x0 , y0) outside a circle of radius 0, i.e. outside the origin and interior of a circle of radius δ . The circle of radius δ is a very small circle.

But that is not all. |f (z )−l|<ε is also the interior of the region of f (z) centred at l of radius ε . Why I said interior is because |f (z )−l|≤� ε.So as the annulus 0<|z−z0|<δ shrinks, the region |f (z )−l|<ε shrinks and vice-versa because of the ‘whenever’.Remarks: 0<|z−z0|<δ also means 0<z−z0<δ and 0<z0−z<δ. That is limit can be right handed and left handed. For 0<|z−z0|<δ, we are referring to limz→z0

+¿ ❑¿¿ and for

0<z0−z<δ, we are referring to limz→z0−¿❑¿

¿. z→ z0+¿¿ means right of z0 and z→ z0

−¿¿ means left of z0.We can take an example.Example3

(i) Show that limz→ i

f (z )=−1, where f ( z )={ z2 , z ≠i¿0 , z=i.

Solution:f ( z )={ z2 , z ≠i¿0 , z=i

means f ( z )={z2 , z>i∨z<i¿0 , z=i

So limz→z0

+¿ f (z )= limz→i+¿ z2=i2=−1

¿¿ ¿¿ and lim

z→z0−¿ f ( z )= lim

z→i−¿ z2=i2=−1¿ ¿¿

¿

Since limz→z0

+¿ f (z )=¿¿¿¿ lim

z→z0−¿ f ( z ) ,¿

¿then the limz→z0

f (z) exists and limz→ z0f ( z )=−1.

(ii) Let f :C→C be defined by f ( z )=2 z. Show that limz→z0

f (z)=2i.Solution: Let 0<|z−z0|<δ , i.e. 0<|z−i|<δ. Then we check ¿ f ( z )−l∨¿ if it is less than ε . |f (z )−l|=|f ( z )−2 i|=|2 z−2 i|=2∨z−i∨¿.Since |z−i|<δ, 2|z−i|<2δ. Hence,

|f (z )−2i|<2δ

Let 2δ=ε since preferably δ<ε. Hence,|f (z )−l|<ε where ε=2δ, as required.

(iii) Let f :C→C be defined by f ( z )=zsin 1z where |z|≤1. Show that lim

z→0f ( z )=0.

Proof. Let 0<|z−z0|<δ, i.e. 0<|z−0|<δ, i.e, 0<|z|<δ… (1). Then |f (z )−l|=|f ( z )−0|=|zsin 1z−0|=¿ zsin 1

z∨¿. From the Property of Absolute Value (

¿ab∨≤∨a∨¿b∨¿), ¿ zsin 1z∨≤∨z∨¿ sin 1z∨¿

Now, |f (z )−l|≤∨z∨¿ sin 1

z∨¿, but since sine is sinusoidal with highest independent variable 1 and lowest -1;

|sinθ|≤1

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Then |f (z )−l|≤|z||sin 1z|≤∨z∨¿ and |z|<δ.Thus, |f (z )−l|<δ. Let δ=ε since where δ comes from is where ε comes from.Thus, |f (z )−l|<ε, as required.

(iv) Letf :C→C be defined by f ( z )={¿ z−a∨ ¿z−a

, z≠ a¿

0 , z=a. Show that lim

z→af ( z ) does

not exist.Proof. From Tricotomy f ( z )={¿ z−a∨ ¿

z−a, z<a∨z>a¿

0 , z=a

limz→z0

+¿ f (z )= limz→ a+ ¿¿z−a∨ ¿

z−a= lim

z →a+¿ z−az−a= lim

z→a+¿1=1¿¿

¿¿¿¿¿¿

¿

Since limz→z0

+¿ f (z )=1≠−1= limz→ z0

−¿f ( z ) ¿¿¿

¿, then limz→z0

+¿ f (z )≠ limz→ z 0

−¿f (z ) ¿¿ ¿

¿. Hence, limz→z0

f (z) does not exist since it is not unique.This set us to prove the following theorem.Theorem. The limit of a complex function, if it exists, is unique.Proof. This proof is by reducio ad absurdum. That is a proof by contradiction. Suppose on the contrary that it is not unique, i.e. f (z) has loimits l1 and l2. Then|f (z )−l1|<ε whenever 0<|z−z0|<δ and|f (z )−l2|<ε whenever 0<|z−z0|<δ

Now,|l1−l2|=¿ l1−f ( z )+f ( z )−l2∨¿ since – f ( z )+ f ( z )=0

≤∨l1−f (z )∨+¿ f (z )−l2∨¿ from Triangular Inequality¿∨f ( z )−l1∨+¿ f ( z )−l2∨¿ Property of Absolute Value5

¿ε+ε=2 ε

That is, ¿ l1−l2∨¿2 ε. Since ε is a very small number, then 2 ε is also a very small number.This shows that the difference between l1 and l2 is insignificant (irrelevant). Hence ¿ l1−l2∨¿0 which implies l1=l2, certifying that l is unique.Q) Let f :C→C be defined by f ( z )=2 z2. Show that lim

z→z0f ( z )=2 z0

2.Proof. Let 0<|z−z0|<δ . Then

|f (z )−l|=|2 z2−2 z02|=2|z2−z02|=2∨z−z0∨¿ z+z0∨¿

¿2δ∨z+ z0∨¿ since ¿ z−z0∨¿ δ.Now, |z+z0|=¿ z+z0−z0+z0∨¿ since −z0+z0=0

¿∨z−z0+z0+ z0∨¿ since z0−z0=−z0+z0 ∵ commutativity¿∨z−z0+2 z0∨¿

≤|z−z0|+|2 z0|∵Triangular inequality¿1+2∨z0∨¿ since ¿ z−z0∨¿ δ which is less than 1.Thus,

|f (z )−l|<2δ|z−z0|<2δ ¿

Let 2δ (1+2|z0|)=ε. Thenδ= ε2+4∨z0∨¿¿ , i.e. δ is depending on ε and z0 which is also allowed. Hence,

¿ f ( z )−l∨¿ε, as required.Theorem. Limit forms an algebraic field.

(i) limz→ z0

( f (z )+g ( z ) )=limz→z0

f (z )+ limz→z0

g(z ).Proof. lim

z→z0( f (z )+g ( z ) )≡|f ( z )−l1|< ε

2 and |g ( z )−l2|< ε2 whenever 0<|z−z0|<δ.

Hence, ¿ f ( z )−l1∨+¿g ( z )−l2∨¿ ε

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But |f (z )+g ( z )−l1−l2|≤∨ f ( z )−l1∨+¿ g ( z )−l2∨¿ ε implies|f (z )+g ( z )−(l1+l2 )|<ε≡ lim

z→z 0( f (z )+g( z)).

ii.limz→z0

( f (z ) . g ( z ) )=limz→z0

f ( z ) . limz→z0

g(z )

Proof. limz→z0

( f (z )×g ( z ) )≡|f ( z )− l1||g ( z )−l2|<ε where |f (z )−l1|<√ε and |g ( z )−l2|<√ε

|( f ( z )−l1 ) (g ( z )−l2 )|≤∨f ( z )−l1∨¿g ( z )−l2∨¿

|f (z ) g ( z )−l2 f (z )−l1g ( z )+ l1 l2|<ε

l2 f ( z )+l1 g (z )=l2l1+l2l1 since f (z)→l1 and g ( z )→l2

Hence,|f (z ) g ( z )−2l2 l1+l1 l2|=¿ f (z ) g ( z )−l1l2∨≡ lim

z→z0( f ( z ) . g (z)) where l1l2=l3.

(ii) limz→z0

f (z )g (z)

=limz→z0

f (z )

limz→z 0

g (z)

Proof. limz→z0f (z )

limz→z0

g (z)≡ ¿ f ( z )−l1∨

¿

¿ g (z )−l2∨¿=¿ f ( z )−l1∨| 1g ( z )−l2|¿

¿

≤|( f (z )−l1) .1

g ( z )−l2|¿| f (z )g (z )−l2

−l1

g ( z )−l2|Now, |g ( z )−l2|<ε

g ( z )−ε< l2

| f (z )g (z )−ε

−l1

l2+ε|≤| f (z )g ( z )−l2

−l1

g ( z )−l2|<ε since – ε<g ( z )−l2<ε. l2−ε<g (z )<l2+ε

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| f (z )g (z )−ε

−l1

l2+ε|which implies | f ( z )g (z )

−l1l2|<ε ≡ limz→z0

f (z)g (z)

Corollaryi) lim

z→z0(cf (z))=c lim

z→z0f (z)

This is from the fact that limz→z0

¿¿ since limz→z0

( f (z )+g ( z ) )=limz→z0

f (z)+ limz→z0

g( z).ii) lim

z→z0( f (z )¿−g(z ))= lim

z→z0f ( z )−lim

z→z 0g(z)¿

This is also from limz→z0( f (z )¿+g(z))=lim

z→z0f ( z )+ lim

z→z0g( z)¿

limz→z0

( f (z )¿−g(z ))= limz→z0

¿¿¿

¿ limz→ z0

f ( z )+ limz→z0

(−g ( z ) )

limz→z0

f (z)+ limz→z0

−1×g(z )

From corollary (i).limz→z0

(−1×g ( z ) )=−limz→z0

g( z)

Hence, limz→z0( f (z )¿−g(z ))= lim

z→z0f ( z )−lim

z→ z 0g(z)¿

iii) limz→z0

¿¿=limz→z0

f 1( z)+ limz→ z0

f 2(z )+ limz→ z0

f 3( z)+…+ limz→z0

f n(z)

iv) limz→z0

¿¿=limz→z0

f 1( z)× limz→ z0

f 2(z )× limz→z 0

f 3(z )×…× limz→z0

f n(z)

Q) Find limz→ i ( z

3+6 z2+i2 z−i ).

Solution: Using the Theorem and Corollaries,limz→i ( z

3+6 z2+i2 z−i )=

limz→ i

❑z3+6 limz→ i

z2+ limz→i

i

2 limz→i

z−limz→ i

i

8

¿ i3+6 i2+i2 i−i

=−i−6+i2 i−i

=−6i

=6 i

Theorem. If f (z) is continuous, then limz→z0f (g ( z ) )=f ( limz→z 0

g(z)) .

Proof. Let limz→z0

g(z )=g(z0). Since f (z) is continuous, |f (z )−f ( z0 )|<ε whenever |z−z0|<ε

. Hence,|f (g (z ) )−f (g ( z0 ))|<ε, i.e. limz→z0

f ¿

Example: Find limz→ i

(z2+1)3.Solution: Since polynomials are all continuous, (z ')3 is continuous and limz→i

( z2+1 )3=( limz→ i(z2+1))3= ((−i)2+1)3

¿(−1+1)3=0

Infinite Limits and Limits at InfinityOur definition 2.1 of limit subsequently was for limz→z0

f ( z )=l, but we can have: limz→z0

f (z)=∞, limz→z0

f (z)=−∞ , limz→∞

f ( z)=l , limz→−∞

f (z )=l.The victims of lim

z→∞f (z )=l are the sequence, which we will see. Mind you, sequence are also functions only that with special domain.

Definition(limz→z0

f (z)=∞): ∀M>0 ,∃ δ>0∋ f ( z )>M whenever 0<¿ z−z0∨¿δ. M is no matter how large if not infinity. M is replacing our definition. So we suppose to have |f (z )−∞|<ε whenever 0<¿ z−z0∨¿δ. But ε∈(0,1), so ε is positive, i.e. f (z) is a little bigger than ∞. But we are not working on extended complex plane. So we use M, where M is no matter how large rather than ∞. Since M is large and δ is small, we have inverse proportion M α 1δ .

Definition(limz→z0

f (z)=−∞): ∀M>0(no matter how large),∃δ>0∋ f ( z )←M whenever 0<¿ z−z0∨¿δ.9

Here also we suppose to have |f (z )+∞|<ε whenever 0<|z−z0|<δ . So, we have: f ( z )←M .Definition(lim

z→∞f (z )=l): ∀ ε>0 ,∃M>0∋|f ( z )−l|<ε, ∀ z>M .

This one also we suppose to have |f (z )−l|<ε whenever |z−∞|<δ.Definition( lim

z→−∞f ( z )=l): Given ε>0 ,∃M>0∋|f ( z )−l|<ε ,∀ z←M .

Q) Let f :C→C be defined by f ( z )=1z . Show that lim

z→∞f ( z )=0.

Proof. limz→+∞

f (z )= limz→+∞

1z= 1

+∞=0and lim

z→−∞f ( z )= lim

z→−∞

1z= 1

−∞=0.

Hence, limz→+∞

f (z )= limz→−∞

f (z )=0.ContinuityLet f :C→C be a complex valued function ( f is defined and single valued). Then f is said to be continuous at z0 in a region if given ε>0 ,∃ δ>0∋|f ( z )−f ( z0 )|<ε whenever |z−z0|<δ.This is like saying f (z) converges to f (z0) rather than l. We will see its analogue under sequences. Cauchy Sequence is called!Continuity is at z0. The fact that |z−z0|<δ not 0<|z−z0|<δ means at z=z0 , f (z ) is defined. That is, we are having a disk rather than annulus.Q) Let f :C→C be a single valued function and defined. Then prove that f ( z )=2c is continuous in C.Proof. Since f ( z )=2c , f ( z0 )=2c and |f (z )−f ( z0 )|=|2c−2c|=|0|=0, by definition of ε; 0<ε. Thus,|f (z )−f ( z0 )|<ε whenever |z−z0|<δ.Q) Prove that f ( z )=2 z is continuous in a region of C.Proof. Let |z−z0|<δ. Then f ( z0 )=2 z0 and |f (z )−f ( z0 )|=|2 z−2 z0=2|z−z0∨¿<2δ since |z−z0|<δ. Hence, |f (z )−f ( z0 )|<ε, where ε=2δ.Theorem. f (z) is continuous at z0 if and only if the limit of f (z) is f (z0).10

Proof(Neccesity): Let f (z) is continuous at z0. Then ε>0 ,∃ δ>0∋|f ( z )−f ( z0 )|<ε whenever |z−z0|<δ. At z≠ z0, i.e. z−z0≠0 ,0<¿ z−z0∨¿. Hence 0<|z−z0|<δ. Thus, limz→ z0

f ( z )=f (z0).(Sufficiency):Let limz→z0

f ( z )=f (z0). Then, by definition 2.1,|f (z )−f ( z0 )|<ε whenever 0<|z−z0|<δ which implies ¿ z−z0∨¿ δ when z≠ z0.This means the statement “except possibly” may be impossible.Q) Let f :C→C be defined by f ( z )={sin zz , z≠0

¿1 , z=0. Show that f is continuous at 0.

Solution: If limz→0

f ( z )=00∨∞∞ , we use L’Hospital Rule (Differentiate the numerator and the denominator). So

limz→z0

f ( z )=limz→0

sin zz

=limz→0

cos z1

=cos0=1 and f (0 )=1.Hence, lim

z→0f ( z )= f (0). Thus, f is continuous.

Theorem. Continuity forms an algebraic field.Proof. Let f (z) and g(z ) and h(z ) are continuous at z=z0. Then

i. Closure(+): f +g is continuous.ii. Associatyivity(+): f +(g+h )= (f +g )+h shifting bracket is possible over continuous function.iii. There is a continous function f ( z )=0.iv. The additive inverse of f (z) is – f (z ) or f (−z).v. Closure(×): f ×g is continuous.vi. Associatyivity(×): f × (g×h )= (f ×g )×h shifting bracket is possible over continuous function.vii. There is a continous function f ( z )=1.viii. The additive inverse of f (z) is f−1(z) or f (z−1).Examples:i. Exponent is continuous.ii. All polynomials are continuous.iii. Sine and cosine are continuous.

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Theorem. A composite function of continuous function is continuous.Proof. This is similar to the proof of lim

z→ z0f (g ( z ) )=f ( lim

z→ z0g(z )).

Example1:Q) Let g ( z )=ez and f ( z )=zn+i. Then fog ( z )=g( f (z ))

¿ g(zn+i)

¿ez n+i is continuous because ez and zn+i are continuous.Q) Let f ( z )=cos z , g ( z )=ezand h ( z )=z3+3. Then

f (g (h (z ) ) )=¿

f (g ( z3+3 ))=¿

f (ez3+3 )=¿

cos ez3+3

which is continuous.Theorem(Boundedness Theorem). If f (z) is continuous on a closed and bounded region, then f (z) is bounded.Proof. Let f (z) is continuous, we show that f (z) is bounded, i.e. ∃M ∈C∋|f ( z )|<M . Since f (z) is continuous, |f (z )−f ( z0 )|<ε whenever |z−z0|<δ. But|f (z )−f ( z0 )|<|f ( z )−f ( z0)|<ε from Triangular Inequality.Hence

|f (z )|<|f ( z0 )|+ε

Since f (z0) is a particular number, then |f (z )|<M where M=max {f ( z ) ,|f ( z0 )|+ε }.Uniform ContinuityLet f :C→C be a complex valued function defined by f . Then f is uniformly continuous on I if ∀ ε>0 ,∃ δ>0∋|f ( z1 )−f ( z2 )|<ε whenever |z1−z2|<δ .

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z1 and z2 are two points of a region. Hence, uniform continuity is not at a point, but at the region.Examples:

1) Show that f :C→C defined on |z|≤1 by f ( z )=z2 is uniformly continuous.Proof. Let |z1−z2|<δ . Then |f ( z1 )−f ( z2 )|=|z12−z2

2|=¿ z1+z2∨¿ z1−z2∨¿

≤2∨z1−z2∨¿

Since ¿ z1∨+¿ z2∨≤∨z1+z2∨¿ and |z|≤1implies ¿ z1∨≤1, and |z2|≤1.Hence, |f ( z1 )−f ( z2 )|≤2|z1−z2|<2δ since |z1−z2|<δ .Thus, |f ( z1 )−f ( z2 )|<ε whereδ= ε

2 .2) Prove that f :C→C defined by f ( z )=1

z is not uniformly continuous in the region ¿ z∨¿1.Proof. |z|<1 contains uncountably infinite many complex numbers and real numbers (−1 ,0 )∪(0,1). Let z=δ since |z|<1 and |δ|<1. We need |z−z0|<δ. Let z0= δ

1+ε since we need |z−z0|<δ. Then

|z−z0|=|δ− δ1+ε|= ε

1+εδ<δ

Hence, |z−z0|<δ. But |1z− 1z0|=|1δ −1+εδ |= ε

δ .Let δ= ε

2 . Then εδ= εε2

=2 εε=2>ε .

Thus,|1z− 1

z0|>ε and |f (z )−f ( z0 )|>ε making f not uniformly continuous in the region.Sequence and Series of Complex NumbersA sequence uz of complex numbers is a function f : A⊆N→B⊆C.13

The idea of saying domain subset of natural numbers is from the fact that one cannot have -1st term or zeroth term or down the road. If A is finite, then uz is finite. If A is infinite, then uz is infinite.An example of a finite sequence is f : {1,2,3,4 }→C defined by f ( z )=zi and an example of infinite sequence is f :N→C defined by f ( z )= i

z .We are more interested in the infinite sequences.Definition. A sequence uz is said to be:

i. Increasing if uz≤uz+ 1,∀ z∈Nii. Decreasing uz≥uz+ 1 , ∀ z∈Niii. Constant uz=uz+1, ∀ z∈Niv. Strictly Increasing uz<uz+1 ,∀ z∈Nv. Strictly Decreasing uz>uz+1 ,∀ z∈N

(i)and (ii) are called monotone and (iv) and (v) are called strictly monotone. When we want to avoid the mention of increasing or decreasing regarding sequences, we say the sequence is monotone.Definition(Bounded Above Sequence). A sequence uz is said to be bounded from above if ∃M ∈C∋uz≤M , ∀ z∈N .Definition(Bounded Above Sequence). A sequence uz is said to be bounded from above if ∃M ∈C∋uz≥M , ∀ z∈N .Definition(Bounded Sequence). A sequence uz is said to be bounded from above if ∃M ∈C∋|uz|≤M ,∀ z∈N .Examples: The sequence uz=3 z is bounded from below by 1 and it is not bounded from above.The sequence uz=

iz is bounded from below by I and from above by 0.

Convergence of a SequenceA sequence uz is said to converge to a complex number l(or it has limit l) if ∀ ε>0 ,∃N∈N (N is no matter how large)∋|uz−l|<ε , ∀ z≥ N .That is, limz→∞

uz=l. Compare with the definition of limz→∞

f ( z )=l because a sequence is a function!14

N depends on ε and possibly z0. For uniform convergence, N depends on only ε . Since N depends on ε , and N is no matter how large and ε is a very small number; then N α 1ε . A sequence that does not converge, limn→∞

uz=∞, is said to diverge. We will treat divergence later.Examples

1) Prove that limz→∞

1z=0 , z>0.

Proof. uz=1z . |uz−l|=|1z−0|=|1z|<ε. Since z>0, 1z <ε implies 1ε <z. Let N=1

ε . Then N<z and z>Nand z≥ Nand ∀ z>−N is implied as required.2) Prove that lim

z→∞

1(1+ p)n

=0 , where z>0.Proof. Now (1+z)0=1

(1+z)1=1+z

(1+z)2=1+2 z+z2

Hence (1+z)2>1+2 z by dropping z2 where z>0.Thus, changing 2 with n, (1+z)n>1+nz. Since (1+z)n>1+nz , 1

(1+z)n< 11+nz which is

less than ε since n>0 and z>0.Thus1

1+nz<ε implies 1<ε (1+nz )=ε+nεz and 1−ε

εn<z implies 1εn− ε

εn<z which further

implies 1εn−1n<z. Set N= 1

εz−1n since N is no matter how large and ε>0 , z>0.

Hence, | 1(1+z )n|<ε implies ∀ z≥ N.

3) Show that limz→∞uz=1, where uz=

2 z−i2 z+i .

15

Proof. |uz−l|=|2 z−i2 z+i

−1|=|2 z−i−2 z−i2 z+i |=|−2i

2 z+i|= 2i2 z+i

<ε, i.e. 2 i<2 zε+εi, i.e. 2i−εi2 ε

< 2 zε2 ε , i.e. z> i

ε− i2.

Since iε is on the par bigger than i2, then we can call N= iε− i2 . Hence, |uz−l|<ε which implies ∀ z≥ N.

Hence, |uz−l1|<ε2, ∀n≥N and |uz−l2|<

ε2, ∀ z≥ N .

Hence, |l1−l2|=|l1+uz+uz−l2|≤∨l1+uz∨+¿uz−l2∨¿ ε2+ ε2

Thus, |l1−l2|<ε means the difference between l1 and l2 is irrelevant or insignificant.Hence, complex sequence converges to a unique limit.Theorem. Every convergent sequence is bounded.Proof. Suppose uz is convergent, then |uz−l|<ε ,∀ z ≥N . Since 0<ε<1 ,|uz−l|<1.|uz|−|l|<¿uz−l∨¿1 from Triangular Inequality.Thus, ¿uz∨−¿ l∨¿1, i.e. ¿uz∨¿1+¿ l∨¿. Let M=max {u1, u2 ,…,uz−1 , uz }=max ¿ since ¿uz∨¿1+¿ l∨¿ and z≥ N .

Therefore |uz|<M , where M∈C.Definition(Subsequence). Let uz is a sequence, the subsequence of uz is uzk where z1< z2<z3<….Examples:

1) Let uz=zi. Then uz=i ,2i ,3 i ,4 i ,…. Let vz={2 zi }. Then vz=2i ,4 i ,6 i ,… and vz is a subsequence of uz.2) Let uz=12iz and vz= 1

2 iz . Then vz is a subsequence of uz.3) uz+1 is a subsequence of uz.Theorem. Any subsequence of a convergent complex sequence converges to the same limit point on the super sequence.16

Proof. Let uzk is a subsequence of uz, let uz converges to l, i.e. |uz−l|<ε ,∀ z ≥N . Then |uzk−l|=|uz−uzk+uzk−l|≤∨uz−uzk∨+¿uzk−l∨¿ ε

2+ ε2=ε.

Theorem. Limit of complex sequence forms an algebraic field.i. lim

z→∞{uz+v z }=lim

z→∞uz+ lim

z→∞vz

ii. limz→∞

{uz−vz }= limz→∞

uz− limz→∞

v z

iii. limz→∞

{uz×v z }= limz→∞

uz× limz→∞

vz

iv. limz→∞

{uz÷v z }=limz→∞

uz÷ limz→∞

v z

Examples1) Use classical method to show that lim

z→∞

3 z2+5 z+4z2+10

=3.

Show. limz→∞

3 z2+5 z+4z2

z2+10z2

=limz→∞

3+ 5z+ 4z2

1+10z2

=3+ 5

∞+ 4∞2

1+ 10∞2

=3

2) Use algebraic method to show that limz→∞

3 z2+5 z+4z2+10

=3 , z>0.Proof. |3 z2+5 z+4z2+10

−3|=|3 z2+5 z+4−3 z2−30z2+10 |=|5 z−26z2+10 |=5 z−26z2+10< 5 zz2

=5z<ε, i.e. 5ε <z. Set

N=5ε , then |3 z2+5 z+4z2+10

−3|<ε, which implies the ∀ z≥ N.Theorem. If f (z) is continuous, then lim

z→∞f ( zn )= f ( lim

z→∞zn).

Proof. Let limz→∞zn=z0. Then

¿ zn−z0∨¿ δ ,∀ z ≥N which implies ¿ f ( zn )−f (z0)∨¿ ε since f is continuous.Examples:

1) Show that limz→∞

e2z2=1.

17

Proof. Since ez is continuous, limz→∞

e2z2=e

limz→ ∞

2z 2

¿e2∞2

¿e0=1

2) Show that limz→∞

sin( 3π2 + 32 z2 )=−1.

Proof. Since sin z is continuous on C, thenlimz→∞

sin(3π2 + 32 z2 )=sin( limz→∞

3 π2

+ limz→∞

32 z2 )

¿ sin( 3 π2 +0)¿−1

Divergent SequenceA sequence is said to be divergent if it does not converge. That is, its limit does not exist. When limz→∞

uz=∞∨−∞, uz diverges.A bounded sequence which does not converge is said to oscillate. An example is uz=i2 z.Definition. A sequence uz is said to diverge if ∀ A>0(A is no matter how large), ∃N>0 ( N is no matter how large) ∋ uz≥ A (¿uz≤−A ) , ∀ z≥ N. A is directly proporsional to N.Example: Show that uz=z diverges.Proof. Let N=A+1. Then uz=z ≥N because of the ∀ z≥ N

uz=z ≥N=A+1≥ A, i.e. uz≥ A , ∀ z≥ N.Definition(limit Superior). Let uz is a sequence of complex numbers which is bounded from above. Let

M n=lub {uz , uz+1 ,…}

18

Then M 1≥M 2≥M 3≥…. Hence M n is a decreasing sequence bounded from above. Since M n is not possibly bounded from below, M n may converge or diverge.Case I: If M n converges, limit superior of M n=lim

z→∞M n

Case II: If M n diverges, limit superior of uz=−∞= limz →∞

M n

Definition(limit Inferior). Let uz is a sequence of complex numbers which is bounded from below. LetM n=glb {uz , uz+1 ,…}

Then M 1≤M 2≤M 3≤…. Hence M n is an increasing sequence bounded from below. Since M n is not possibly bounded from above, M n may converge or diverge.Case I: If M n converges, limit inferior of M n=lim

z→∞M n

Case II: If M n diverges, limit inferior of uz=∞=limz→∞

M n

Examples1) For uz=(i)2 z , ∀ z>0; M n=glb{−1 ,1 }.Lim inf uz=−1 convergesLim sup uz=1 converges

SeriesLet uz be a sequence. Then ∑

z=1

∞

uz is called an infinite series associated with the sequence uz. ∑

z=1

n

uz is called a partial sum of ∑z=1

∞

uz. Let sz=∑z=1

n

uz, called partial sum where sz is a sequence of its partial sum.We will presently see that limn→∞

sn=s, then ∑z=1

∞

uz=s.If sz converges, ∑

z=1

∞

uz converges and if sz diverges, ∑z=1

∞

uz diverges. Just like we have convergence of sequences, we have convergence of series.Examples:19

1) Show convergence of ∑k=0

∞

zk, where z ,r>0.Proof. Let sn=∑

k=0

n

zk. Then sz=1+z+…+zk−1

z sz=z+z2+…+zk

z sz−sz=zk−1

sz( z−1)z−1

= zk−1z−1

if z>1

sz=1−zk

1−zif 0<z<1

Now, limz→∞

sz=∑k=0

∞

zk and limz→∞

sz=limz→∞

zk−1z−1

=∞∞

which is undefined. So,limz→∞

zn−1z−1

= k zk−1

1=∞, i.e. divergence.

limz→∞ ( 1−zk

1−z )=limz→∞

11−z

− z k

1−z= 1i−z

−0 since zk=( 1a )k

= 1ak and 1

a∞= 1∞

=0 from the fact that z<1, which is divergence.Q) Show that ∑

z=1

∞ 2 iz (z+1)

=2 i.Proof. ∑

z=1

∞ 2 iz (z+1)

= iz− iz+1

.Hence,

s1=( i1− i2 )

s2=( i1− i2 )+( i2− i

3 )s3=( i1− i

2 )+( i2− i3 )+( i3− i

4 )…

20

sz=( i1− i2 )+( i2− i

3 )+…+( iz−1

− iz )+( iz− i

z+1 )=i− iz+1

limn→∞

s z=limn→∞ (i− i

z+1 )=i− i∞+1

=i−0=i

Hence, ∑z=1

∞ iz (z+1)

=i.Theorem. The necessary condition for a series ∑

z=1

∞

az to converge is limz→∞az=0.

Proof. limz→∞

sz=∑z=1

∞

az=s, let ∑z=1

∞

az converges. Then sz converges and sz=a1+a2+…+az−¿

sz−1=a1+a2+…+az−1

sz−sz−1=az

limz→∞

(sz−s z−1)= limz→∞

sz− limz→∞

sz−1

¿ limz→∞

az=s−s=0=limz→∞

az

Q) ∑z=1

∞ iz converges since the sequence iz converges.

Q) ∑z=0

∞

( 12i )n converges since 0<r= 1

2i<1

Convergence Tests1) Comparison Test: If uz<v z and uz diverges, then vz diverges. If vz converges, then uz converges.

Q) Show that ∑1

∞ 15z+i

converges.Proof. ∑

1

∞ 15z+i

<∑1

∞ 15z

=∑1

∞

( 15 )zwith 0< 15=r<1.

21

Thus, it converges to 11−r= 1

1−1s

= 145

=54 .

2) Ratio Test: If limz→∞ |uz+1

uz|=ρ and ρ<1, then ∑ uz converges, otherwise ∑ uz

diverges.Q) Show that ∑

1

∞ z3z

converges.Proof. uz=

z3z

and uz+1=z+13z+1

= z+13z .3

Then uz+1uz

=

z+13.2z

z3z

= z+13 z

=13 ( z+1z )=13 (1+1z )

limz→∞

13 (1+ 1z )=13 (1+0 )=1

3

So ∑1

∞ z3 z

converges to 13 .Differentiation of Complex FunctionsThe word differentiation comes from the root word difference. Difference between two points in a plane mean slope between them. Limit of slope is differentiation. So the reason for starting this chapter with limit is this reason.

f ' ( z )= limz→z0

f ( z )−f (z0)z−z0

Let ∆ z=z−z0. Then: z=∆z+z0 and z→ z0 implies z−z0→0 which further implies ∆ z→0.

Hence, f ' ( z )=lim

z→z0

f ( z0+∆ z )−f (z0)∆ z

Analytic Function22

f (z) is said to be analytic, holomorphic or regular if its derivative exists.Cauchy-Riemann EquationA necessary condition that w= f (z )=u (x , y )+iv(x , y) be analytic in a region R is that, in R; u and v satisfy the Cauchy-Riemann Equations: dudx=dv

dy , dudy=−dvdx .

Harmonic FunctionThe function u+ivwhich satisfy d2u

dx2+ d2udy2

=0, and d2 vdx2

+ d2 vdy2

=0 laplace equations is called harmonic function.Let ∇=i d

dx+ j d

dy . Then ∇2=(i ddx

+ j ddy )( i ddx + j d

dy )= d2

dx2+ d2

dy2

So, harmonic functions f satisfy ∇ f =0.Theorem. Every analytic function is harmonic.Proof. The Cauchy-Riemann Equations:dudx

=dvdy

… (1), dudy=−dvdx …(2)

Differentiate (1) by dx and (2) by dy, we have:d2udxdy

=d2 vdy2

…(3) and d2udxdy

=−dvd x2

…(4)Subtracting (4) from (3), we have:

d2udx2

+ d2udy2

=0

Differentiate (1) by dy and (2) by dx, we have:d2udxdy

=d2 vdx2

…(5) and d2udxdy=−d2 v

d y2…(6)

Adding (5) from (6), we have:d2 vdx2

+ d2 vdy2

=0

23

Rules of DifferentiationLet f ( z ) , g ( z ) and h(z ) are analytic functions of z. Then the following differentiation rules (identical with those of elementary calculus) are valid.

1. ddz { f (z )+g(z )}= d

dzf ( z )+ d

dzg ( z )=f ' ( z )+g' (z)

2. ddz { f (z )−g (z)}= d

dzf ( z )− d

dzg ( z )=f ' ( z )−g' (z )

3. ddz { f (z ) g(z )}=g(z ) d

dzf (z )+ f (z ) d

dzg ( z )=g(z ) f ' ( z )+ f (z) g' (z)

4. ddz {cf ( z ) }=c d

dzf ( z )=c f ' ( z )

5. ddz { f (z )÷g (z)}=

g ( z ) ddz

f ( z )− f ( z ) ddz

g (z)

(g (z))2=

g (z ) f ' ( z )− f (z ) g' (z)(g (z))2

if g(z )≠0These rules are easily proved by first principle.Derivative of Elementary Functions

1. ddz

(c )=0, where c is a constant2. d

dzzn=n zn−1

3. ddz

ez=ez

4. ddz

az=az ln a

5. ddzcos z=−sin z

6. ddzsin z=cos z

7. ddztan z=sec2 z

Higher Order Derivative24

If w=f (z) is analytic in a region, its derivative is given by f '( z) or w’ or dwdz . If f '( z) is also analytic in the region, its derivative is denoted by f’’, w’’ or d

dz (dwdz )=d2wd z2

. Similarly, the nth derivative of f (z), if it exists, is denoted by f (n ) ( z ) ,w(n ) or dnw

dzn

where n is called the order of the derivative.Theorem. Suppose f (z) is analytic in a region R. Then so also are f ' ( z ) , f ' ' ( z ) ,…analytic in R, i.e, all higher derivatives exist in R.Examples: Find the derivative of w=f (z )=z3−2 z at the point where

i) z=z0ii) z=−1

Solution:i) f ' ( z0 )= lim

∆z→ 0

f ( z0+∆ z )−f ( z0)∆ z

= lim∇ z→0

(z0+∆z )3−2 ( z0+∆ z )− {z03−2 z0 }∆ z

¿ lim∆ z→0

z03+3 z0

2∆z+3 z0(∆z )2+(∆z )3−2 z0−2∆ z−z0

3+2 z0∆ z

¿ lim∆ z→0

3 z02+3 z0∆ z+(∆z )2−2

¿3 z02−2

In general, f ' ( z )=3 z2−2 ,∀ z.ii) f ' ( z=−1 )=3(−1)2−2=1

Example: Show that ( ddz ) z does not exist anywhere.Solution:f ' ( z )= lim

∆ z→0

f ( z+∆ z )−f (z)∆ z

if this limit exists independently of the manner in which ∆ z=∆ x+i ∆ y approaches zero. Then:

ddz

z= lim∆ z→ 0

z+∆z− z∆ z

= lim∆ x→0∆ y→0

x+iy+∆ x+i ∆ y−x+iy∆x+i ∆ y

25

¿ lim∆x→0∆ y→0

x−iy+∆ x−i ∆ y−(x− y )∆ x+i ∆ y

¿ lim∆ x→0∆∆ y→0

∆ x−i ∆ y∆ x+i ∆ y

If ∆ y=0, the required limit is lim∆ x→0

∆ x∆ x

=1.If ∆ x=0, the required limit is lim

∆ y→0

−i ∆ yi∆ y

=−1.Then, since limit depends on the manner in which ∆ z→0, the derivative does not exist, i.e; f ( z )=z is non-analytic anywhere.Example: Q) Find ddz ( 1+z1−z ).Solution:

dwdz = lim

∆ z→ 0

f ( z+∆ z )−f (z)∆z

¿ lim∆ z→0

1+( z+∆ z )1−( z+∆ z )

− 1+z1−z

∆ z

¿ lim∆ z→0

2(1−z−∆ z ) (1−z )

= 2(1−z )2

Indepent of the manner in which ∆ z→0, provided z≠1.b) The function f (z) is analytic for all finite values of z except z=1 where the derivative does not exist and the function is non-analytic. The point z=1 is a singular point of f (z).Example: Prove that u=e−x (xsin y− y cos y ) is harmonic.Solution:dudx

=(e−x ) (sin y )+(−e−x ) ( xsin y− ycos y )=e− xsiny−x e−x s iny+ y e−x cosy …(1)26

d2udx2

= ddx

(e−xsin y−xe− x siny+ y e−x cos y )=−2 e−xsin y+x e−x sin y− y e−x cos y

dudy

=e− x ( x cos y+ ysiny−cos y )=x e− xcos y+ y e−x sin y−e− xcos y

d2udy2

= ddx

(x e−x cos y+ y e−x siny−e− xcos y )=−x e−x siny+2e− x siny+ y e− xcos y…(2)Adding (1) and (2) yields d2u

dx2+ d2udy2

=0 and u is harmonic.Example: Prove that ddz ez=ez.Proof. Let w=e z, i.e. w=e x+iy=ex ( cosy+isin y )=u+iv. Then:u=ex cos y , v=ex sin y.Since dvdx=ex cos y=dv

dy and dvdx=ex sin y=−dudy , the Cauchy-Riemann Equations are satisfied. Then the required derivative exits andis equal to:

dvdx

+i dvdx

=−i dvdy

+ dvdy

=ex cos y+i exsin y=ez

Example: Prove that ddz sin z=cos z.Proof. Since sin z= e iz−e−iz

2i, we have:

ddzsin z= d

dz ( eiz−e−iz

2i )= 12 i

ddz

e iz− 12 i

ddz

e−iz

¿ 12e iz+ 1

2e−iz

¿cos z

Example: Find ddz (w3−3 z2w+4 ln z=0 ).Solution: Differentiating with respect to z, considering w as an implicit function of z, we have:27

ddz

(w3 )−3 ddz

( z2w )+4 ddz

( ln z )=0 or 3w2 dw

dz−3 z2 dw

dz−6 zw+ 4

z=0

Then solving for dwdz , we obtain dwdz

=6 zw−4

z3w2−3 z2

.Q) Find ddz (√sin z+√sin z+√sin z+…∞ ).Solution: Let w=√sin z+√sin z+√sin z+…∞ . Then w=√sin z+w and w2=sin z+w, i.e. w2−w=sin z. Then, 2w dw

dz−dw

dz=cos z.

dwdz

(2w−1 )=cos z. Hence, dwdz = cos z2√sin z+√sin z+√sin z+…∞−1

Complex IntegrationJust as we have limit of Riemann Sum, the Riemann Integral; we have limit of ∑m=1

n

f (zm)∆ zm, where ∆ zm=zm−zm−12 as ∮C

❑

f (z)dz, i.e.|∑m=1

n

f (zm)∆zm−∮C

❑

f (z )dz|<ε whenever ¿ zm−zm−1∨¿ ε

Example:∫0

1+i

z2dz=13z2 ¿ ]0

1+i=13(1+i)3=−2

3+ 23i

Example:∫−πi

πi

cos z dz=sin z ¿ ]−πiπi =2 sinπi

Example:∫8+ πi

8−3πi

ez2 dz=2 e

z2 ¿ ]8+πi

8−3 πi=2(e4−3πi2 −e

4+ πi2 )=0

Integration by the Use of the Path28

Let C be a piecewise smooth path represented by z=z (t), where a≤ t ≤b. Let f (z) be a continuous function on C. Then∮C

❑

f (z)dz=∫a

b

f (z (t )) dzdt

dt=∫a

b

(u+iv ) ddt

(x+iy )dt=∫C

❑

(udx−vdy )+i∫C

❑

(udy+vdx )

Integral of 1z Around the Unit Circle∮C

❑ dzz

=2πi (C is the unit circle, read counter clockwise).Proof. Let z (t )=cos t+i sin t=e¿.Differentiating, we have: dzdt =ie¿.Now, f ( z ( t ) )= 1

z (t)= 1e¿ =e−¿.

Using ∮C

❑

f (z)dz=∫a

b

f (z (t )) dzdt

dt , we have: ∫0

2π

f ( z (t ) ) dzdt

dt=∫0

2π

e−¿ i e¿ dt=i∫0

2π

dt=i (2π−0 )=2 πi

Integral of ∮C

❑

(z−z0)mdz

Let f ( z )=(z−z0)m , where m is the integer and z0 a constant.

Proof. Let z−z0=zr. Then z−z0=r e¿.Squaring both sides m times,(z−z0)

m=rme imt, dz=ir e¿dt and obtain∮C

❑

(z−z0)mdz=∫

0

2π

rmeimt ir e¿dt=irm+1∫0

2π

ei (m+1)tdt

By Euler Formula,i rm+1[∫0

2π

cos (m+1 ) tdt+ i∫0

2π

sin (m+1 )tdt ]29

If m=−1 , rm+1=1 ,cos0=1 , sin 0=0.If m≠−1, each of the two integrals is zero because we integrate over an interval of length 2π , equal to a period of sine and cosine. Hence, the result is

∮c

❑

(z−z0)mdz={ 2 πi ,m=−1

¿0 ,m≠−1∧integer

Cauchy Integral TheoremIf f (z) is analytic in a simply connected domain D, then for every simple closed path C in D, ∮

c

❑

f (z )dz=0.Proof.

∮C

❑

f (z )dz=∮C

❑

(udx−vdy )+i∮C

❑

(udy+vdx )

By Green Theorem,¿∬

R

❑

(−dvdx

+−dudy )dxdy+ i∬

Q

❑

(dvdy + dudx )dxdy=0

By Cauchy-Riemann Equations,¿∬

R

❑

(−dvdx

+−dudy )dxdy=0∧i∬

Q

❑

(dvdy + dudx )dxdy=0

Remark. In a multiple connected region,

∮C

❑

f (z )dz=0

Example is doubly connected∮C

❑

f (z )dz=∮C1

❑

f (z )dz−∮C2

❑

f (z )dz=0

30

Example: Find ∮C

❑ e z

z−2dz for any contour enclosing z0=2

.Solution: ∮C

❑ e z

z−2dz=2πi ez ¿ ] z=2

❑ =2πi e2≈46.4.Example: Find ∮

C

❑ z3−62 z−i

dz=∮C

❑ z3−62 z−i

dz

¿2πi [ z3−6 ]z= i2=2πi(( i2 )

3

−6)¿2πi(−i

8−6)= π

4−12πi

¿ π8−6πi

Evaluation of IntegralsLet f (z) be single-valued and analytic inside and on a circle C except at the point z=a chosen as the centre of C. Then ∮

C

❑

f (z )dz=2 πia−1 ,i.e. 2πi times a residue. If the function f (z) has many residues, then the integral will be 2πi times sum of residues.The general for residues is a−1=lim

z→a

1(k−1 )!

dk−1

dzk−1{(z−a)k f ( z)}.

Hence, ∮C

❑

f (z )dz=2 πi limz→a

1( k−1 ) !

dk−1

dzk−1{(z−a)k f (z)}

Examples1) Find the residues of f ( z )= z2−2 z

(z+1)2(z2+4) at all its poles in the finite plane.

Solution: f (z) has a double pole at z=−1 and simple poles at z=±2i.Res(z=−1)¿ lim

z→−1

11!

ddz {(z+1)2 z2−2 z

(z+1)2(z2+4) }31

¿ limz→−1

( z2+4 ) (2 z−2 )−( z2−2 z ) (2 z )(z2+4)2

¿−1425

Res( z=2 i )= limz→2 i {( z−2i ) z2−2 z

(z+1 )2 ( z−2 i) ( z+2i ) }¿ −4−4 i(2 i+1)2(4 i)

= 7−i25

2) Find the residues of f ( z )= ez

z (z+1) .

Solution: f ( z )= ez

z (z+1) has poles at z=0 and z=−1. Therefore,

Res(z=0¿=limz→0

zf ( z )=limz→0

ez

z+1=1∧¿¿

Res(z=−1¿= limz→−1

¿

3) Compute the residues at the singularities of f ( z )= cos zz2(z−π )3

.Solution: The function has a pole of order 2 at z=0 and a pole of order 3 at z=π.

Res ( z=0 )=limz→ 0

11!

ddz (z2 f (z ))=lim

z→0

ddz ( cos z( z−π )3 )

¿ limz→0 [− (z−π )sin z−3cos z

(z−π )4 ]=−3π 4

Res (z=π) = 12! limz→π

d2

dz2[(z−π)3 f (z)]=1

2limz→π

d2

dz2 [ cos zz2 ]¿ 12limz →π [ (6−z2 )cos z+4 z sin z

z4 ]=−(6−π 2)2π4

=π 2−62 π4

4) Evaluate 12πi∮C❑ ezt

z2(z2+2 z+2)dzaround the centre C with equation |z|=3.

32

Solution: The integrand ezt

z2(z2+2 z+2) has a double pole at z=0 and two simple

poles at z=−1± i, roots of z2+2 z+2=0. All these poles are inside C.Res( z=0 )=lim

z→0

11 !

ddz {z2 ezt

z2(z2+2 z+2) }=limz→ 0

( z2+2 z+2 ) (t ezt )−(ezt )(2 z+2)(z2+2 z+2)2

= t−12

Res( z=−1−i )= limz→−1−i {[z−(−1−i)] ezt

z2(z2+2 z+2)}= limz→−1+i {e

zt

z2 } limz→−1+i

z+1−iz2+2 z+2

¿ e(−1+i)t

(−1+i)2. 12 i

= e(−1−i)t

4

By the residue theorem,∮C

❑ ezt

z2( z2+2 z+2)dz=2πi(sum of residues)

¿2πi {t−12 + e(−1+i)t

4+ e(−1−i)t

4 }¿2πi {t−12 + 1

2e−t cos t}

That is 12πi∮C

❑ ezt

z2(z2+2 z+2)dz= t−1

2+ 12e−t cos t

5) Show that ∫−R

R x2

(x2+1 )2(x2+2x+2)+∫

τ

❑ z2

( z2+1 )2(z2+2 z+2)=7 π50 as R→∞.

Solution: The poles of z2

(z2+1)2 ( z2+2 z+2 ) enclosed by the contour C are z=i of order 2 and z=−1+i of order 1.Res( z=i )=lim

z→i

ddz {(z−i)2 z2

(z2+1)2 ( z2+2 z+2 ) }=9i−12100

Res( z=−1+i )= limz→−1+i

( z+1−i ) z2

( z2+1 )2 ( z2+2 z+2 )=3−4 i

25

Then 33

∮ z2

(z2+1)2 ( z2+2 z+2 )=2 πi{9i−12100

+ 3−4 i25 }=7π50 and

∫−R

R x2

(x2+1 )2(x2+2x+2)+∫

τ

❑ z2

( z2+1 )2(z2+2 z+2)=7 π50

Taking the limit as R→∞ and noting that the second integral approaches zero, we obtain the required result.6) Show that ∫

0

2π dθ3−2cosθ+sinθ

=π

Solution: Let z=eiθ. Then sinθ= eiθ−e−iθ

2 i= z−z−1

2 i and cosθ= eiθ+e−iθ

2= z+z−1

2, dz=iz dθ, so that

∫0

2π dθ3−2cosθ+sinθ

=∮C

❑dziz

3−2(z+z−1)

2+ z−z−1

2 i

¿∮C

❑ 2dz(1−2i ) z2+6 iz−1−2 i

Where C is the circle of unit radius with center at the origin.The poles of 2dz

(1−2 i ) z2+6 iz−1−2i are the simple poles

z=−6 i±√(6 i)2−4 (1−2 i)(−1−2 i)2(1−2 i)

=−6 i±4 i2(1−2 i)

=2−i , 2−i5

Only 2−i5 lies inside C.

Res(z=2−i5 )= lim

z→ 2−i5

( z−(2−i)5 )( 2dz

(1−2i ) z2+6 iz−1−2i )¿ lim

z→2−i5

22 (1−2i ) z+6 i

= 12i by L’Hospital Rule.

34

∮C

❑ 2dz(1−2 i ) z2+6 iz−1−2i

=2 πi( 12 i )=π, as required.7) Show that ∫

0

2π dθa+b sinθ

= 2 π√a2−b2

.Solution: Let z=eiθ. Thensinθ= eiθ−e−iθ

2 i= z−z−1

2 i and cosθ= eiθ+e−iθ

2= z+z−1

2, dz=iz dθ, so that

∫0

2π dθa+b sinθ

=∮C

❑dziz

a+b z−z−1

2 i

=∮C

❑ 2dzb z2+2aiz−b

Where C is the circle of unit radius with centre at the origin.The poles of 2dz

b z2+2aiz−b are obtained by solving b z2+2aiz−b=0 and are given byz=−2ai±√−4 a2+4 b2

2b=−ai±√a2−b2i

b

¿−ai+√a2−b2ib

,−ai−√a2−b2 ib

Only −ai+√a2−b2ib

lies inside C since|−ai+√a2−b2i

b |<1 by Conjugation and setting a>¿b∨¿.Res(z1=−ai+√a2−b2i

b)¿ lim

z→z1(z−z1)

2b z2+2aiz−b

¿ limz→z1

22bz+2ai

= 1b z1+ai

= 1√a2−b2i

by L’ Hospital Rule.∮C

❑ 2dzb z2+2aiz−b

=2πi( 1√a2−b2 i )= 2π

√a2−b2

Hence,35

∫0

2π dθa+b sinθ

= 2 π√a2−b2

8) Show that ∫0

∞ cosmxx2+1

dx= π2e−m ,m>0.

Solution: The integrand has simple poles at z=± i, but only z=i lies inside C.Res( z=i )=lim

z→i( z−i ) e imz

( z−i ) ( z+i )= e−m

2i

And,∫−∞

∞ cosmxx2+1

dx=2 πi( e−m

2 i )=π e−m

That is,∫0

∞ cosmxx2+1

dx= π2e−m ,m>0

QUESTIO PAPER1(a) Define a sequence of complex numbers. (b) Prove that lim

z→ i

3 z4−2 z3+8 z2−2 z+5z−i

=4+4 i. (c) Find the ddz ( z3−2 z ) at z=−1. (d) Use Green Theorem and Cauchy - Riemann Equation to show that ∮

C

❑

f (z )dz=0, where C is a closed path.2(a) Define convergence of a sequence. (b) Prove that lim

n→∞(an+bn )=lim

n→∞an+ lim

n→∞bn. (c) Show that f '( z), where f ( z )=z and z is the conjugate of z, is not differentiable. (d) Find the residues of ez

z( z+1).

3(a) Draw: (i)0<¿z−z0∨¿δ ; and (ii)¿ f ( z )−l∨¿ε. (b) Prove that f ( z )=z2 is uniformly continuous in the region ¿ z∨¿1. (c) If w2−3 z2w+4 ln z=0, find dwdz .36

(d) Find ∮C

❑ e z

z−2dz .

4(a) Why is the number of complex numbers in |z|<1 uncountable? (b) Prove that ∑

n=1

∞ zn

n(n+1) absolutely converges for ¿ z∨≤1.

(c) Prove that ddz sin z=cos z. (d) Define subsequence of a complex sequence.5(a) Define limz→z0

f ( z )=−∞. (b) Show that every Cauchy Sequence is bounded. (c) Show that ∫

−R

R x2dx(x2+1 )2 (x2+2 x+2 )

+i∫Q

Q y2

( y2+1 )2 ( y2+2 y+2 )=7 π50

as R ,Q→∞. (d) Find ddz (√ tan z+√ tanz+√ tanz+…∞).6(a) State the Boundedness Theorem. (b) Show that if a function is analytic, it is harmonic. (c) Evaluate ∫

0

∞ cosmxx2+1

dx. (d) Show that f ( z )=c is continuous.7(a) Define limit superior. (b) Prove that the sequence { 1

1+nz } is uniformly convergent to zero for all z such that ¿ z∨≥2. (c) Show that ∮C

❑

(z−z0)mdz={2 πi,m=−1

¿0 ,m≠−1 . (d) Relate |z−z0|<δ and n≥ N.MTH 3307(Complex Analysis I) Marking SchemeQ1a) Define a sequence of complex numbers.Answer: A sequence of complex numbers is a function whose domain is a subset of natural numbers and its range is a subset of complex numbers: f :N→C.Q1b) Prove that lim

z→i

3 z4−2 z3+8 z2−2 z+5z−i

=4+4 i.Answer: Given ε>0, we find δ>0 such that ¿ whenever 0<|z−i|<δ.Since z≠ i, we can write37

3 z 4−2 z3+8 z2−2 z+5z−i

=(3 z3−(2−3 i ) z2+(5−2i ) z+5 i ) ( z−i )

z−i¿3 z3−(2−3 i ) z2+(5−2i ) z+5iHence, given |3 z3−(2−3i ) z2+ (5−2 i ) z−4+i|<ε, we find δ>0

|3 z3−(2−3i ) z2+ (5−2 i ) z−4+i|=|z−i|||3 z3−(2−3i ) z2+(5−2 i) z−1−4 i||=¿|z−i||3 ( z−i+i )3−(2−3i ) z2+ (5−2 i ) ( z−i+i )−4+i|=¿

|z−i||3 ( z−i )2−(12 i−2 ) ( z−i )−10−6 i|=¿δ ¿

¿28δ=ε , δ= ε28

.Q1c) Find the derivative of z3−2 z at z=−1.Answer: f ' ( z0 )= lim

∆z→ 0

f ( z0+∆ z )−f ( z0 )∆ z

=lim∆ z→0

( z0+∆ z)3−2 ( z0+∆ z )− {z03−2 z0}∆z

=¿

lim∆ z→0

3 z02+3 z0∆ z+ (∆ z )2−2=¿

3 z02−2

Hence, f ' ( z )=3 z2−2.At z=−1 , f ' (−1 )=3 (−1 )2−2=3−2=1

Q1d) Use Green Theorem and Cauchy-Riemann Equation to show that ∮C

❑

f (z )dz=0, where C is a closed path.Answer: ∮

C

❑

f (z )dz=∮C

❑

(udx−vdy )+i∮C

❑

(udy+vdx) and ∮

C

❑

(udx−vdy )=∬(−dvdx

−dudy )dxdy

R by Green Theorem.

¿0 by Cauchy-Riemann Equation.Also, i∮

C

❑

(udy+vdx )=0.Q2a) Define convergence of a sequence.Answer: A sequence {xn } is said to converge to a complex number l if for any ε>0 ,∃ N∈ N (with N depending on ε) such that ¿ xn−l∨¿ ε ,∀ n≥ N .

38

Q2b) Prove that limn→∞

(an+bn )= limn→∞

an+ limn→∞

bn.Answer: By definition of limit of a sequence, given ε>0; we can find N such that |an−A|< ε

2,∨bn−B∨¿ ε

2 for n>N .Then for n>N ,|(an+bn )−( A+B )|=|(an−A )+(bn−B )|≤|an−A|+|bn−B|<ε which proves the result.Q2c) Show that f '( z), where f ( z )=z and z is the conjugate of z, is not differentiable.Answer: d

dzf ( z )=

lim∆ z→0

f ( z0+∆ z )− f ( z0 )∆ z

and ∆ z=∆ x+i ∆ y→0

Then ddz

z=lim∆ z→0

z+∆z− z

∆ z=lim∆ z→0

x+iy+∆ x+ i∆ y−x+iy

∆ x+i ∆ y

¿lim∆z→0

x−iy+∆ x−i ∆ y−( x−iy )

∆ x+i ∆ y

¿

lim∆ x→0∆ y→0

∆ x−i ∆ y

∆x+i ∆ y=1if ∆ y→0

¿

lim∆ x→0∆ y→0

∆ x−i ∆ y

∆ x+i ∆ y=−1if ∆ x→0

If limit exists, it must be unique. Hence, f '( z) does not exists.Q2d) Find the residues of ez

z( z+1).

Answer: Res ( z=0 )=limz→ 0

zf ( z )=limz→0

ez

z+1=1

Res ( z=−1 )= limz→−1

( z+1 ) f ( z )= limz→−1

ez

z=−e−1

39

Q3a) Draw: (i)0<¿z−z0<δ∨¿

(ii)¿ f ( z )−l∨¿ε

Answer:

Q3b) Prove that f ( z )=z2 is uniformly continuous in the region ¿ z∨¿1.

Answer: We must show that given any ε>0, we can find δ>0 such that ¿ z2−z02∨¿ε whenever |z−z0|<δ, where δ depends only on ε and not on the particular point z0 of the region.

If z and z0 are any points in ¿ z∨¿1, then |z2−z0

2|=¿ z+z0∨¿z−z0∨≤{|z|+¿ z0∨}∨z−z0∨¿2∨z−z0∨¿ δ which implies ¿ z−z0∨¿ δ

2=ε. Hence, f ( z )=z2 is uniformly continuous in the region.

Q3c) If w2−3 z2w+4 ln z=0, find dwdz .Answer: Differentiating with respect to z, considering w as an implicit function of z; we have ddz (w2 )−3 d

dz( z2w )+4 d

dz( ln z )=0

3w2 dwdz

−3 z2 dwdz

−6 zw+ 4z=0

dwdz

=6 zw−4

z3w2−3 z2

Q3d) Find ∮C

❑ e z

z−2dz .

40

Answer: ∮C

❑ e z

z−2dz=2πi ez ¿z=2=2πi e

2≈46.4

Q4a) Why is the number of complex numbers in |z|<1 uncountable?Answer: |z|<1 is uncountable because it is a unit circle that contains (0,1 ) , (0 ,i ) , (0 ,−i ) ,(0 ,−1) and so much more. By Cantor Diagonalization, (0,1) is not countable, hence |z|<1 is not countable.Q4b) Prove that ∑

n=1

∞ zn

n(n+1) converges(absolutely) for ¿ z∨≤1.

Answer: If |z|≤1, then | zn

n(n+1)|= |z|n

n(n+1)≤ 1n(n+1)

≤ 1n2

. But ∑ 1n2

converges and by comparison test, ∑

n=1

∞ zn

n(n+1) converges.

Q4c) Prove that ddz sin z=cos z.Answer: Since sin z= e iz−e−iz

2i and cos z= e iz+e−iz

2,

ddz ( e

iz−e−iz

2i )= 12 i

ddz

e iz− 12 i

ddz

e−iz=¿

12e iz+ 1

2e−iz=cos z.

Q4d) Define subsequence of a complex sequence.

41

Answer: Given a complex sequence {un }, consider a sequence {unk } of positive integers such that n1<n2<n3<…, then the sequence {unk } is called a subsequence of {un }.

Q5a) Define limz→z0f ( z )=−∞.

Answer: Let f :C→C be single valued function and defined except possibly at z0 in a region. Then f is said to tend to negative infinity (−∞) as z tends to z0 if ∀M>0(no matter how large), ∃ δ>0∋ f ( z )←M whenever 0<|z−z0|<δ.Q5b) Show that every Cauchy sequence is bounded.Answer: Let {un } is a Cauchy sequence. We show that {un } is bounded. Since {un } is Cauchy, given ε>0 ,∃ N∋|un−um|<ε ,∀n ,m≥ N. Let ε=1. |un−um|<1 ,∀ n ,m≥N . If M=N , we have |un|−|uN|≤|un−uN|<1 from triangular inequality.|un|<1+|uN|,∀n≥ N

Let M be maximum of {u1, u2 ,…,uN−1 ,1+|uN|}: M=max {1+|un∨,…,1+|uN||}. Then: |un|≤M ,∀n

Q5c) Show that ∫−R

R x2dx(x2+1 )2 (x2+2 x+2 )

+i∫Q

Q y2

( y2+1 )2 ( y2+2 y+2 )=7 π50

as R ,Q→∞.Answer: The poles of z2dz

( z2+1 )2 ( z2+2 z+2 ) enclosed by the contour C are z=i of order 2

and z=−1+i of order 1.Res ( i )=lim

z→ i

ddz {(z−i)2 z2dz

( z2+1 )2 ( z2+2 z+2 ) }=9i−12100

Res (−1+i )= lim❑

z→−1+i ( z+1−i ) z2dz( z2+1 )2 ( z2+2 z+2 )

=3−4 i25

∮ z2dz( z2+1 )2 ( z2+2 z+2 )

=2πi{9 i−12100+ 3−4 i25 }=7 π50

42

Hence, ∫−R

R x2dx(x2+1 )2 (x2+2 x+2 )

+i∫Q

Q y2

( y2+1 )2 ( y2+2 y+2 )=7 π50

as R ,Q→∞

Q5d) Find ddz (√ tan z+√ tanz+√ tanz+…∞).Answer: Let w=(√ tan z+√tanz+√tanz+…∞). Then w=√ tan z+w and w2−w=tan z

2w dwdz

−dwdz

=sec2 z, dwdz (2w−1 )=sec2 z, dwdz = sec2 z

2(√ tan z+√tanz+√tanz+…∞ ) .Q6a) State the Boundedness Theorem.Answer: If f (z) is continuous on a closed and bounded region, then f (z) is bounded.

Q6b) Show that if a function is analytic, it is harmonic.Answer: A function f ( z )=u ( x , y )+ iv (x , y ) is analytic if dudx=dv

dy… (1), dudy=−dv

dx… (2)

Differentiating (1) by x and (2) by y, respectively, we have:d2ud x2

= d2vdxdy

… (3 ) , d2udy2

=−d2 vdxdy

…(4)

Adding (3) and (4), we have: d2ud x2

+ d2udy2

=0.The analogous d2 v

d x2+ d2vdy2

=0 follows immediately.

Q6c) Evaluate ∫0

∞ cosmxx2+1

dx.Answer: 43

Res( i )=limz→i {(z−i) e imz

( z−i )(z+ i)}=e−m

2 i and ∫∞

−∞ cosmxx2+1

dx=2 πi( e−m

2 i )=π e−m

That is, ∫0

∞ cosmxx2+1

dx= π2e−m

Q6d) Show that f ( z )=c is continuous.Answer: Let ¿ z−z0∨¿ δ. Then |f (z )−f ( z0 )|=c−c=0<ε<1. Hence, |f (z )−f ( z0 )|<ε.Q7a) Define limit superior.Answer: Let {un } is a sequence of complex numbers which is bounded above. Then ∃M ∋un≤M ,∀ n. The set {un , un+1,…} is bounded above ∀ n and it has lub M n, where M n=lub {un ,un+1 ,…}≥M n+1=lub {un+1 ,un+2 ,…}. {M n } is a decreasing sequence bounded by M. If {M n } converges, then limit superior of un=lim

n→∞M n.

Q7b) Prove that the sequence { 11+nz } is uniformly convergent to zero all ∀ z such

that ¿ z∨≥2.

Answer: We have | 11+nz

−0|<ε when 1¿1+nz∨¿<ε¿ or ¿1+nz∨¿ 1

ε . Now, ¿1+nz∨≤∨1∨+¿nz|¿1+n|z∨¿ and 1+n|z|≥∨1+nz∨¿ 1

ε for n>¿. Thus, the sequence converges to zero for ¿ z∨¿2.Q7c) Show that ∮

C

❑

(z−z0)mdz={2 πi,m=−1

¿0 ,m≠−1 .44

Answer: Let f ( z )=(z−z0)m where m is an integer and z0 is a constant, let

z−z0=zr=r e¿. Then squaring both sides m times, (z−z0)m=rme imt and dz=ir e¿dt.

And, ∮C

❑

(z−z0)mdz=∫

0

2π

rm eimt ir e¿dt=irm+1∫

0

2 π

e i(m+1)t dt

By Euler Formula, i rm+1[∫02π

cos (m+1) tdt+i∫0

2π

sin (m+1) tdt ]If m=−1 ,then rm+1=1 ,cos0=1, sin 0=0.If m≠−1 each of the integrals is zero because, the integration is over an interval of length 2π , equal to a period of sine and cosine. Hence, the result is ∮C

❑

(z−z0)mdt=¿ {2πi ,m=−1

¿0 ,m≠−1¿.

Q7d) Relate |z−z0|<δ and n≥ N.Answer: ¿ z−z0∨¿ δ, then ¿ z−∞∨¿δ implies ¿n−∞∨¿δ which further implies that n>∞ since δ∈(0,1). Hence, n>N , where n is a very big number (if not infinity).Complex Analysis IIJust like the real analysis which has to do with analysis of real numbers which its study has not gone beyond the x-axis; complex analysis has to do with analysis of complex numbers which involves both the x- and the y-axes.The set of real numbers is embedded in the set of complex numbers and we have seen that some properties of real number that hold for complex number, viz: sequences and series, limits and converges, etc.The question that usually arises is why will a number be called complex or imaginary? And, if a+ ib=c+id, does that imply a−c=i(d−b), i.e. real number equals imaginary numbers?We will find out why these needs some explanations in the next section.Complex Numbers are applicable both in the Pure and the Applied Mathematics. For example, we can think of the position (x , y ), a transformation of x+iy of a velocity or calculate workdone W by a Force F as W=∫

−∞

∞

F dz or find the equation of the curve as |z|=1, in the applied realms of mathematics and in the pure aspect of 45

mathematics, we may think of the solution of the polynomial x4−1 forming one of the most important groups, the group of 4th roots of unity:( {1 ,−1 , i ,−i },× ), where × is the complex multiplication. This group satisfies most of group theory theorems. It can have at least one normal P-Sylow Subgroup and it satisfies the Lagrange-, Cauchy, and Sylow Theorems and the subsequent theorems.Also, vector space a group acting on a set is the generalization of complex numbers. This is easy to establish, but no space was provided for it in the course outline. What are obtainable is some time the usual conventional wisdom of astronomical proofs. I conclude with the Jewish Proverb: For example is not a proof.2.0. PreliminariesWhat are the roots of the polynomial x2+2 x+2 ? We use

x=−b±√b2−4ac2a

=−2±√22−4 (1)2

2(1)

¿ −2±√4−82

¿ −2±√−42

=−2±2√−12

¿−1±√−1=−1± i

This is written formally asx2+2 x+2= (x+1−i )(x+1+i)

Where x=−1+1 i or x=−1−1i.If x=a+bi, we say a is the real part of x and b i is the imaginary part of x. We can change x to z, and a and b with x , y respectively.This is the standard. That is,z=x+iy and w=u+iv, where u , v are functions of x and y.The conjugate of z is z=x− yi, now z−z=2 yi which implies y= z−z

2 i andz+z=2 x, i.e. x= z+ z

2

46

Also recall that |z|=|x+ iy|=√x2+ y2, where ¿ z∨¿ is called the modulus or absolute value of z. Thenz . z= (x+iy )(x−iy)

¿ x2+ y2

¿¿ z∨¿2¿

That is, ¿ z∨¿2=z . z¿.Also, recall the algebra of complex numbers

(a+ ib)+( c+id )=(a+c )+ (b+d ) i

(a+ ib)−(c+id )=(a+c )−(b+d )i

(a+ ib)× ( c+id )=(ac−bd )+(ad+bc )i

a+bic+di

=ac+bdc2+d2

+ bc−adc2+d2

i

The distance between two points on the real axis is¿a−b∨¿ and |z−2|<3 is standing for the interior of a circle of radius 3 centred on 2, |z−2|=3 is the boundary of the circle and |z−2|>3 is the exterior of the circle.Now, consider the graph of x+iy on the X−Y axes:Then sinθ=O

H= y

r and cosθ= xr implies x=r cosθ and y=r sinθ, i.e.

z=x+iy=r (cosθ+isinθ), where θ is the direction. So we have gotten a magnitude of z, ¿ z∨¿ and its direction. Thus, it is a vector which can form a vector space.We use the Euler Formula e iθ=cosθ+i sinθ to have z=r e iθ, now, zn=(r eiθ )n=rn e inθ, i.e.

zn=r n(cosnθ+i sin nθ)

This is called De Moivre Theorem.What will be ez? It will be

ex +iy=exe iy=ex (cos y+i siny )

Now, if n= 1m, we have:47

z1m=r

1m( cos θm

+i sinθm )

If θ=2πk where k=0 ,1 ,2 ,…,n−1 and zm=1; then r 1m=11m=1 and z 1m= cos2πk

m+i sin 2πk

m

Where m is called the mth root of unity.Given that z=x+iy , z=x−iy and we have

z=r (cosθ+i sinθ)¿ z=r (cosθ−i sinθ)}⇒¿

Also,e iz=cos z+i sin zand e−iz=cos z−isin z. Then:sin z= e iz−e−iz

2i…(1)

cos z= e iz+e−iz

2…(2)

And, w=f (z) , i.e. u+iv=f (x+ iy). If w=e z, then |w|=|ez|=|ex +iy|=|ex eiy|

¿|e x(cos y+isin y )|

¿|e x||cos y+ isin y|

¿ex √cos2 y+sin2 y

¿ex

Similarly,|e iy|=1

From (1) and (2),sin (iz )=e z−e− z

2i

¿

cos ( iz )= ez+e− z

2

48

Definition. A symbol, such as z, which can stand for any one element of a set of complex numbers is called a complex variable.3.0. Taylor and Laurent Series ExpansionsTaylor Theorem. If a function f (z) is analytic inside a circle C with centre at z0, then ∀ z∈C , we have:

f ( z )=∑n=0

∞ f n(z0)n !

(z−z0)n

Proof. Let z be any point inside C. Construct a circle C1 with centre at z0 and enclosing z. Then by Cauchy Integral Formula,f ( z )= 1

2πi∫C1

❑ f (w)w−z

dz…(1)

We have:1

w−z= 1

w+z0−z0−z= 1

(w−z0 )−(z−z0)

¿1

w−z0 [ 1

1−z−z0w−z0 ]

1w−z=

1w−z0 [1+ z−z0

w−z0+( z−z0

w−z0 )2

+…+( z− z0w−z0 )

n−1

+( z−z0w−z0 )

n 1

1−z−z0w−z0 ]

1w−z

= 1w−z0

+1+z−z0

(w−z0)2+

(z−z0)2

(w−z0)3 +…+

(z− z0)n−1

(w−z0)n +( z−z0

w−z0 )n 1w−z

From (1), we have:f ( z )= 1

2πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw

Using Cauchy General Integral Formula,f n ( z0 )= n!

2πi∫C❑ f (z)

(z−z0)n+1 dz ,n=0 ,1 ,2,…; equation (3) becomes:

49

f ( z )= f ( z0 )+ f ' ( z0 )( z−z0 )1 !

+f ' ' (z0)21

(z−z0)2+…+

f n−1(z0)(n−1 )!

(z−z0)n−1+Rn

The proof will be complete if we can show that limn→∞Rn=0.

Now, since w is on C1; we have:| z−z0w−z0|=γ <1, γ is a constant and we know that |f (z )|≤M , ∀ z from the Boundedness

Theorem and|w−z|=|(w−z0 )−( z−z0 )|≥r1−|z−z0|

Hence,|Rn|=

12π |( z−z0

w−z0 )n f (w)w−z

dw|← 12 π

γ nMr1−¿ z−z0∨¿2π r1¿

, i.e.|Rn|≤

γnMr1r1−¿ z−z0∨¿→0 ¿

as n→∞ because γ<1.limn→∞

Rn=0, as required.Laurents TheoremSuppose that f (z) is analytic inside and on a closed circular annulus

R :r ≤|z−z0|≤ ρ

Then there is a series of positive and negative powers of (z−z0) that converges to f (z) at every point w in the open annulus r<|z−z0|<ρ and

f ( z )=∑n=0

∞

an ( z−z0 )n+∑

n=1

∞

bn ( z−z0 )−n

Wherean=

12πi∫C1

❑ f (w)(w−z0)

n+1 dw ,n=1 ,2 ,3 ,…

bn=12πi∫C2

❑ f (w)(w−z0)

−n+1 dw ,n=1 ,2 ,3 ,…

50

And C1:∨z−z0∨¿ ρ, C2:|z−z0|=r both are positively oriented.Proof. Let z be an arbitrary fixed point such that r<|z−z0|<ρ. Then, by the Cauchy Formula;f ( z )= 1

2πi∫C1

❑ f (w)w−z

dw− 12πi∫C2

❑ f (w)w−z

dw…(1)We can consider the first integral in (1) as follows:

1w−z

= 1

(w−z0 ) [1− z−z0w−z0 ]

¿ 1w−z0

+z−z0

(w−z0)2+…+

(z−z0)n−1

(w−z0)n +( z−z0

w−z0 )n 1w−z0

So that,12πi∫C1

❑ f (w)w−z

dw

¿12πi∫C1

❑ f (w )w− z0

dw+z−z02 πi ∫C1

❑ f (w )

(w−z0 )2dw+…+

( z−z0 )n−1

2πi ∫C1

❑ f (w )w−z0

dw+Rn

That is,12πi∫C1

❑ f (w)w−z

dw=a0+a1 ( z−z0 )+…+an−1(z−z0)n−1+Rn…(2)

Where a0=

12πi∫C1

❑ f (w)w−z0

dw ,a1=z−z02 πi ∫C1

❑ f (w)(w−z0)

2 dw

an−1=12 πi∫C1

❑ f (w)(w−z0)

n dw and Rn=12πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw

Let us consider the second integral in (1). Then; we have:−1w−z

= 1

(z−z0)(1−w−z0z−z0 )

= 1z−z0 (1−

w−z0z−z0 )

−1

51

¿ 1z−z0 {1+w−z0

z−z0+(w−z0

z−z0 )2

+…+(w−z0z−z0 )

n−1

+(w−z0z−z0 )

n 1

1−w−z0z−z0 }

¿ 1z−z0

+w−z0

(z−z0)2+

(w−z0)2

(z−z0)3 +…+

(w−z0)n−1

(z−z0)n +(w−z0

z−z0 )n 1z−w

−12πi∫C2

❑ f (w)w−z

dw= 12πi∫C2

❑ f (w)z−z0

dw+ 12 πi∫C2

❑ w−z0( z−z0)

2 f (w )dw+…+ 12πi∫C2

❑ (w−z0)n−1

(z−z0)n dw

¿b1

z−z0+

b2(z−z0)

2+…+bn

(z−z0)n+Pn

¿b1(z−z0)−1+b2(z−z0)

−2+…+bn(z−z0)−n+Pn…(3)

Whereb1=

12 πi∫C2

❑

f (w )dw ,b2=12πi∫C2

❑

(w−z0 ) f (w )dw ,…bn=12πi∫C2

❑

(w−z0 )n−1 f (w)dwand pn=

12πi∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw

From (1), (2) and (3), we have:f ( z )=a0+a1 ( z−z0 )+…+an−1 ( z−z0 )n−1+b1(z−z0)

−1+…+bn(z−z0)−n+Rn+Pn

If we can show that limn→∞Rn=0and limn→∞

Pn=0, we are done.But we haved earlier that limn→∞

Rn=0 .Weonly need to show that limn→∞Pn=0 as follows:

|w−z0z−z0 |= γ

σ<K<1 since w is on C2.

By boundedness theorem,|f (z )|≤ M , ∀w . And, |z−w|=|( z−z0 )−(w−z0 )|≥∨z−z0∨−r

Therfore,|Pn|=

12 π |∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw|≤ 12π

KnM¿ z−z0∨−r

2πr

52

That is, limn→∞Pn=0 and equation (4) becomes:f ( z )=a0+a1 ( z−z0 )+…+an−1 ( z−z0 )n−1+b1(z−z0)

−1+…+bn(z−z0)−n+…

f ( z )=∑n=0

∞

an(z−z0)n+∑

n=1

∞

bn(z−z0)−n as required.

The part ∑n=0

∞

an(z−z0)n is called the Laurent Series, while the remainder of the series

which consists of inverse powers of z−z0 is called the principal part. If the principal part is zero, the Laurent Series reduces to a Taylor Series. In that sense, a Taylor Series is a special case of the Laurent Series.3) Find the Laurent Series for f ( z )= 1

z (z−1) for 0<|z|<1.Solution: Write f (z) in partial fractions as:

f ( z )= 1z−1

−1z

We then get the geometric series:1

z−1= −11−z

=−(1+z+z2+z3…)

Where f ( z )=−1z

−1− z−z2−z3−…

If |z|>1, then: 1z−1

=1z ( 1

1− 1z )=1z (1+ 1z+ 1z2+ 1z3 +…)

Where f ( z )= 1z2

+ 1z3

+ 1z4

+…

Therefore, it is important to note that given a function f (z) and a point z0 in the plane, it is possible that f (z) may have more than one Laurent Series with centre at z0 depending on the annulus of convergence on which the Laurent Series is to represent f .In general, for a given centre z0, the number of distinct Laurent Series that a function f will admit will depend on the location of the centre z0 and the number of singularities of f .53

4) The function f ( z )= 3z (z−i) has three distinct series expansion with centre at

z0=−i:a) A Taylor Series converging on the open disk: |z+i|<1b) A Laurent Series converging over the annulus: 1<|z+i|<2c) A Laurent Series converging over the annulus 2<|z+i|<∞:The three regions of convergence are shown shaded.Note:For(a): The radius of the disk of convergence is the distance between the centre and the singularity z=0 of f that is nearest the centre z0=−i.For(b): The annulus of convergence extends between the singularities z=0 and z=i.For(c): The annulus of convergence extends between the singularities z=i and infinity.This is the typical way in which the region of convergence of the Laurent Series Expansion of a function are determined, and it is extremely helpful in geometrical realization of subsequent supporting theory.Example: Find the Laurent Series Expansion of the function f ( z )=1

z with z0=1 and annulus of convergence A: 1<|z−1|<∞.Using 1

1+w=∑

n=0

∞

(−1)nwn ,where |w|<1,1z= 1

( z−1 )+1= 1z−1

1

1+ 1z−1

, 1¿ z−1∨¿<1¿

¿ 1z−1∑n=0

∞

(−1)n( 1z−1 )

n

,1<¿ z−1∨¿

¿∑n=0

∞

(−1)n 1(z−1)n+1

¿ 1z−1

− 1(z−1)2

+ 1(z−1)3

− 1(z−1)4

+…1<|z−1|<∞

Singularities and Zeros of Analytic Function54

A point z0 is a singularity of a function f (z) provided that f fails to be analytic at z0 while every neighborhood of z0 contains at least one point at which f is analytic.There are two types of singularities:1)Non-Isolated Singularity, and2) Isolated SingularityDefinition. A point z0 is a non isolated singularity of a function f iff z0 is a singularity of f and every deleted neighborhood of z0 contains at least one singularity of f .For example, f ( z )=log z has a non isolated singularity at every point of the non p[ositive real axis. Since by definition, every deleted neighborhood of a non isolated singularity pof a function contains at least one other singularity of f ; it follows that if a function has one non-isolated singularity, then it has infinitely many singularities, although thet may not all be necessarily non isolated.Definition. Supposenow that z0 is a singularity of a function f (z). Then z0 will be called Isolated Singularity of f , provided that there exist a deleted neighborhood of z0 throughout which f is analytic.For example, the function f ( z )= 4 i

z2+1 has two Isolated Singularities, one at i and the

other at – i for a deleted neighborhood of radius less than 2 can be drawn about either of these two points throughout which f is analytic.Classifications of Isolated SingularitiesSuppose that z0 is an isolated singularity of a function f (z). Then f (z) is analytic throughout a deleted neighborhood N ' (z0 , ρ), i.e. 0<|z−z0|<ρ. That is, f has Laurent Series Expansion f ( z )= ∑

n=−∞

∞

cn(z−z0)n…(1) which converges over this annulus region.

The following 3 cases are possible:1. No negative powers of ( z−z0 ) appear in the Laurent Expansion os (1). That

is, z0 is called a removable singularity. For example, the function f ( z )= sin zz has a removable singularity at the origin since its series expansion at the origin is

sin zz

=1− z2

31+ z4

5 !− z6

7 !+… …(2) which contains no negative powers of z (i.e.

the series is actually a Taylor Series)55

In general, if a function f (z) has a removable singularity at z0, then we re-define the function to have the value c0 at z0.2. Only a finite number of negative powers of (z−z0) with non- zero coefficients appear in (1). In this case, (1) takesf ( z )= ∑

n=−N

∞

cn(z−z0)n

¿C−N

(z−z0)N +

C−N+1

(z−z0)N−1+…+

C−1

z−z0+c0+c1(z−z0)+….

Where N is a positive integer and C−N ≠0. Then z0 is called a pole of order N.For example, f ( z )= 1

z3 has a pole of order 3 at z=0 and g ( z )= 1

( z+2 i )5 has a pole of order 5 at

z=−2 i and h ( z )= ez−1z2

has a pole of order 1 at z=0 since ez=∑

n=0

∞ zn

n!=1+∑

n=1

∞ zn

n !−1

z2

Suppose now that f (z) has a pole of order N at z0. Then f has a Laurent Series Expansion given in (3) with C−N≠0.That isf ( z )=

C−N

(z−z0)N +

C−N+1

(z−z0)N−1+…+

C−1

z−z0+C0+… positive powers

Multiplying both sides by (z−z0)N to obtain

( z−z0 )N f ( z )=C−N+C−N+1 ( z−z0 )+C−N+2(z−z0)2+……(4)Taking limit as z tends to z0, we have:

limz→z0

[ ( z−z0 )N f (z)]=CN≠0Hence, the following theorem.Theorem. Suppose that f (z) is analytic throughout a deleted neighborhood 0<¿ z−z0∨¿ ρ of a point z0. Then f has a pole of order N at z0 if ana only if ( z−z0 )N f (z) has a removable singularity at the point z0 and

limz→ 0

( z−z0 )N f (z )≠0 .3. The principal part of (1) contains an infinite number of negative powers of (z−z0) with non-zero coefficients. In this case, z0 is called an essential singularity of the function. For example, the function f ( z )=sin 1

z has an essential singularity at z=0 since56

sin 1z=1z− 13! z3

+ 15 ! z5

− 17 ! z7

+…

Examples:i. The function f ( z )= z+1

z2+1 has two poles at z=i and z=−i each of order 1.

We have:( z−i ) f ( z )=( z−i ) z+1

z2+1=( z−i ) z+1

( z−i ) ( z+i )= z+1

z+i and limz→ i

( z−i ) f (z)=limz→i

z+1z+i

=i+12i

≠0

Since these 2 conditions are satisfied, the theorem asserts that f has a pole of order 1.ii. The function f ( z )= e z−1

z2 appears to have a pole of order 2 at z=0.

However, a careful examination reveals that the pole is, in fact, of order 1 since the series expansion of f yieldsez−1z2

= 1z2 (∑n=0

∞ zn

n !−1)=1z + z

3 !+ z2

4 !+…

Had we chosen to use the to use the criterion provided by the theorem under the assumption that the pole has order 2, we would have to find out that limz→ 0

z2 f ( z )=0 ,whichwill indicate that our assumption is wrong.iii) Although the function f ( z )= sin hz

z3 appears to have a pole of order 3 at z=0

, its pole is actually of order 2.f ( z )= sin hz

z2= 1

z2+ 13 !

+ z2

5 !+…

These last two examples demonstrate the fact that appearances must be examined critically before assumption are to be made regarding order of a pole.Definition(Zero). If φ (z) is an arbitrary function, then a number z1 is called a zero of φ if and only if φ ( z1 )=0.Suppose that f (z) is analytic at z0. Then, f has a Taylor Series Expansion of the form57

f ( z )=∑n=0

∞

an(z−z0)n

¿a0+a1 ( z−z0 )+… converging to the function over some circular region centred at z0 .Cauchy Integral Formula and Related TheoremsLet f (z) be analytic inside and on a simple closed curve C and let a be any point inside C. Then

f (a )= 12πi∮C

❑ f (z )z−a

dz

Where C is transverse in the positive (counter clock wise) sense. Also, the nth derivative of f (z) at z=a is given by:

f n (a )= n !2πi∮

f (z)(z−a)n+1 dz , n=1,2,3 ,…

Morera TheoremLet f (z) is continuous in a simply connected region R and suppose ∮

C

❑

f (z )dz=0 around every simple closed curve C in R. Then f (z) is analytic in R.Then, by reasoning identical with that used in F ( z )=∫

a

z

f (u)du and F ' ( z )=f (z), it follows that F (z) is analytic in R and F ' ( z )=f (z).It follows that F '(z ) is also analytic if f (z) is. Hence, f (z) is analytic in R.Fundamental Theorem of AlgebraEvery polynomial equation P ( z )=a0+a1 z+a2 z

2+…+an zn=0, where the degree n≥1 and

an≠0 has at least one root.Proof. Suppose on the contrary, P ( z )=0 has one root, the f ( z )= 1

P(Z) is analytic for all z. Also, |f (z )|= 1

¿P (z)∨¿¿ is bounded (and in fact approaches zero) as |z|→∞. Then it follows that f (z) and thus P(Z) must be a constant. Thus, we are led to a contradiction and we must conclude that P (Z )=0 must have at least one root or P(Z) has at least one zero.58

Maximum Modulus TheoremSuppose f(z) is analytic inside and on a simple closed curve C. Then the maximum value of ¿ f (z)∨¿ occurs on C, unless f (z) is a constant.Proof. By Cauchy Integral Formula,f (a )= 1

2πi∮C❑ f (z )z−a

dz, i.e. f (a )= 12πi∫0

2π f (a+r e iθ)ir eiθ

r eiθdθ= 1

2π∫02π

f (a+r eiθ)dθ

That is,|f (a )|≤ 1

2π∫02π

|f (a+r e iθ)|dθ…(1)

Let us suppose that ¿ f (a)∨¿ is a maximum so that ¿ f (a+r e iθ)∨≤∨f (a)∨¿. If |f (a+r e iθ )|<¿ f (a)∨¿ for one value of θ, then by continuity of f , it would hold for a finite arc, say θ1<θ<θ2. But, in such case, the mean value of ¿ f (a+r e iθ)∨¿ is less than ¿ f (a)∨¿ which would contradict (1).It follows that in any δ neighborhood of a, f (z) must be a constant. If f (z) is not a constant, the maximum value of ¿ f (z)∨¿ must occur on C.Minimum Modulus TheoremLet f (z) be analytic inside and on a simple closed curve C and if f (z)≠0 inside C, then ¿ f (z)∨¿ must assume its minimum value on C.Proof. Since f (z) is analytic inside and on C and since f (z)≠0 inside C, it follows that 1

f (z ) is analytic in inside C. By the maximum modulus theorem, it follows that 1

¿ f (z )∨¿¿ cannot assume its maximum value inside C and so ¿ f (z)∨¿ cannot

assume its maximum value inside C. Then, since ¿ f (z)∨¿ has a minimum. This minimum must be attained on C.Argument TheoremLet f ( z ) be analytic inside and on a simple closed curve C except for a poles z=α of order (multiplicity) P inside C. Suppose also that inside C, f (z) has only one zero at z=β of order (multiplicity) n and no zer on C. Then:

12πi∮C

❑ f ' (z)f (z )

dz=n−p

59

Proof. Let ❑❑ and ❑❑ be non overlapping circleslying inside C andenclosing z=α and z=β respectively. Then12πi∮

f ' (z)f (z )

dz= 12πi∮C1

❑ f ' (z)f (z )

dz+ 12πi∮❑

❑ f ' (z )f (z)

dz…(1)

Since f (z) has a pole of order p at z=α, we have:f ( z )= F (z)

(z−α)p

Where F (z) is analytic and different from zero inside and on C1. Then taking logarithm in (2) and differentiating, we have:f ' (z )f (z)

= F '(z )F (z )

− pz−α

…(3) so that12πi∮C1

❑ f ' (z)f (z )

dz=¿ 12 πi∮C1

❑ F '(z )F (z )

dz− p2πi∮C1

❑ dzz−α

dz=0−p=−p…(4)¿

Since f (z) has a zero of order n at z=β, we have:f ( z )= ( z−β )nG ( z )… (5)

Where G(z ) is analytic and different from zero inside and on γ1. Then Logarithmic Differentiation yields:f ' (z )f (z)

= nz−β

+G' (z )G(z)

…(6)

12πi∮C1

❑ f ' (z)f (z )

dz=¿ 12 πi∮C1

❑ f ' (z)f (z )

dz+ 12πi∮C1

❑ f ' (z)f (z)

dz=n−p¿

Rouches TheoremSuppose f (z) and g(z ) are analytic inside and on a simple closed curve C and suppose ¿ g(z )∨¿∨f ( z)∨¿ on C. Then f ( z )+g(z ) and f (z) have the same number of zero inside C.Proof. Let F ( z )= g(z )

f ( z) so that g ( z )= f ( z ) F(z ) or briefly g= fF. Then if N1 and N 2 are the number of zeros inside C of f +g and f , respectively,we have from Argument Theorem, using the fact that these functions have no poles inside C,60

N1=12πi∮C

❑ f '+g 'f+g

dz , N2=12πi∮C

❑ f 'fdz

ThenN1−N2=

12 πi∮c

❑ f '+ f 'F+fF 'f + fF

dz− 12πi∮C

❑ f 'fdz= 1

2πi∮c❑ f ' (1+F)+fF '

f +fFdz− 1

2πi∮C❑ f '

fd

N1−N2=12 πi∮c

❑

( f'

f+ F '1+F

)dz− 12 πi∮C

❑ f 'fdz= 1

2πi∮c❑ F '1+F

dz

¿ 12πi∮C

❑

F ' (1−F+F2−F3+…)dz=0

This from the fact |F|<1 on C so that the series is uniformly convergent on C and term by term integration yields the value zero. Thus, N1=N2 , as required.Example, by Rouches Theorem, that all the roots of z7−5 z3+12=0 lie between the circles |z|=1 and |z|=2.Proof. Consider the circle C1:|z|=1. Let f ( z )=12, g ( z )=z7−5 z3. On C2, we have:

|g ( z )|=|12−5 z3|≤|12|+|5 z3|≤60<27=¿ f (z)∨¿

By the Rouche Theorem, f ( z )+g ( z )=z7−5 z3+12 has the same number of zeros inside |z|=2 as f ( z )=z7, i.e. all the zeros are inside C2.Hence, all the roots lie inside |z|=2, but outside |z|=1, as required.Conformal and Bilinear MappingsThe set of equations

u=u( x , y)¿v=v (x , y)}…(1)

¿

defines in general a transformation or mapping which establishes a correspondence between points in the uv and xy planes. The equation (1) are called transformation equations. If to each point of the uv plane, there corresponds one and only one point of the xy plane, and conversely, wespeak of a one-one transformation or mapping. In such a case, a set of points in the xy plane (such as a curve or region) is mapped into a set of points in the uv plane (curve or region) and conversely. The corresponding sets of points in the two planes are often called images of each other.61

Suppose that under the transformation (1), point (x0 , y0) of the xy plane is mapped into point (u0 , v0) of the uv plane while curves C1 and C2 are mapped, respectively, into curves C1' and C2

' intersecting at (u0 , v0). Then, if the transformation is such that the angle at (x0 , y0) between C1 and C2 is equal to the angle at (u0 , v0) betweenC1' and C2

' both in magnitude and direction, the transformation or mapping is said to be conformal at (x0 , y0). A mapping that preserves the magnitudes of angle but not direction is called isogonal.Theorem. If f (z) is analytic and f '( z)≠0 in a region R, then the mapping w=f (z) is conformal at all points of R.For example, the fixed or invariant points of the transformation w=z2 are solutions of z2=z, i.e. z=0,1.Some General TransformationsIn the following α ,β are given complex constants while a ,θ0 are real constants.

1. Translation. w=z+β, z are displaced or translated in the direction of vector β.2. Rotation. w=e i θ0 z, z is rotated through angle θ0. If θ0>0, the rotation is counter clockwise while if θ0<0, the rotation is clockwise.3. Stretching. w=az , z are stretched (or contracted) in the direction z if a>1 (or 0<a<1). We consider contraction as a special case of stretching.4. Incversion. w=1

z

Successive TransformationIf w= f 1(τ ) maps region Rτ of the τ plane into Rw of the w plane while τ=f 2( z) maps region R z of the z plane into region Rτ, then w= f 1(f 2(z )) mapps R z into Rw. The functions f 1 and f 2 define successive transformations from one plane to another, which are equivalent to a single transformation. These ideas are easily generalized.Linear TransformationThe transformationw=αz+ β where α and β are given complex constants, is called a linear transformation. Letting α=aei θ0, we see that a general linear transformation is a combination of the transformation of translation, rotation and stretching.Bilinear Transformation62

The transformationw=αz+β

γz+σ,αδ−βγ ≠0 is called a bilinear transformation. This transformation can be considered as a combination of transformations of translation, rotation, stretching and inversion.

Mapping of a Half Plane onto a CircleLet z0 be any point P in the upper half of the z plane denoted by R. Then, the transformation w=e i θ0( z−z0

z−z0 ) maps this upper half plane in a one-one manner onto the interior R ' of the unit circle |W|=1. Each point of the x-axis is mapped to the boundary of the circle. The constant θ0 can be determined by making one particular point of the x axis correspond to a given point on the circle.Examples

1) Let the rectangular region R in the z plane be bounded by x=0 , y=0 , x=2 , y=1. Determine the region R’ of the w plane into which R is mapped under the transformations:(a)w=z+(1−2 i ) (b)w=√2 e

πi4 z (c)w=√2e

πi4 z+(1−2 i )

Solution(a)Given w=z+(1−2i). Then u+iv=x+iy+1−2 i=( x+1 )+i( y−2) and u=x+1 , v= y−2. Line x=0 ismapped into u=1; y=0 into v=−2; x=2 into u=3; y=1 into v=−1.The transformation or mapping accomplishes a translation of the rectangle. In general, w=z+β accomplishes a translation of any region.b) Given w=√2e

π i4 z. Then u+iv=(1+i ) ( x+iy )=x− y+i(x+ iy) and u=x− y , v=w+ y. Line

x=0 is mapped into u=− y , v= y or u=−v; y=0 into u=x or u=v; x=2 into u=2− y, v=2+ y or u+v=4; y=1 into u=x−1 , v=x+1 or v−u=2.The mapping accomplishes a rotation of R (through 4 50) and a stretching of length √2. In general, the transformation w=az accomplishes a rotation and stretching of a region.

d) Given w=√2eπi4 z+(1−2i). Then u+iv=(1+i ) ( x+iy )+1−2i and u=x− y+1 ,

v=x+ y−2. The lines x=0 , y=0 , x=2 , y=1 are mapped, respectively, into u+v=−1 , u−v=3 , u−v=1.

63

The mapping accomplishes a rotation and stretching and a subsequent translation. In general, the transformation w=αz+ β accomplishes a rotation, stretching and translation. This can be considered as two successive mappings w=αz (rotation) and z=z+ βα (translation)

2) Determine the region of the w plane into which each of the following is mapped by the transformation w=z2 bounded by x=1 , y=1 and x+ y=1.Solution. Since w=z2 is equivalent to u+iv=(x+iy)2=x2− y2+2 ixy, we see that u=x2− y2 , v=2xy . Then line x=1 mapped u=1− y2, v=2 y or u=1− v2

4; line y=1 into

u=x2−1, v=2x or u= v2

4−1; line x+ y=1 or y=1−x into

u=x2−(1−x )2=2 x−1 , v=2 x (1−x )=2x−2 x2 or v=12 (1−u2) on eliminating x.3) Find a bilinear transformation that maps points z=0 ,−i ,−1 into w=i ,1 ,0 respectively.

Solution. Since w=αz+βγz+δ , we have:

i=α (0 )+ βγ (0 )+δ

…(1)

1=α (−i )+βγ (−i )+δ

…(2 )

0=α (−1 )+βγ (−1 )+δ

…(3)

From (3), β=α. From (1), δ= βi=−iα. From (2), γ=iα . Then

w= αz+αiαz−iα

=1i ( z+1z−1 )=−i( z+1z−1 )

4) Find the fixed or invariant points of the transformation w=2 z−5z+4 .

Solution. The fixed points are solutions of 2 z−5z+4 , i.e. z (z+4 )−(2 z−5 )=0, i.e. z2+2 z+5=0 ,i.e. z=−1±2i.64

Analytic ContinuationLet F1(z ) be a function of Z which is analytic in a region R1. Suppose that we can find a function F2( z) which is analytic in a region R2 and which is such that F1(z )=F2(z) in the region common to R1 and R2. Then we say that F2( z) is analytic continuation of F 1(z ). This means that there is a function F (z) analytic in the combined regions R1 and R2 such that F ( z )=F1(z) in R1 and F ( z )=F2(z) in R2. Actually, it suffices for R2 and R2 to have only a small arc in common, such as LMN as follows:Example

1) Let F (z) be analytic in a region R and suppose that F ( z )=0 at all points on an arc PQ inside R. Prove that F ( z )=0 throughout R.Proof. Choose any z0 in the circle C. At z0, the function F (z) with Taylor Expansion,F ( z )=F ( z0 )+F' ( z0 ) ( z−z0 )+12 F

' ' ( z0 ) ( z−z0 )2+… forms F ( z0 )=F' ( z0 )=F ' ' ( z0 )=…=0. Hence, F ( z )=0 inside C. This process continuous throughout R.

Stereographic ProjectionLet P be the complex plane and consider a sphere tangent to P at z=0. The diameter NS is perpendicular to P and we call points N and S the north and south poles of Q. Corresponding to any point A on P, we can construct line NA intersecting Q at point A’. Thus to each point of the complex plane P , there corresponds one and only one point of the sphere Q, and we can represent any complex number by a point on the sphere. For completeness, we say that the point N itself corresponds to the “point at infinity” of the plane. The set of all points of the complex plane including the points at infinity is called the entire complex plane, the entire z plane, or the extended complex plane. The above method for mapping the plane onto the sphere is called stenographic projection. The sphere is sometimes called the Riemann Sphere. When the diameter of the Riemann Sphere is chosen to be unity, the equator corresponds to the unit circle of the complex plane.

QUESTION PAPER OF COMPLEX ANALYSIS II1(a) Define a complex function.(b) Use Euler Formula and De Moivre Theorem to find the 8 roots of unity.(c) If f (z) is analytic inside a closed circular annulus C1:r ≤|z−z0|<ρ and w is in the open annulus C2:r<|z−z0|<ρ, show that:65

limn→∞ [ 12πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw+ 12πi∫C2

❑

( w−z0z−z0 )

n f (w)z−w

dw ]=0.2(a) Define a pole of order N and a zero of order N of a complex valued function.(b) i. Find the Laurentz Series Expansion of f ( z )= 1

z (z−1) for 0<|z|<1and for ¿ z∨¿1. ii. Write a Taylor Series Expansion of sin zz .(c) Let f(z) is single-valued and analytic inside and on a circle C except at the point z=a. If f ( z )= ∑

n=−∞

∞ 12πi∮C

❑ f (z)(z−a)n+1 dz and ∮ f (z )dz=2 πia−1 when n=1, show that

a−1=limz→a

1(m−1 ) !

dm−1

dzm−1 [(z−a)m f (z)] .

3(a) State and prove the maximum modulus theorem.(b) Show that∫

−R

R x2dx(x2+1)2 (x2+2 x+2 )

+∫Γ

❑ z2dz(z2+1)2 ( z2+2 z+2 )

+i∫−Q

Q y2dy( y2+1)2 ( y2+2 y+2 )

=7 π50 ,

where Q ,R→∞.(c) Find the nth derivative off (a )= 12πi∮C

❑ f (z )z−a

dz.4(a) State Morera Theorem and prove Argument Theorem.(b) Use Argument Theorem to prove Rouche Theorem.(c) Use Rouche Theorem to show that all the roots of z7−5 z3+12=0 lie between the circles |z|=1 and |z|=2.5(a) i. Define conformal and isogonal mappings. ii. Why is the bilinear transformation w=αz+β

γz+σ a combination of translation, rotation, stretching and inversion? Otherwise, show that ασ−βγ ≠0.(b) Find the fixed or invariant points of w=z9.(c) Show that ∫

0

∞ cosmxx2+1

dx= π2e−m ,m>0.

6(a) Let the rectangular region R in the z-plane be bounded by x=0 , y=0 , x=2 , y=1. Determine the region R ' of the w-plane into which R is mapped under the transformationw=√2 e

πi4 z+(1−2i .)(b) Explain the concept of analytic continuation of the functions F1(z ) and F2( z) in the regions R1 andR2.(c) Use Riemann Sphere to explain why complex plane C does not require a new infinity besides the aleph naught of counting numbers, N , and the continuum hypothesis of R, the x-axis.

66

7(a) State and prove the Fundamental Theorem of Algebra.(b) Find a bilinear transformation that maps points z=0 ,−i ,−1 into w=i ,1 ,0 respectively.(c) Show that ∫

0

2π

[ 1a+bsinθ +

13−2cosθ+sinθ ]dθ=π ( 2+√a2+b2

√a2+b2 ) .MARKING SCHEME OF MTH 43121(a) A complex variable is a function from C to C, i.e. f :C→C.1(b) Euler Formula is given as e iθ=cosθ+i sinθ and De Moivre Theorem states that zn=r n(cosnθ+i sin nθ). Thus, z 1n=r

1n (cos θn+isin θ

n ). If θ=2πk where 2π is the circumference of the unit circle and k=0 ,1 ,2 ,…n−1 . e2πki=cos 2 πkn +isin 2πk

n since r1n=1

1n=1. Since we are looking for the 8-roots of unity, we divide 360 by 8 to get: 45.00, 90.00, 135.00, 180.00, 225.00, 270.00, 315.00, 360.00 and the roots are:

e45 .00 i=cos 45 .00+isin45 .00=√ 12 (1+i ) , e90 .0

0 i=cos90 .00+isin90 .00=i ,

e135.00 i=cos135 .00+isin135.00=√ 12 (−1+i), e180.0

0 i=cos180 .00+isin180 .00=−1

, e225.00i=cos225 .00+isin225 .00=√ 12 (−1−i ) , e270 .0

0 i=cos 270.00+isin270 .00=−i

, e315.00 i=cos315 .00+isin315.00=√ 12 (1−i) , e360 .0

0 i=cos360 .00+isin360 .00=1

1(c) Proof: Since w is on C1; we have: | z−z0w−z0|=γ<1, where γ is a constant. By

Boundedness Theorem [since f is analytic (continuous) in a closed circular annulus (closed interval)]: |f (w)|<M. And, |w−z|=|(w−z0 )−(z−z0)|≥r 1−|z−z0|. Hence, |Rn|=

12π |( z−z0

w−z0 )n f (w)w−z

dw|≤ 12π

γ nMr1−|z−z0|

.2π r 1=γn Mr1

r 1−|z−z0|→0as n→∞.

That is, limn→∞

12 πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw=0.

67

Also, since w is on C2; we have: |w−z0z−z0 |=k<1. By Boundedness Theorem, |f (w)|≤ M

for all w on C2 and |z−w|=|( z−z0 )−(w−z0 )|≥|z−z0|−r. Therefore, |Pn|=

12 π |∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw|≤ 12π

k nM|z−z0|−r

.2πr= kn M|z−z0|−r

→0 as n→∞.That is, lim

n→∞

12 πi∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw=0

Therefore, limn→∞

12 πi∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw+ limn→∞

12 πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw=0

2(a) Consider the Laurentz Series Expansion, f ( z )= ∑

n=−N

∞

cn ( z−z0 )n=c−N

( z−z0 )N+

c−N+1

( z−z0 )N−1 +…+c−1

z−z0+c0+c1 ( z−z0 )+…, where N is a

positive integer and c−N≠0. Then z0 is called a pole of order N.Consider the Laurentz Series Expansion, f ( z )=∑

n=N

∞

an ( z−z0 )n. Then z0 is called a zero of f of order N.2(b)i. For 0<|z|<1, f (z )= 1

z (z−1)= 1z−1

−1z= −11−z

−1z=− (1+ z+z2+…)−1

z . For |z|>1 , 1

z−1=1

z ( 1

1−1z )=1z (1+1z + 1

z2+ 1z3

+…)= 1z+ 1z2+ 1z3 +…

ii. sin zz

=1− z2

3 !+ z4

5!− z6

7 !+…

2(c)68

Proof [a−1=limz→a

1(m−1 ) !

dm−1

d zm−1 {( z−a )m f (z )}]. Suppose f (z) has a pole a of order m. Then the Laurentz Series of f (z) is f ( z )=

a−m

(z−a)m+

a−m+1

(z−a)m−1+…+a−1

z−a+a0+a1 ( z−a )+a2(z−a)2+…(1)

Multiplying (1) by (z−a)m from both sides, we have:(z−a)m f (z )=a−m+a−m+1 ( z−a )+…+a−1(z−a)m−1+a0(z−a)m+…(2)Differentiating both sides of (2), m−1׿ , with respect to z❑; we have:dm−1

d zm−1 {( z−a )m f (z)}=(m−1 ) !a−1+m (m−1 )…2a0 ( z−a )+…(3)Let z→a in (3). Then lim

z→a

dm−1

dzm−1 {( z−a )m f (z) }=(m−1 ) !a−1, i.e.a−1=lim

z→a

1(m−1 ) !

dm−1

d zm−1 {( z−a )m f (z )}

3(a) Maximum Modulus Theorem: Suppose f (z) is analytic inside and on a simple closed curve C. Then the maximum values of ¿ f (z)∨¿ occurs on C, unless when f (z) is a constant.Proof: By Cauchy Integral Formula,

f ( a )= 12πi∮C

❑ f (z )z−a

dz=¿ f (a )= 12π∫0

2 π f (a+r e iθ) ir e iθdθr e iθ

= 12π ∫0

2π

f (a+r e iθ)dθ

|f (a)|≤ 12π∫0

2π

|f (a+r e iθ)|dθ… (1)

Let us suppose that ¿ f (a)∨¿ is a maximum, so that ¿ f (a+r e iθ)∨≤∨f (a)∨¿ and|f (a+r e iθ )|<¿ f (a)∨¿ for one value of θ, then by continuity of f , it will hold for a finite arc, say θ1<θ<θ2. But |f (a+r eiθ )|<|f (a)| contradicts (1). Therefore, in any σ neighbourhood of a, |z−a|<σ ; f (z) must be a constant. If f (z) is not a constant, the maximum value of |f(z)| must occur on C.69

3(b)Solution: The poles of z2

( z2+1 )2 ( z2+2 z+2 ) enclosed by the contour C are z=i of order 2

and z=−1+i of order 1. Res (i )=lim

z→ i

ddz {( z−i )2 z2

( z+i )2 ( z−i )2 ( z2+2 z+2 ) }=9 i−12100

Res (−1+i )= limz→−1+ i

( z+1−i ) z2

( z2+1 )2 ( z+1−i ) ( z+1+i )=3−4 i

25

∮ z2dz( z2+1 )2 ( z2+2 z+2 )

=2πi{9 i−12100+ 3−4 i25 }= 7 π50

Hence, ∫−R

R x2dx(x2+1 )2 (x2+2 x+2 )

+∫Γ

❑ z2dz( z2+1 )2 ( z2+2 z+2 )

+i∫−Q

Q y2dy( y2+1 )2 ( y2+2 y+2 )

=7 π50 , where

R ,Q→∞.3(c) Given that f (a )= 1

2πi∮C❑ f (z )z−a

dz, its first derivative is f ' (a )= 12πi∮C

❑ f ( z)( z−a )2

dz, the second derivative is f ' ' (a )= 1

2πi∮C❑ f (z)

( z−a )3dz and the nth derivative is f n (a )= n!

2πi∮C❑ f (z )

(z−a )n+1 dz since ddz (z−a)−1=−(z−a)−2 , d

dz(−( z−a )−2 )=−1×−2 (z−a)−3=2! (z−a)−3

4(a) Morera Theorem: Let f (z) be a continuous function in a simply connected region R and suppose ∫

C

❑

f ( z )dz=0around every simple closed curve C in R. Then f (z) is analytic in R.4(b)Proof [Argument Theorem]: Let C1 and Γ1 be non-overlapping circles lying inside C and enclosing z=α and z=β respectively. Then: 12πi∮

f ' (z)f (z )

dz= 12πi∫C1

❑ f ' (z)f (z )

dz+¿ 12πi∫Γ 1

❑ f '(z )f (z)

dz ¿…(1)

70

Since f (z) has a pole of order p at z=α❑, we have: f ( z )= F (z )( z−α )p

…(2). Taking logarithmic differentiation of (2), we have: f ' (z )

f (z)=

F '(z )F (z )

− pz−α

…(3), so that 12πi∫C1

❑ f ' (z)f (z)

dz= 12πi∫C1

❑ f ' (z)f (z)

dz−¿ p2πi∫C1

❑ 1z−α

dz=0−p=−p… (4)¿

Since f (z) has a zero of order n at z=β, we have: f ( z )= ( z−β )nG ( z )…(5)

By logarithmic differentiation of (5), we have: f ' (z )f (z)

= nz−β

+G' (z )G(z)

…(6)

So, 12πi∫Γ1❑ f ' (z)

f (z)dz= n

2πi∮Γ 1

❑ dzz−β

+ 12πi∮

G' (z)G (z)

dz=n…(7)

From (1), (4) and (7), we have the required result as:12πi∮

f ' (z)f (z )

dz= 12πi∫C1

❑ f ' (z)f (z )

dz+¿ 12πi∫Γ 1

❑ f '(z )f (z)

dz=n−p ¿

4(c)Proof [Rouche Theorem]: Let F ( z )= g(z )f ( z) , so that g ( z )= f ( z ) F(z ) or briefly g=fF.

Then if N1 and N 2 are the number of zeros inside C of f +g and f respectively, from Argument Theorem using the fact that these functions have no poles inside C, N 1=

12πi∮C

❑ f '+g 'f+g

dz,N 2=12πi∮C

❑ f 'fdz. Then:

N 1−N2=12 πi∮C

❑ f '+ f ' F+fF 'f + fF

dz− 12πi∮C

❑ f 'fdz= 1

2πi∮C❑ f ' (1+F)

f (1+F)dz− 1

2πi∮C❑ f '

fdz= 1

2πi∮C❑

{ f 'f + F '1+F }dz− 1

2 πi∮C❑ f '

fdz= 1

2πi∮C❑ F '1+F

dz= 12πi∮C

❑

F ' (1−F+F2−F3+…)dz

Using the fact that |F|<1 on C, the series is uniformly convergent on C because term by term integration yields the value zero. Thus, N1=N2.4(d)Solution: Consider the circle C1:∨z∨¿1. Let f ( z )=12 andg ( z )=z7−5 z3. On C, we have: |g ( z )|=|12−5 z3|≤|12|+|5 z3|≤60<27=¿ f (z)∨¿

By the Rouche Theorem, f ( z )+g ( z )=z7−5 z3+12 has the same number of zeros inside |z|=2 as f ( z )=z7, i.e. all the zeros are inside C2. Hence, all the roots lie inside |z|=2 but outside |z|=1, as required.5(a) 71

i. Conformal Mapping: Suppose that under a transformation, point (x0 , y0 ) of the xy-plane is mapped into point (u0, v0 ) of the uv-plane while curves C1 and C2 are mapped, respectively, into curves C1' and C2

' intersecting at (u0 , v0 ). Then, if the transformation is such that the angle at (x0 , y0 ) between C1' and C2

' both in magnitude and direction, the transformation or mapping is said to be conformal at (x0 , y0 ). Isogonal: A mapping that preserves the magnitude of angle but not direction is called isogonal.ii. w=αz+β

γz+σ is a translation because it is of the form αzγz+σ

+ βγz+σ , it is a rotation

because α may equal e i θ0 , it is a stretch because α may equal az, it is an inversion of 1

γz+σ . Now, if w= z+ βz+ β

=1, it is not a bilinear transformation. It is so if α=γ and β=σ ⇒ α

γ=1orσβ=1. Only negation works, that is, αγ ≠1 ,and

σβ≠1⇒ ασ

γβ≠1⇒ ασ ≠ γβ⇒ασ−βγ ≠0.

5(b) The fixed or invariant points of w=z9 are:

z=0 , z=√ 12 (i+1 ) , z=i , z=√ 12 (−1+i ) ,

z=−1 , z=√ 12 (−1−i ) , z=−i , z=√ 12 (1−i )andz=1.5(c) Solution: The integrand has simple poles at z=± i ,but only z=i lies inside C.Res ( i )=lim

z→i {( z−i ) e imz

(z−i )( z+i)}= e−m

2i and ∫

−∞

∞ cosmxx2+1

dx=2 πi( e−m

2 i )=π e−m

That is, ∫0

∞ cosmxx2+1

dx= π2e−m.

72

6(a) Given w=√2 e14 z+(1−2 i ) ,u+ iv=(1+i ) ( x+iy )+1−2 i and u=x− y+1 , v=x+ y−2. The lines x=0 , y=0 , x=2 , y=1 are mapped respectively into

u+v=−1 , u−v=3 , u+v=3 ,u−v=1.

6(b) Analytic Continuation: Let F1(z ) be a function of z which is analytic in a region R1. Suppose that we an find a function F2( z) which is analytic in a region R2 and which is such that F1(z )=F2(z) in the region common to R1 and R2. Then we say that F2( z) is an analytic continuation of F1(z ). This means that there is a function F(z) analytic in the combined regions R1 and R2 such that F ( z )=F2(z) in R1 and F ( z )=F2(z) in R2. It suffices for R1 and R2 to have only a small arc in common such as LMN as follows:

6(c) Let P be the complex plane and consider a sphere Q tangent to P at z=0. The diameter NS is perpendicular to P and we call points N and S the north and south poles of Q. Corresponding to any point A on P, we can construct line NA intersecting Q at point A’. Thus, to each point of the complex plane P, there corresponds one and only one point of the sphere Q, and we can represent any complex number by a point on the sphere. For completeness, we say that the point N itself corresponds to the point at infinity of the plane. The set of all points of the complex plane including the point at infinity is called the entire complex plane, the entire z-plane, or the extended complex plane. The sphere is sometimes called the Riemann Sphere. When the diameter of the Riemann Sphere is chosen to be unity, the equator corresponds to the unit circle of the complex plane.

73

7(a) Fundamental Theorem of Algebra: Every polynomial equation p ( z )=a0+a1 z

1+a2 z2+…+an z

n=0, where the degree n≥1 and an≠0, has at least one root.Proof: Suppose on the contrary, p ( z )=0 has no root, then f ( z )= 1p(z ) analytic for all

z. Also, |f (z)|= 1|p(z)| is bounded (and in fact approaches zero) as |p ( z )|→∞. Then it

follows that f (z) and p(z ) must be constants. This is a contradiction and we thus conclude that p ( z )=0 must have at least one root.7(b)Solution: Haven known that w=αz+β

γz+σ , we have: i=α (0 )+ βγ (0 )+σ

… (1 ) ,1=α (−i )+βγ (−i )+σ

…(2),and 0=α (−1 )+β

γ (−1 )+σ …(3). From (3), β=α. From (1) and (3), σ= β

i=−iα . From (2), γ=iα .

Therefore, w= αz+αiαz−iα

=1i ( z+1z−1 )=−i( z+1z−1 ).

7(c)Solution: Let z=eiθ. Then sin θ= z−z−1

2i, cosθ= z+z−1

2, dz=izdθ, so that

∫0

2π dθ3−2cosθ+sinθ

=∮C

❑ 2dz(1−2i ) z2+6 iz−1−2 i

. The poles of 2dz(1−2 i ) z2+6 iz−1−2i

are 2−i and 2−i

5 , but only 2−i5 lies inside C.

74

Res( 2−i5 )= lim

z→ 2−i5

{z−2−i5 }{ 2dz

(1−2 i) z2+6 iz−1−2i }=lim

z→ 2−i5

2

2 (1−2 i ) z+6 i= 12 i

by L-Hospital Rule.Therefore, ∮

C

❑ 2dz(1−2 i ) z2+6 iz−1−2i

=2 πi( 12 i )=π. And, ∫0

2π dθa+bsinθ

=∮C

❑ 2dzb z2+2aiz−b

. The poles of 2

b z2+2aiz−b are ( – a+√a2−b2

b )iand( – a−√a2−b2

b ) i. only ( – a+√a2−b2

b )i lies inside C since |( – a+√a2−b2

b ) i|=|√a2−b2−ab

√a2−b2+a√a2−b2+a|<1. Whenever a>¿b∨¿,

Res(z1=( –a+√a2−b2

b )i)=limz→z1

2b z2+2aiz−b

= limz→z2

22bz+2ai

= 1b z1+ai

= 1√a2−b2i

by L’ Hospital Rule. Therefore, ∮

C

❑ 2dzb z2+2aiz−b

=2πi( 1√a2−b2 i )= 2π

√a2−b2. And, ∫

0

2π 1a+bsinθ

= 2π√a2−b2

. Thus, ∫0

2π

( 1a+bsinθ+

13−2cosθ+sinθ )dθ=π ( 2+√a2+b2

√a2+b2 ).

Real AnalysisReal Analysis is the analysis of real numbers, which can summarily be described as the x-axis. Real numbers have two dichotomous properties: Algebraic Property and Topological Property. The we only delve into the algebraic property.PreliminariesThe algebraic property of R, the denotation of real number system or set of real numbers, explicates that R is a complete ordered field. This means R is a field, it is complete and it is partially ordered set.75

We show that R satisfies properties of a field as follows: Let +¿ and × be the ordinary addition and multiplication of binary operation. Then:i. R is closed with +: a+b is a real number if a and b are real numbers;ii. +¿ is associative. Shifting bracket is possible over +¿ in R,

a+ (b+c )= (a+b )+c , ∀a ,b , c∈R ;iii. Zero is a real number, and it behaves as:0+a=a=0+a;iv. If a is a real number, then – a is also a real number and a+ (−a )=(−a )+a=0 ;v. ∀ a ,b∈R ,a+b=b+a

That is, R is an abelian group with respect to addition. We now show that R is also an abelian group with respect to × as follows:vi. R is closed with respect to ×. That is, a×b is a real number if a and b are real numbers;vii. × is associative. Shifting bracket is possible over × in R,

(a×b )×c=a× (b×c ) , ∀a ,b , c∈R;viii. 1 is a real number and it behaves as 1×a=a=a×1 ;ix. If a is a real number, then1a is also a real number and a× 1a=1a×a=1 ;

x. a×b=b×a , ∀a ,b∈R

R is a field if the following is satisfied:xi. Left Distributivity: a× (b+c )= (a×b )+(a×c) and Right Distributivity:

(b+c )×a= (b×a )+ (c×a ) ,∀a ,b , c∈R.Theorem.

(a) If v and a are elements in R with v+a=a, then v=0.(b) If u and b≠0 are elements in R with u×b=b, then u=1.(c) If a∈R, then a×0=0.Proof.

(a) v+a=a implies v+a+b=a+b (comfortability of addition). Let b=−a. Then, v+a+ (−a )=a+(−a). By (iv), v=0(b) u×b=b implies u×b×c=b×c (comportability of addition)

Let c=1b . Then, u×b× 1b=b× 1b . By (ix), u=1.

(c) a×0=a× (0+0 )=a×0+a×0. That is,a×0=a×0+a×0

76

Add −a×0 to both sides,−a×0+a×0=−a×0+a×0+a×0

By(xi),(−a+a )×0=(−a+a )×0+a×0

By (iv), 0×0=0×0+a×00=a×0, that is; a×0=0Theorem. If a×b=0, then either a=0 or b=0.Proof. Since a×b=0, 1a × (a×b )=1

a×0, i.e. b=0. Also,1b× (a×b )=1

b×0

1b× (b×a )=0× 1

bby (x)

( 1b ×b)×a=0by (vii)

a=0

Theorem. There does not exist a rational number r such that r2=2.Proof. Suppose, on the contrary, that p and q are integers such that ( pq )

2

=2 where gcd ( p ,q )=1. Then p2

q2=2, i.e. p2=2q2. This implies that p2 is even which further

implies that p is even. This is from the fact that the square of odd number is odd. And, p=2m where m∈N . This means the gcd ( p ,q )=2contradicting gcd ( p ,q )=1. Because of this, it contradicts pq is not a rational number.Remark. This theorem says that irrational number do exist. And, R={irrational number }∪ {0 }∪{rational number }. We will use it in proving that Q is not complete.Theorem. The set of real numbers is an ordered set.Proof. An order relation is “<”. For all x , y∈R; exactly one of the following 3 relations hold:

(i) x= y (ii) x< y (iii) y<x77

These are called trichotomy. It is conventional to agree that x< y means y>x. Also, we write x≤ y for x< y or x= y. Numbers x∈ R such that x>0 are called positive, and numbers y∈R such that y<0 are called negative.Proposition. Foa all x , y∈R ,

(i) x< y if and only if x− y<0 iff y−x>0(ii) If x< y and y<z, then x<z.Proof. x< y iff x+(− y )< y+(− y) whence comportability of addition.That is, x< y iff x− y<0.Moreover,x− y<0 iff x− y+( y−x )<0+( y−x ) because comportability of addition of ( y− x)

That is, x+0−x<0+ ( y−x )

That is, 0< y−x since x−x=0 and y− y=0

ii.) x< y and y<z mean y−x>0∧z− y>0. That is y−x+( z− y )>0, i.e. z−x>0 which implies x<z.R is an ordered field.Absolute ValueThe absolute value or numerical value or modulus of a real number a∈R is denoted by ¿a∨¿ and defined by

|a|={ aifa>00 if a=0

−a if a<0

For example, |−3|=3 ,|3|=3, and so on. Intuitively, the absolute value of a∈R represents the distance from 0 to a.Some properties of absolute values are given in the following theorems.Theorem.

i. |x|≥0 ,∀ x∈ R ;|x|=0 iff x=0ii. x≤∨x∨¿iii. |−x|=¿ x∨¿iv. |−x|=|x|=xv. ¿−x∨¿2=x2=¿x∨¿2¿¿

78

vi. |xy|=|x|∨ y∨¿

Theoremi. |x|≥0∀ x∈ R; |x|=0 iff x=0ii. x≤|x|iii. |−x|=|x|=xiv. −¿ x∨¿2=x2=¿ x∨¿2¿¿v. ¿ xy∨¿∨x∨¿ y∨¿vi. |x+ y|≤∨x∨+¿ y∨¿vii. ¿ x∨−¿ y∨≥∨x− y∨¿

The last two are called |Triangular Inequalities.Proof(i). ∀ x∈R, from trichotomy, x>0 , x<0 or x=0. By definition, |0|=0. When x>0 ,|x|=x which is greater than 0. When x<0 ,|x|=−x. Since x<0 ,−x>0(when you divide both sides of −1, the sign changes). Hence, |x|=−x>0. Therefore, |x|≥0 ,∀ x∈ R.Proof(ii). From trichotomy, x>0 , x<0 or x=0. If x=0 ,∨x∨¿0=x…(1). When x>0 ,|x|=x…(2). When x<0 ,|x|=−x>x…(3). From (1), (2) and (3), ¿ x∨≥ x.Proof(iii). When x=0 ,|x|=0=−x. This implies |x|=¿−x∨¿. When ¿0 ,|x|=−x , where – x>0 and |−x|=x, where x>0, thus 0← x=x>0. Hence, ¿ x∨¿∨−x∨¿.Proof(iv). From (iii), |x|=¿−x∨¿. That is, x=|x|=|−x|=x. This iimplies, |x|=|−x|=x.Proof(v). From (iv) , |x|=|−x|=x. Applying square all through, we get the required result.Proof(vi). If x , y<0 ,|xy|=(−x ) (− y )=¿ x∨¿ y∨¿. If x , y>0 ,|xy|=xy=¿ x∨¿ y∨¿. O=If x>0 and y<0, |xy|=−xy=( x ) (− y )=¿ x∨¿ y∨¿. So far the 3 cases, |xy|=|x|∨ y∨¿.Proof(viii). If x , y>0 ,| xy|= x

y=¿ x∨ ¿

¿ y∨¿¿¿. If x , y<0 ,| xy|= x

y=¿ x∨ ¿

¿ y∨¿¿¿ becomes

−x− y

= xy . If x<0 , y>0 ,|xy|=−x

y=−x

y=¿x∨ ¿

¿ y∨¿¿¿. For the 3 cases, |xy|=¿ x∨ ¿

¿ y∨¿¿¿.

Proof(viii). ¿ x∨¿2=x2 ¿. So ¿ x+ y∨¿2=(x+ y )2=( x+ y )(x+ y)¿ from ¿ x∨¿2=x2 ¿

¿ x2+ xy+xy+ y2

¿ x2+2xy+ y2

79

≤ ¿ x∨¿2+2|x||y|+¿ y∨¿2 since xy ≤∨x∨¿ y∨¿¿¿

¿¿¿

Hence, ¿ x+ y∨¿2≤ ¿¿¿

Proof(ix). ¿ x− y∨¿2=(x− y)2=( x− y )(x− y )¿ from (v) ¿ x∨¿2=x2 ¿.¿ x2−xy−xy+ y2

¿ x2−2 xy+ y2

¿¿ x∨¿2−2|x|∨ y∨+¿ y∨¿2¿¿

xy ≤∨x∨¿ y∨¿ which implies – xy ≥−¿x∨¿ y∨¿ and −2 xy≥−2∨x∨¿ y∨¿

That is, ¿ x− y∨¿2≥ ¿¿¿

|x− y|≥∨x∨−¿ y∨¿

Density Property of RBetween any two real numbers, there is a rational and irrational numbers. That is, Q is dense in R, making Q an un complete, as we shall see very soon.Countability of QThe set of rational numbers is countable since it is the union of {0n , 1

n,−1n

, 2n,−2n

,…}and {m1 , m

−1, m2, m−2

, m3,…} where n ,m∈Z.

Finite (or infinite) union of countable sets is countable.Just that we have seen that Q is countable, we will see that (0 ,1), a subset of R is not countable.Theorem. The interval (0 ,1) is not countable.Proof. This proof is by contradiction. Suppose on the contrary (0 ,1 ) is countable. Then we can find a function, say f (n) that is in one-one correspondence with the set of natural numbers. f (n )=0.n will certainly do and f (n )=0.0n will also do, but f (n )=0.n+0.0n

2 has not been taking cared of as well as f (n )=0. ¨n , f (n )=0.0 ¨n , f (n )=0.00 n, etc. Hence, (0,1) is not countable is a contradiction.80

Theorem. R is not countable.Proof. Since R=(−∞ ,0 )∪ {0 }∪ (0 ,1 )∪ {1 }∪(1,∞ ) and (0 ,1) is not countable from the theorem above, R is not countable.IntervalA subset I of R is called an interval if it contains to two distinct elements and every element (number) lying between the 2 elements (numbers) belongs to I.An interval of the type I={a≤x<b :a<b } or ¿ or ¿ is semi open or semi closed interval.An interval of the form I={a≤x ≤b : a<b } or [a ,b] is a closed interval.Bounded and Unbounded SetsA set A⊆R is said to be bounded from above if ∃ k∈ R∋ x≤ k ,∀ x∈ A. K is the upper bound of A.A set A⊆R is said to be bounded from below if ∃ k∈ R∋ k ≤x ,∀ x∈ A. K is the lower bound of A.Example

1) The upper and the lower bounds of {1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 } are 9 and 1 respectively.2) N is bounded from below by 1, it is not bounded from above.3) {1n :n∈N }={1 , 12 , 13 , 14 , 15 ,…} is bounded from above by 1 and from below by 0.

Definition. A set A⊆R is said to be bounded if it is both bounded from above and below.Definition. The set of upper bounds of A is the set of all k∈R∋ x≤ k ,∀ x∈ A, i.e. {k :k∈R∋ x≤ k ,∀ x∈ A⊆R }

Definition. The set of lower bounds of A⊆R is the set of all k∈R∋ k ≤x ,∀ x∈ A, i.e. {k :k∈R∋ k ≤ x ,∀ x∈ A⊆R }.Example : The set of lower bounds of N is ¿.Definition (Maxima Element) . Let A⊆R. M is said to be the maximal element of A if :

i) m∈ Aii) m is the supremum of A81

Definition (Minima Element) . Let A⊆R. M is said to be the minima element of A if : i) m∈ Aii) M is the infimum of A

Non Example: 0 is not the minima element of {1n :n∈N } despite 0 is the lower bound.Theorem. A set cannot have more than one supremum.Proof. Let m1 and m2 are 2 suprema of A⊆R. If m1 is the supremum and m2 is an upper bound, then m1≤m2…(1)If m2 is the supremum and m1 is an upper bound, then m1≥m2…(2)

From (1) and (2), m1=m2. That is, the supremum is unique.Completeness of RThe set of real numbers is complete. This means that every non-empty subset of R bounded from above has its supremum(lub) in R or every non-empty subset of R boundedfrom below has its infinum (glb) in R.Theorem . Q is not complete.Relations, Mappings and FunctionsWe are familiar with many relations such as “less than”, “is parallel to”, “is a subset of”, and so on. In a certain sense, these relations consider the existence or nonexistence of a certain connection between pairs of objects taken in a definite order. Formally, we define a relation in terms of these “ordered pairs”.Definition. Let A and B be two sets. Then the set of all ordered pairs (a, b) with a∈ A and b∈B is called the catesian product of A and B, denoted by A×B (read A cross B).That is A×B={(a ,b ) :a∈ A∧b∈B }

Example: Let A={1,2 } and B={a ,b , c }. Then A×B={(1 , a ) , (1 , b ) , (1, c ) , (2, a ) , (2 , b ) , (2, c ) , }

B× A={(a ,1 ) , (b ,1 ) , (c ,1 ) , (a ,2 ) , (b ,2 ) , (c ,2 ) , }

Also, 82

A× A={(1 ,1 ) , (1 ,2 ) , (2 ,1 ), (2 ,2 )}

The idea of points in a plane is an example of Cartesian product.Let A, B and C be sets. ThenA×B×C={ (a ,b , c ) :a∈B ,∧c∈C } are triplesA1× A2×…× An={ (a1 , a2 ,…,an ): an∈ An } is called n-turples.An means A× A× A…n׿

Definition. A subset of Cartesian product is called a relation. Let A and B be sets and A×B is the Cartesian product of A and B. Then, any subset R of A×B is called a relation between A and B. If (a ,b)∈R, we write aRb, a is related to b.Example: Let A={1 ,2,3 } and B={x , y , z } and let R={(1 , y ) , (1 , z ) ,(3 , y)}. Then R is a relation from A to B since R is a subset of A×B.Types of RelationsThe following are types of relation:

a) Reflexive RelationA relation R on a set A is reflexive if aRa for every a∈ A. A relation such as R={(1 ,1 ) , (2 ,2 ) ,(3 ,3)} is reflexive relation defined on A={1 ,2,3 }.Other examples are a≤a ,¿a, where a is a line, a¿, a divides a, a is a subset of a, etc.

b) Symmetric RelationA relation R on a set A is symmetric if whenever aRb ,then bRa.Examples: If a¿b, then b¿a.c) Antisymmetric RelationIf R is not symmetric it is called antisymmetric.

c) Transitive RelationA relation R on a set A is transitive if whenever aRb and bRc, then aRc.Examples: If a¿band b¿b, then a¿c

If A⊆B and B⊆C , then A⊆C.A relation R on a set A is called a partial order on S if it is reflexive, antisymmetric and transitive.83

Equivalence RelationA relation R on a set A is said to be an equivalence relation if it is: reflexive, symmetric and transitive.Example: Let A is a non-empty set. Then the relation “”is equal to’’ is an equivalence relation.

i. ∀ x∈ A ,x=x (Reflexive)ii. ∀ x , y∈ A , x= y implies y=x (Symmetric)iii. ∀ x , y , z∈ A , x= y∧ y=z⇒ x=z(Transitive)K. F. Gauss proposes an equivalence relation that generalizes the notion of equality. It is congruence relation. a≡bmod ciffa−b

c∈Z.

i. It is clear that a≡amod c since a−ac0∈Z(Reflexive)

ii. If a≡bmod cthen a−bc

∈Z which implies b≡amod csinceb−ac

∈Z.iv. If a≡bmod dand b≡cmod d, then: a−b

d∈Z and b−c

d∈Z which implies

a−bd

+ b−cd

∈Z which further implies a−b+b−cd

∈Z which implies a−cd

∈Z

That is a≡cmod d( Transitivity)Definition (Equivalence Class). Let be an equivalence relation on a set A. Equivalence Class of a, denoted by a is the set {x∈ A : x a}. That is, a={x∈ A : x a }.Theorem. Let be an equivalence relation on a set A. Then a b implies a=b , ∀a ,b∈ A.Proof. Since a b , then x∈a implies x∈b ,∀ x∈ A, i.e. a⊆b…(1)Since is an equivalence relation, b a and x∈b implies x∈a ,∀ x∈ A, i.e. b⊆a…(2)From (1) and (2), a=b.Theorem. Let be an equivalence relation on A. Then a=b implies a b , ∀a ,b∈ A.Proof. Given that a=b, if a∈a, then a∈b= {x : x b }, i.e. {a ,a b }, i.e. a b

84

Theorem. a=b≡a b.Proof. This result follows immediately from the two theorems above.Theorem. Let is an equivalence relation on A. For any, a ,b∈ A ,a=b or a∩b=∅ .Proof. Let a ,b∈ A. Then either a b or a≁ b. If a b, then a ,b∈ A ,a=b. If a≁ b, then suppose a∩b≠∅ , ∃ c∈a∩b which implies c∈a and c∈b which imply c a and c b. Because of symmetry of , a c. Hence, a c and c b. This implies a b which further implies a=b by transitivity of . This is a contradicting a≁ b. RemarkThis theorem says that equivalence classes are either equal or no common element between them. That is equivalence classes partition set into disjoint equivalence classes. We take example of how equivalence relation ≡ partitions Z into disjoint equivalence classes.Examples:

i) a≡bmod 4 has the following classes:a−04

∈Z=0={…,−12 ,−8 ,−4 ,0 ,4 ,8 ,12,…}

a−14

∈Z=1= {…,−11 ,−7 ,−3 ,1,5 ,9 ,13 ,… }

a−24

∈Z=2= {…,−10 ,−6 ,−2 ,2 ,6 ,10 ,14 ,…}

a−34

∈Z=3={…,−9 ,−5 ,−1,3 ,7 ,11 ,15… }

a−44

∈Z=4={…,−12 ,−8 ,−4 ,0 ,4,8,12 ,…}

ii. The equivalence relation a≡b mod 7 has the following equivalence classes:a−07

∈Z=0= {…,−21 ,−14 ,−7 ,0 ,7 ,14 ,21 ,28 ,…}

a−17

∈Z=1= {…,−20 ,−13 ,−6 ,1,8 ,15 ,22,… }

a−27

∈Z=2= {…,−19 ,−12 ,−5 ,2 ,9 ,16 ,23 ,… }

a−37

∈Z=3={…,−18 ,−11 ,−4 ,3 ,10 ,17 ,24 ,… }

85

a−47

∈Z=4= {…,−19 ,−10 ,−3 ,4 ,11 ,18 ,25 ,32 ,…}

a−57

∈Z=5={…,−16 ,−9 ,−2 ,5 ,12 ,19 ,26 ,33 ,…}

a−67

∈Z=6={…,−15 ,−8 ,−1 ,6 ,13 ,20,27 ,34…}

FunctionOne of the most important concepts in mathematics is that of a function. The terms “map”, “mapping”, “transformation” and many others mean the same thing. Mapping is the generalization of a function where for the sets A and B, the subset of A×B is a mapping if all the elements of A are involve and no two elements in B have one pre-image is a mapping.A function is a relation (or rule) f which assigns to each element x∈ A unique f (x)∈B.Examples: If f :R+¿→R¿ defined by f ( x )=√x is a function because it is single-valued, then f (±√ x )=x is multiple valued, therefore it is not a function. f ( x )=1

x is a function except at x=0.Operations on FunctionsLet f : A→B and g :A→B are two functions defined by f and g respectively. Then:

i. ( f ± g ) ( x )=f ( x )±g ( x ) ,∀ x∈ A and f ± g is a functionii. fg ( x )=f (x ) . g ( x ) ,∀ x∈ A and fg is a function.iii. f

g( x )= f (x )

g ( x ), ∀ x∈ A where g ( x )≠0 , f

g is a function.Definition. Let f : A→B is a function. Then f is said to be one-one or injective if f (x1 )=f (x2) implies x1=x2 ,∀ x1 , x2∈ A. Example: If f :R❑→R is defined by f ( x )=2x−3, then f is one-one whence f (x1 )=f (x2) implies 2 x1−3=2 x2−3 which implies x1=x2.Definition. Let f : A→B be a function. Then f is said to onto or surjective if ∀ y∈B ,∃ x∈ A∋ f ( x )= y. That is, range of f is B.Example: f ( x )=sin x

86

Definition. Let f : A→B be a function. Then f is said to be bijective if f is both one-one and onto.Example:f ( x )=sin x+x

Composition of FunctionsLet f : A→B and g :B→C are two functions. Then gof : A→C is a function given as (gof ) (x )=g ( f ( x ) ) , x∈ A.Example: Let f :R❑→R be defined by f ( x )=3 x−2 and g :R❑→R be defined by g ( x )=2 x2−5x+1. Then

( fog) ( x )=g ( f ( x ) )

¿ g (3 x−2 )

¿2(3 x−2)2−5 x+1

¿18 x2−39x+19

Definition. Let f : A→A be a function. Then the inverse f−1:B→A of f is also a funvtion iff f is one-one and onto. Definition(One-One Correspondence). Let f : A→B be bijective function. Then A and B are said to be one-one correspondence.Graph of a FunctionThe graph G of f : A→B has given rise to a relation as follows:

G={(x , f ( x )) : x∈ A }

Sequence and Series of Real NumbersA sequence of real numbers is a function whose domain is a subset of natural numbers and its range is a subset of real numbers,. That is f : A⊆N→B⊆R.If the subset of the domain is finite, the sequence is finite, other wise , it is an infinite sequence.The idea of saying subset of N is from the fact that no zeroth term or -10th term or so in a sequence. A sequence is having countable number as its terms.We are more interested in the infinite sequences. Since sequence is a function, most of the property that hold in function holds in sequences: Monotonicity, Convegence, etc.87

Definition. A sequence xn is said to be:i. Increasing if xn≤ xn+1 ,∀n∈Nii. decreasing if xn≥ xn+1 ,∀n∈Niii. Constant if xn=xn+1 ,∀ n∈Niv. Strictly Increasing if xn<xn+1 ,∀n∈Nv. Strictly decreasing if xn≤ xn+1 ,∀n∈N

The first two are refered to as Monotne and the last two are called strictly monotone.Definition (Bounded Above Sequence). A sequence xn is said to be bounded from above if ∃m∈R∋ xn≤M ,∀n∈N

Definition (Bounded Below Sequence). A sequence xn is said to be bounded from below if ∃m∈R∋M ≤xn ,∀n∈N

Definition (Bounded Sequence). A sequence xn is said to be bounded if ∃m∈R∋∨xn∨≤M , ∀n∈N .A sequence is bounded if its range is bounded.Examples: xn defined by xk=3k is bounded from below and it is not bounded from above.Example: xn defined by xn=

1n is both bounded from above and below.

xn=11, 12, 13, 14,…, 1

10000,…, 1

∞=0

xn=1 ,12, 13, 14, 15,…, 1

10000,…,0

It is bounded from below by 0 and from above by 1.Convergence of a SequenceA sequence xn is said to converge to a real number l (or is said to have limit ) if given ε>0 ,∃ N∈N(with N depending on ε and possibly x0 and N is no matter how large) ∋|xn−l|<ε , ∀n≥N .This is the definition of limn→∞

xn=l which compares with limn→∞

f ( x )=l. The relation between N and ε is Nα 1ε .

88

SinceR is x-axis, ¿ xn−l∨¿ ε means l−ε<xn< l+ε. limn→∞xn=±∞ means xn diverges.

Examplesi) Prove that lim

n→∞

12n

=0.Proof. We assume | 12n−0|<ε and show ∀ n≥N (i.e. n→∞). | 12n−0|<ε implies | 12n|<ε. Suppose n∈N , n is positive; thus 12n <ε. And, 12 ε<n. Let N= 1

2 ε<n, i.e.

n>N , i.e. n≥ N, i.e. ∀n≥N .ii) Prove that 2

(1+ p)3n=0 ,where p>0.

Proof. We assume | 2(1+ p)3n

−0|<ε and show ∀ n≥N (ie n→∞). | 2(1+ p)3n

−0|<ε implies | 2

(1+ p)3n|<ε. Since p>0, n is positive. Thus, 2(1+ p)3n

<ε. And, (1+ p)3=(1+ p ) (1+ p ) (1+ p )= (1+ p )(1+ p+ p+ p2)

¿ (1+ p ) (1+2 p+ p2 )=1+2 p+ p2+ p+2 p2+p3=1+3 p+3 p2+ p3

Thus, (1+ p)3>1+3 p since 3 p2+p3>0.(1+ p)3n>1+3np

Hence 1(1+ p)3n

< 11+3np which is less than ε since n , p>0.

Thus, 11+3np

<ε which implies 1<ε (1+3np )=ε+3 εnp which implies 1−ε3 εp

<n which implies 13 εp− 1

3 p<n. Set 13 εp− 1

3 p=N . Hence, N<n which finally implies ∀n≥N .

iii) Prove limn→∞

3n2

n2+1=3.

89

Proof. Assume | 3n2n2+1|<ε which implies |3n2−3n2−3n2+1 |<ε which implies | −3n2+1|<ε

which implies 3n2+1

<ε since n>0 which implies n2>0.3<ε n2+ε e which implies 3−ε

ε<n2 which implies √ 3ε −1<n which implies N=√ 3ε−1. Hence, N<n which implies ∀n≥N , as required.Like convergence of function, sequences converges to a unique limit.

Theorem. A sequence converges to a unique limit.Proof. This proof is by reducio ad absurdum (by contradiction). Suppose on the contrary xn→l1 and l2 where l1≠ l2. Then xn→l1 means |xn−l1|<ε ,∀n≥N andxn→l2 means |xn−l2|<ε ,∀n≥N We check ¿ l1−l2∨¿: |l1−l2|=¿ l1−xn+xn−l2∨¿

≤|l1−xn|+|xn−l2| from Triangular Inequality≤|xn−l1|+|xn−l2|<

ε2+ ε2=ε

Hence, |l1−l2|<ε, but only 0<ε. Hence, ¿ l1−l2∨¿0 which implies l1−l2=0 which implies l1=l2.Theorem. Every convergent sequence is bounded.

Proof. Let xn converges. That is, ¿ xn−l∨¿ ε ,∀n≥ N. Since 0<ε<1, ¿ xn−l∨¿ ε<1 which implies |xn−l|<1 which implies |xn|−|l|<1 which implies |xn|<1+¿ l∨¿.Let M=max {x1 , x2 ,…, xn−1 , xn}

¿max {x1 , x2 ,…, xN−1 ,1+¿l∨} since n≥ N. Then |xn|≤M ,∀ n∈ N.Definition(Subsequence). Given a sequence xn, consider a sequence xnkof positive integers such that n1<n2<n3…; then the sequence xnkis called a subsequence of xn.Examples

90

i) { 12n }, { 12n }, { 12n } , …, are subsequences of {1n }ii) {n+1 }, {n+2 } , {n+3 },… are subsequence of {n}iii) xn+1 , xn+2 , xn+3 are thus sequences of xn

Theorem. Any subsequence of a converging sequence converges to the same limit point.Proof. Let xn converges. Then |xn−l|< ε

2,∀n≥N and every Cauchy Sequence

converges, i.e. |xn−xnk|<ε2 which implies |xnk

−xn|< ε2 from property of absolute value. Thus,

¿ xnk−xn∨+¿ xn−l∨¿ ε

2+ ε2=ε and ¿ xnk

−l∨≤∨xnk−xn∨+¿ xn−l∨¿ ε which further implies

|xnk−l|<ε ,∀n≥ N, as required.Algebraic Operations on Sequences and Their LimitsLet xn and yn are two real sequences . Then:

i) limn→∞

{xn+ yn }=limn→∞

{xn }+ limn→∞

{yn }

ii) limn→∞

{xn− yn }=limn→∞

{xn}− limn→∞

{ yn}

iii) limn→∞

{xn× yn }=limn→∞

{xn }× limn→∞

{ yn}

iv) limn→∞

{xn÷ yn }= limn→∞

{xn}÷ limn→∞

{ yn}

This proofs are easy and we have seen their analogue in complex analytic part. The theorem portrays that limit forms an algebraic group.Examples

1) Prove that limn→∞

3n3+4n3+3

=3.Proof. We assume |3n3+4n3+3

−3|<ε which implies |3n3+4−3n3−9n3+3 |<ε which implies | −5n3+3|<ε which implies 5

n3+3<ε which implies 5

n3<ε which implies 5ε <n3 which

3√ 5ε <n , ∀ n≥N .91

2) Classically show limn→∞ ( 3n3+4n3+3 )

3

=27.

Proof. limn→∞ ( 3n3+4n3+3 )

3

=( limn→∞

3n3+4n3+3 )

3 since polynomials are continuous.

¿( limn→∞

3+ 4n3

1+ 3n3

)3

=(3+01+0 )3

=33=27

3) If 0<a<1, prove that {an } converges.Proof. Since 0<a<1, 0n<an<1n which implies 0<an<1. Hence, {an } is bounded. an+1−an=an (a−1 )<0 since a−1<0 where a<1. Hence, an+1−an<0 which implies an+1<an.Theorem (Sandwish Theorem). If yn→0 and 0≤ xn≤ yn, ∀ n∈N ; then xn→0.Proof. Since 0≤ xn≤ yn,|0|≤∨xn≤|yn|, yn→0 means ¿ yn∨¿ ε ,∀n≥ N. Hence, 0≤|xn|≤|yn|<ε which implies |xn|<ε ,∀ n≥N . That is, xn→0.Corollary. If xn→a, zn→a and xn≤ yn≤ zn, then yn→a.Proof. xn≤ yn≤zn implies xn−xn≤ yn− xn≤ zn−xn , ie 0≤ yn−xn≤ zn−xn

limn→∞

¿

By the Sandwish Theorem, limn→∞

( yn−xn )=0 which implies limn→∞

yn−limn→∞

xn=0 which implies limn→∞

yn−a=0 which implies limn→∞yn=a. Hence, yn→a.

Cauchy SequenceA sequence xn is called Cauchy Sequence if given ε>0 ,∃ N∋|xn−xm|<ε ,∀n ,m≥ N.Theorem. Every Cauchy Sequence is bounded.Proof. Let xn is Cauchy. That is, |xn−xm|<ε ,∀n ,m≥ N . By definition of ε , ε<1 implies |xn−xm|<1. From Triangular Inequality,

92

|xn|−|xm|<|xn−xm|<1 implies |xn|−|xm|<1which further implies |xn|<1+|xm|. Let m=N , since m≥N . Then|xn|<1+|x N|

Let M = max {x1 , x2 ,…,xn−1 , xn }

¿max {x1 , x2 ,…, xN−1 ,1+|xN| since n≥ N.Then ¿ xn∨≤M , ∀n∈N , ie boundedness.Lemma. Every sequence has a monotone subsequence.Proof. Every sequence is either monotone or constant. If it is a constant sequence, it has a constant subsequence. If it is monotone, say {xn+1}. Sine n is monotone, nk is monotone and xnkis motone subsequence of xn if xn is not a constant sequence.Bolzano-Weierstrass TheoremEvery infinite and bounded sequence converges.Proof. From the lemma, the sequence has a monotone subsequence. Since the super sequence is bounded, the subsequence is also bounded. Any sequence that is increasing or decreasing and bounded converges. Hence, the sequence converges to the limit poit of its subsequence.Corollary. Every infinite bounded sequence of real number has a convergent subsequence.Proof. Since every sequence has a monotone subsequence, then the subsequence is also bounded since the super sequence is bounded. Thus, the subsequence is monotone and bounded. By Bolzano-Weierstrass Theorem, it converges.Cauchy Sequence Convergence TheoremEvery Cauchy Sequence converges.Proof. Let xn is Cauchy. That is, |xn−xm|<

ε2, ∀n≥m≥N . That is,

|xn−xnk|<ε2, ∀ n ,m≥ N

93

From Bolzano-Weierstrass Theorem, it has a convergent subsequence, that is, |xnk−l|< ε

2, ∀ k ≥N . Then

¿ xn−xnk∨+¿ xnk

−l∨¿ ε2+ ε2 , ie ¿ xn−xnk

∨+¿ xnk−l∨¿ ε

From Triangular Inequality,|xn−l|≤|¿ xn−xnk

+xnk−l∨¿ ε|≤∨xn−xnk

∨+¿ xnk−l∨¿ε

This implies ¿ xn−l∨¿ ε ,∀n ,m≥ N .

Example: Show that {sinnπn } is Cauchy and it converges.Proof. |xn−xn|=|sin nπn

−sinmπm |=|m sin nπ−nsin nπ

nm |≤|msinnπmn |+|nsinmπmn |≤| mnm+ n

nm|=m+nnm

Suppose M≥ N ,|xn−xm|≤

m+nnm

≤ 2mnm

=2n

Let 2n<ε. Then 2ε <n. Set N=2ε . Then N<n implies ∀ n≥N .

Divergent SequenceA sequence is said to be divergent if it does not converges. If it does not converges mean if it is limit does not exist.A bounded sequence which does not converge is said to oscillate. An example is xn=(−1)n.We take the formal definition of divergence.Definition. A sequence xn is divergent if ∀ A>0 ,∃N>0∋ xn≥ A∨xn≤−A ,∀ n≥ N

That is, limn→∞xn=∞ or limn→∞

xn=−∞

94

Example: Set of natural number diverges.Proof. Let N=A+1, since ε>δ (most prefarably). Then N>A. If ∀ n≥N exists, then n≥ N>A. Hence, {n }≥ N≥ A and {n} is the set of natural numbers, where n∈N .Definition(Limit Superior). Let xn is a sequence of real numbers which is bounded above. Then ∃M>0∋ xn≤M ,∀ n∈ N. The set {xn , xn+1 ,… } is bounded from above by M and it has lub M n. That is, M n=lub {xn, xn+1 ,…}. Then M n≥M n+1≥M n+2≥…. Hence {M n } is a decreasing sequence bounded from above by M.Case I: If {M n } converges, limit superior of xn=lim

n→∞¿ xn=lim

n→∞M n.

Case II: If {M n } diverges, limit superior of xn=−∞=limn→∞

M n

Definition(Limit Superior). Let xn is a sequence of real numbers which is bounded below. Then ∃M>0∋M ≤xn ,∀ n∈N. The set {xn , xn+1 ,… } is bounded from below by M and it has glb M n. That is, M n=glb {xn , xn+1 ,…}. Then M n≤M n+1≤M n+2≤…. Hence {M n } is a increasing sequence bounded from below by M.Case I: If {M n } converges, limit inferior of xn=lim

n→∞inf xn=lim

n→∞M n.

Case II: If {M n } diverges, limit inferior of xn=∞= limn→∞

M n

SeriesLet xn is a sequence. Then ∑ xn is a series. That is, a series is summation of sequences. If ∑

n=1

∞

xn is an infinite series, then ∑k=1

n

xk is its partial sum. Let sn=∑k=1

n

xn. Then lim

n→∞∑n=1

∞

xn. That is, limn→∞

sn=∑n=1

∞

xn ,the infinite sequence.If {Sn } converges, then ∑

n=1

∞

xn converges. If {sn } diverges then ∑n=1

∞

xn diverges.Example:

95

1) Show that ∑k=0

∞

rk converges to 11−r when 0<r<1 and it is undefined when r=1

and it diverges when r>1.Proof. ∑

k=0

∞

rk=r 0+r1+…+r∞

Sn=∑k=0

n−1

rk=r0+r1+…+r n−1

sn=r0+r1+…+r n−1

r sn=r+r2+…+rn

sn−r sn=1−r n

This imples sn=1−rn

1−r. Hence, sn= 1

1−r− rn

1−r and

limn→∞

sn=limn→∞ ( 1

1−r− rn

1−r )If 0<r<1, then lim

n→∞ ( 11−r

− rn

1−r )= 11−r

−0= 11−r

since r∞=0 since r=1k , ie ( 1k )

∞

= 1k∞= 1

∞=0

If r=1, limn→∞sn is undefined.

If r>1 , limn→∞

sn= limn→∞ ( 1

1−r− rn

1−r )= 11−r

− r∞

1−r

¿ 11−r

−∞=−∞, ie divergence.Example: Show that ∑

n=1

∞ 2n(n+1)

conveges to 2.Proof. ∑

n=1

∞ 2n(n+1)

=∑n=1

∞ 2n− 2n+1

by Partial FractionHence, sn=∑

n=1

n 2n− 2n+1

s1=( 21−22 )96

s2=( 21−22 )+( 22−23 )s3=( 21−22 )+( 22−23 )+(23−24 )…

sn=( 21−22 )+( 22−23 )+(23−24 )+…+( 2n−1

−2n )+( 2n− 2

n+1 )sn=2−

2n+1

limn→∞

sn=limn→∞ (2− 2

n+1 )=2− 2∞+1

=2−0=2

Theorem. The neccsssary condition for the series ∑n=1

∞

an to converge is its series converges to 0 (ie limn→∞

an=0).Proof. Let sn=∑

n=1

n

an. Then sn=a1+a2+…+an andsn−1=a1+a2+…+an−1.

sn−sn−1=an

So limn→∞

an= limn→∞

(sn−sn−1 )= limn→∞

sn− limn→∞

sn−1=s−s=0

Hence, limn→∞an=0.

Example1) Show that ∑

n=1

∞ 1n

converges.Proof. Since lim

n→0

1n=0, then ∑

n=1

∞ 1n

converges.2) Show that ∑

n=0

∞

(12 )nconverges to 2.

97

Proof. ∑n=0

∞

(12 )nis a geometric series with r=12 . Hence, it converges to

11−r

= 1

1−12

= 112

=2.Theorem (Cauchy General Principle of Convergence). The series ∑

n=1

∞

an converges iff given ε>0 ,∃ N∋|sn−sm|<ε , ∀m,n≥N . Proof.Necessity: Let ∑

n=1

∞

an converges. Then ∑n=1

n

an converges, ie {sn } converges. That is, sn converges to sm. Thus, |sn−sm|<ε ,∀m,n≥ N.Sufficiency: Let |sn−sm|<ε ,∀m ,n≥ N. Then {sn } is Cauchy. Thus, {sn } converges. Therefore, ∑

n=1

n

an converges. Hence, ∑n=1

∞

an converges.Theorem. Let ∑ an and ∑ bn be series and c ≠0, a constant. And ∑ an and ∑ bn are convergent series; then:1) ∑ (a¿¿n+bn)=∑ an+∑ bn¿ converges2) ∑ (a¿¿n−bn)=∑ an−∑ bn¿ converges3) ∑ c (a¿¿n)=c∑ an¿ converges

Definition. The series ∑ an is said to be absolutely convergent or that it converges absolutely if the series ∑ ¿an∨¿¿ converges.Definition: If ∑ an converges, but ∑ ¿an∨¿ diverges, then ∑ an is conditionally convergent.Theorem. If ∑ an converges, then ∑ an converges.Proof. Let ∑ ¿an∨¿¿ converges. Then ∑

1

n

|an|=sn converges and sn=|a1|+|a2|+…+|am|+…+¿an∨¿ and sm=|a1|+|a2|+…+|an|

sn−sm=|am+1|+|am+2|+…+|an|= ∑k=m+1

n

|ak|<ε, ie sn converges to sm.

98

But | ∑k=m+1

n

ak|≤ ∑k=m+1

n

|ak|<ε ,∀ n ,m≥N

Hence, | ∑k=m+1

n

ak|<ε, making ∑1

n

an to converge. Hence, ∑n=1

∞

an converges.Convergence TestIf an≤bn and an diverges, then bn diverges. If bn converges, then an converges.Examples:

4) Use comparison test to find the convergence of ∑1

∞ 15n−1+2

.Solution: ∑

1

∞ 15n−1+2

<∑1

∞ 15n−1

=∑1

∞

( 15 )n−1

<∑1

∞

( 15 )n and ∑

1

∞

( 15)n converges since 0<r=15<1

, then ∑ 15n−1+2

converges.Ratio TestSuppose an≠0 and lim

n→∞|an+1

an |<1, then ∑ an converges. If limn→∞|an+1

an|=1, the Ratio Test

has failed. If limn→∞|an+1

an |>1, then ∑ an diverges.Example

5) Use ratio test to show the convergence of ∑ n2n

.Proof. an=

n2n

and an+1=n+12n+1=

n+12n .2

. Hence, an+1

an=

n+12n2n2n

=n+12n

=12 (1+ 1n )

limn→∞

❑ 12 (1+ 1n )=12 (1+ 1∞ )=12

99

QUESTION PAPER REAL ANALYISIS I1(a) Define an Interval. (b) If a×b=0, show that either a=0 or b=0. (c) Show the Triangular Inequality |x+ y|≤|x|+|y|,∀ x , y∈R. (d) Why is R complete ordered field?2(a) Define binary relation. (b) By provision of counter example, testify that (0 ,1 ) is not countable much less R. (c) Why is the set {1n :n∈N } bounded. Is it having maxima element? (d) Show that a set cannot have more than one supremum.3(a) Define equivalence class a of a∈ A. (b) Show that congruence ≡ is an equivalence relation. (c) Show that either a=b or a∩b=∅ ∀a ,b∈ A. (d) Partition Z into two disjoint equivalence classes.4(a) Define graph. (b) Show that lim

n→∞

1(1+ p)n

=0, where p>0. (c) Show that the necessary condition for a series ∑ an to converge is the convergence of an to zero. (d) Use comparison test to show that ∑

1

∞ 1n!

converges.5(a) Define convergence of a sequence. (b) Use ratio test to show that ∑

1

∞ n2n

converges. (c) Use Sandwich Theorem to show that if xn→a, zn→a and xn≤ yn≤ zn, then yn→a. (d) Show that {an }, where 0<a<1, converges.6(a) State and prove the Sandwich Theorem. (b) Show that ∑

n=1

∞ 1n(n+1)

=1. (c) Why is f ( x )=±√x a function and why not f (±√ x )=x? (d) Define graph of a function.7(a) State and prove the Bolzano - Weierstrass Theorem. (b) Show that Q is not complete.

100

MTH 2307 (Real Analysis I) Marking SchemeQ1a) Define an Interval.Answer: A subset I of R is called an interval if it contains two distinct elements and every real number lying between the two elements belongs to I.1b) If a×b=0, show that either a=0 or b=0.Answer: Since a×b=0 , 1

a(a×b )=1

a×0 ∵Comportability of Multiplication and Existence of

1a in R.1a×a×b=0 ∵ Associativity,

b=0 since 1a ×a=1

Also, a×b=0 , 1b

(a×b )=1b×0 ∵ Comportability of Multiplication and Existence of 1b in R

1b (a×b )=1b×0 implies 1b (b×a )=1

b×0 ∵ Cummutativity in R

1b ×b×a=1b×0 ∵ Associativity inR

a=0 since 1b ×b=1 and 1b ×0=0Q1c) Show the Triangular Inequality |x+ y|≤|x|+|y|,∀ x , y∈R.Answer: Since ¿ z∨¿2=z2¿, |x+ y|2=(x+ y)2=x2+2 xy+ y2

≤|x|2+2|x||y|+|y|2

101

¿ (|x|+|y|)2

Hence, |x+ y|2≤(x+ y)2. Thus, |x+ y|≤|x|+|y|,∀ x , y∈R

Q1d) Why is R complete ordered field?Answer: R satisfies properties of a field. ∀a ,b∈R, a<b or a>b or a=b. That is R is ordered and R is complete because any subset of R bounded from above has its supremum in R and any subset of R bounded from below has its infimum in R. That is R is not dense in C.Q2a) Define binary relation.Answer: A binary relation is a subset of the Cartesian product A×B, where A and B are sets.Q2b) By the provision of counter example, testify that (0,1 ) is not countable much less R.Answer: In (0,1), there are

0.10.20.30.40.5…0.9

0.110.220.330.440.55…0.99

0.1110.2220.3330.4440.555…0.999

0.11110.22220.33330.44440.5555…0.9999…

But the number 0.123456789 is not captured as an element of a set that are in one-one correspondence with the set of natural numbers.And, R=(−∞ ,0 )∪ {0 }∪ (0,1 )∪ {1 }(1 ,∞) which is not countable since (0,1) is not.Q2c) Why is the set {1n :n∈N } bounded and is it having maxima element?

102

Answer: The set {1n :n∈N }={1 , 12 , 13 ,…}. It is bounded from above by 1, and from below by 0. Zero is not its maxima element because zero does not belong to the set. So it has no maxima element.Q2d) Show that a set cannot have more than one supremum.Answer: Suppose on the contrary that a set has two suprema and let m1 and m2 are the suprema of A⊆R. If m1 is the supremum and m2 is an upper bound, thenm1≤m2… (1)If m2 is the supremum and m1 is an upper bound, thenm1≥m2 … (2)From (1) and (2), m1= m2 . That is, the supremum is unique.Q3a) Define equivalence class a of a∈ A.Answer: Let be an equivalence relation. Then a={x∈ A : x a }

Q3b) Show that congruence “ ≡” is an equivalence relation.Answer:

i. Reflexivity: a≡amod c since a−ac

=0∈Z

ii. Symmetry: If a≡bmod c, then a−bc

∈Z implies −a−bc

∈Z which further implies b−a

c∈Z. That is b≡amod c

iii. Transitivity: If a≡bmod d and b≡cmod d, then a−bd

∈Z and b−cd

∈Z which implies a−b

d+ b−c

d∈Z which implies a−b+b−c

d=a−c

d∈Z.

103

Q3c) Show that either a=b or a∩b=∅ ∀ a ,b∈ A.Answer: Let is an equivalence relation and let a ,b∈ A. Then either a b or a≁ b. If a b, then a=b. If a≁ b, then suppose on the contrary that a∩b≠∅ and ∃ c∈a∩b. Then c∈a and c∈b which implies c a and c b. Because of symmetry of , a c and c b which implies a b which implies a=b, a contradiction.

Q3d) Partition Z into two equivalence classes.Answer: 0={…,−6 ,−4 ,−2,0,2,4 ,…} and

1= {…,−5 ,−3 ,−2 ,−1,1,3,5 ,…}

Q4a) Define graph.Answer: Let f : A→B be defined by y=f (x ). Then the graph G of a function is G= {(x , f (x)) : x∈ A }.Q4b) Show that lim

n→∞

1(1+ p)n

=0where p>0.Answer: |xn−l|=| 1

(1+ p)n−0|<ε which implies 1

(1+ p)n<ε which implies 1np < 1

(1+ p)n<ε .

Hence, 1np <ε which implies 1εp<n which implies N= 1εp

<n which further implies that n>N exists.Q4c) Show that the necessary condition for a series ∑ an to converge is limn→∞

an=0

.Answer: The lim

n→∞Sn=∑ an=S.

104

Let ∑n=1

∞

an converges. Then {Sn } converges and Sn=a1+a2+…+an, Sn−1=a1+a2+…+an−1 which implies Sn−Sn−1=an. limn→∞

(S¿¿n−Sn−1)= limn→∞

Sn− limn→∞

Sn−1=S−S=0¿, limn→∞

(S¿¿n−Sn−1)=¿ limn→∞

an=0=limn→∞

an¿¿.Q4d) Use comparison test to show that ∑

1

∞ 1n!

converges.Answer: Since n!>2n , ∀n≥N . Then | 1n!|= 1

n!< 12n

and ∑n=1

∞

( 12 )nis a convergent series for geometric series (0< 12 <1).

Hence, ∑n=1

∞ 1n!

converges.

Q5a) Define convergence of a sequence.Answer: A sequence {xn } is said to converge to a real number l (or to have limit l) if given ε>0 ,∃ N∈ N (with N depending on ε) such that ¿ xn−l∨¿ ε ,∀ n≥ N .

Q5b) Use ratio test to show that ∑1

∞ n2n

converges.

Answer: |an+1

an |=1

(n+1 ) !1n!

= n !(n+1 )!

= 1n+1

, 105

|an+1

an |=1n and limn→∞|an+1

an |= limn→∞

1n+1

= 1∞+1

=0. Thus, ρ=0<1 and ∑ 1

n! converges.Q5c) Use sandwich Theorem to show that if xn→a, zn→a and xn≤ yn≤zn, then yn→a.Answer: xn≤ yn≤zn implies xn−xn≤ yn− xn≤ zn−xn, i.e. 0≤ yn−xn≤ zn−xn

limn→∞

( zn−xn )=a−a=0. That is, zn−xn→0. By the Sandwich Theorem, ( yn−xn )→0. That is, lim

n→∞( yn−xn )=0 which implies lim

n→∞yn−lim

n→∞xn=0. That is, lim

n→∞yn−a=0. That is,

limn→∞

yn=a. That is, yn→a.Q5d) Show that {an }, where 0<a<1, converges.Answer: Since 0<a<1, 0n<an<1n which implies 0<an<1. Thus, the sequence is bounded. And, an+1−an=an (a−1 )<0 which further implies an+1<an. The sequence is decreasing. Hence, it is a bounded sequence. Thus, it converges.

Q6a) State and prove the Sandwich Theorem.Answer: Sandwich Theorem: If yn→0 and 0≤ xn≤ yn , ∀n∈N ; then xn→0.

106

Proof. yn→0 means given ε>0 ,∃ N such that ¿ yn−0∨¿∨ yn∨¿ ε ,∀n≥N . 0≤ xn≤ yn implies ¿ xn∨≤∨ yn∨¿ε which implies ¿ xn∨¿ ε , ∀n≥N . That is, xn→0.Q6b) Show that ∑

n=1

∞ 1n(n+1)

=1.Answer: Let xn=

1n(n+1). Then xn=

1n− 1n+1 by partial fraction.

sn=(1−12 )+( 12−13 )+( 13−14 )+…+( 1−n−1

−1n )+( 1n− 1

n+1 )=1− 1n+1

limn→∞

sn=limn→∞ (1− 1

n+1 )=1− 1∞+1

=1−0=1. Hence, ∑n=1

∞ 1n(n+1)

=1

Q6c) Why is f ( x )=±√x a function and why not f (±√ x )=x.Answer: f ( x )=+√x ,−√x, i.e. x∈ A is mapped to +√ x ,−√ x and f (+√x ,−√ x )=x means +√ x ,−√ x is mapped to x. f ( x )=±√x is single valued and f (±√ x )=x is multiple valued.Q6d) Define graph of a function.Answer: Let f : A→B be define by f (x), let G is a graph of f (x). Then G= {(x , f (x ) : x∈ A )}.

Q7a) State and prove the Bolzano - Weierstrass Theorem.Answer: 107

Bolzano - Weierstrass Theorem: Every infinite bounded sequence of real numbers has at least one limit point.Proof. Let {xn } be an infinite bounded sequence of real numbers. Then it has a monotone subsequence. The subsequence will also be bounded since {xn } is bounded. As such it will converge to a limit point.Q7b) Show that Q is not complete.Answer: Let A={x : x∈Q ,0<x∧x2<2 }, a subset of Q bounded from above by 2. Let K, a rational number, is a supremum of A. Then from Law of Tricotomy, one of the following must hold:(i)k 2=2, (ii)k 2<2,(iii) k 2>2But (i) is not possible since there is no rational number whose square is 2. Case II: Let k 2<2 and let y= 4+3k3+2k since y∈ A. Thenk− y=k−4+3k

3+2k=2(k2−2)3+2k

<0, i.e. k< y…(1). And 2− y2=2−(4+3 k3+2k )

2

=2−16+24 k+9k2

9+12k+4k 2= 2−k2

3+2k>0,

i.e. 2− y2>0 which implies 2> y2…(2)(1) and(2) says k is not a supremum of A.Case III: Let k 2>2. Then k− y= 2(k

2−2)3+2k

>0 which implies k− y>0 which further implies k> y…(3). And, 2− y2=2−(4+3 k3+2k )

2

=2−16+24 k+9k2

9+12k+4k 2= 2−k2

3+2k<0 implies 2− y2<0

and 2< y2…(4)(3)and(4) says k is not the lub. Hence A has no supremum in Q.108

Real Analysis III believe you realize fom Real Analysis I that you just need the x-axis of a graph. That is all. But where will the problem comes in is in the continuum property of real numbers and the fact that the set of natural numbers R (as we denote it) is not countable. The set of natural numbers N , the set of integers Z and the set of rationals Q are countable unlike R. For example, just between the interval (0 ,0.00001) is not countable much less R. Any set in one-one correspondence with the set of N is countable. N is vacuously countable, Z is N+¿N−¿¿ ¿ that is union of countable sets, thus it is countable. Q is a fraction of countable integers, thus it is countable, but not R. An interval is a subset of R having two numbers and between them are real numbers.The subset of the catesian product is called a relation. A function f is a relation between two sets (here two subsets of R), say A and B which assigns y∈B for every one and only one x∈ A . A mapping is ageneralization of functions where A and B may not be set of numbers. A is called the domain of f and B is called the codomain of f .Examples:

1) f :R→R defined by f ( x )=x2 is a function2) f :R→R defined by f ( x )=√x is not a real function when x<0.3) f ( x )=1

x of f :R→R is not a function when x=04) f ( x )=3 x2+1 is a function in R

Remark. One-one function is called injective, onto function is called surjective and one-one and onto function is called bijective.Limit of FunctionsYou treat limit of sequence , we will now treat limit of functions whose domain will be not N only, it will be R. So limit of function is the generalization of limit of sequence.Recall,Definition. A sequence xn is said to converge to a real number l or to have a limit l if for each ε>0 ,∃ N∈ N (with N defending on ε) such that |xn−l|<ε ,∀n≥ N.We will now generalize where N will be replaced by an interval(open or closed)in R.109

Defintion(Limit of a Function). Let f be a real valued function (real function) defined on an interval ¿a ,b¿ except possibly at some points x0∈ ¿a ,b¿. We say that the limit of f ( x ) as x tends to x0 exist if ∃ a number l which satisfies: Given ε>0 ,∃ δ>0(depending on ε and x0) such that 0<|x−x0|<δ implies |f (x )−l|<ε.Note. 0<|x−x0|<δ implies |f (x )−l|<ε is the same as saying |f (x )−l|<ε whenever 0<|x−x0|<δ. Note also that f (x) may or may not be defined at x=x0 and even if it is defined, the limit l may not be f (x0).Examples

1) Let f :¿1,3¿→R be defined by f ( x )=c. Show that limx→2

f ( x )=c.Solution: This is a constant function, we assume δ=ε. Then |f (x )−l|=|f ( x )−c|=|c−c|=0<ε since 0<|x−x0|=|x−2|<δ=ε .Here, we assume 0<|x−2|<δ to see if ¿ f ( x )−c∨¿ε which is true.2) Let f : [0 ,4 ]→R be defined by f ( x )=x. Show that lim

x→3f ( x )=3.

Solution: Similarly, we assume δ=ε. And, we assume 0<|x−3|<δ. Then|f (x )−3|=|x−3|<δ

That is, |f (x )−3|<ε since δ=ε

That is, 0<|x−3|<δ implies |f (x )−3|<ε

3) Let f :¿−1 ,1¿ be defined by x sin 1x . Show that limx→0

f ( x )=0.Solution: Assume δ=ε. Always assume what is possible. Also, assume 0<|x−x0|<δ, ie 0<|x−0|=|x|<δ, ie 0<|x|<δNow, |f (x )−l|=|x sin x−0|=|xsin 1x|=|x||sin 1x |≤∨x∨¿ since ¿ sin 1x∨≤1.Thus, |f (x )−l|≤∨x∨¿. But |x| is less than δ=ε. Hence,

|f (x )−l|<ε

So l (the limit exists).One Sided Limit (Limit at +¿ and −¿)

110

Let f be defined on [a ,b] except possibly at x0∈[a ,b]. Definition. f is said to tend to l as x tends to x0 from the right if given ε>0 ,∃ δ>0∋|f ( x )−l|<ε whenever 0<x− x0<δ, ie x0<x<x0+δ.Definition. f is said to tend to l as x tends to x0 from the left if given ε>0 ,∃ δ>0∋|f ( x )−l|<ε whenever 0<x0−x<δ, ie x0−δ<x<x0

Note. We denote limit of f on the right by limx→x0

+¿f ( x)¿¿ or lim

x→x0+ 0f (x ) and on the left

by limx→x0

−¿ f (x)¿¿ or limx→x0

❑−0f (x).

Examples1) Let f :¿ a ,b¿ and x0∈ ¿a ,b¿ defined by f ( x )={1 , x>00 , x<0. Show that lim

x→0f (x ) does

not exist.Solution:lim

x→ 0−¿ f ( x )=0¿¿becouse f ( x )=0 if x<0 and

limx→ 0+¿ f (x )=1¿

¿ becouse f ( x )=1 if x>0And, lim

x→ 0−¿ f ( x )=0≠−1= limx→0+ ¿f ( x )

¿¿ ¿¿

We will see that the limit does not exist from the following theorem.Theorem. The limit of a function exist if and only if the limit on the right of it is equal to the limit on the left of it.2) Let f :¿0 ,1¿ be defined by f ( x )={ 2 , x>1

x+2 ,0<x<13 x2+2x+1 , x<0

. Find the limit of f (x) at 0 and 1.

Solution: limx→0+¿ f (x )=2¿

¿ and limx→0−¿ f ( x )=1¿

¿

By Theorem, limx→0❑

f ( x ) does not exist.Also, lim

x→ 1+¿ f ( x )=2¿¿ and lim

x→1−¿ f ( x )=3 ¿¿

111

Similarly, limx→1+¿ f ( x )≠ lim

x→1−¿f ( x )¿ ¿¿

¿ which implies lim f ( x ) does not exist.3) Let f :¿−1 ,1¿ be defined by f ( x )={¿ x−2∨ ¿

x−2, x ≠2¿

0 , x=2. Find lim

x→2f (x ).

Solution: Let x>2 ,∨x−2∨¿x−2

x<2 ,|x−2|=− (x−2 )

¿2−x

Now, lim

x→2+¿ f (x)= limx →2+¿¿ x−2∨ ¿

x−2= lim

x→ 2❑

x−2x−2

= limx→2+¿1=1

¿ ¿¿¿¿

¿¿¿

And,lim

x→ 2−¿ f (x)= limx →2−¿¿ x−2∨ ¿

x−2= lim

x→ 2−¿ x−2x−2

= limx→ 2−¿1=1

¿ ¿ ¿

¿¿¿¿¿¿

¿

Thus,lim

x→2+¿ f (x)=1≠−1= limx →2−¿f (x) ¿

¿ ¿¿

Hence, limx→2❑

f (x ) does not exist.Theorem. Let f :¿ a ,b [→R, x0∈ ]a ,b¿ and lim

x→x0f (x) exists and is equal to l. Then l is unique.

Proof. We prove this by contradiction. Suppose on the contrary the limit l exist and is not unique. Then we can denote l by l1 and l2 where|f (x )−l1|< ε

2 whenever 0<|x−x0|<δ2 and

|f (x )−l2|< ε2 whenever 0<|x−x0|<

δ2

Let us see is l1−l2 very very small? If this is the case, then l1−l2=0, ie l1 and l2 are equal. Thus, l is unique. Now,|l1−l2|=¿ l1−f ( x )+f ( x )−l2∨¿

112

≤∨l1−f (x )∨+¿ f ( x )−l2∨¿ from Triangular Inequality¿∨f ( x )−l1∨+¿ f ( x )−l2∨¿

But |f (x )−l1|< ε2 and |f (x )−l2|< ε

2 .Thus, ¿ l1−l2∨≤∨f ( x )−l1∨+¿ f ( x )−l2∨¿ ε

2+ ε2=ε

|l1−l2|<ε

Therefore, l1=l2 , certifying that l is unique.Theorem. Let f an g be defined on an interval ¿a ,b¿ except possibly at some points x0∈ ¿a ,b¿ and limx→x0

f (x )=l and limx→x0g ( x )=m. Then:

i) limx→x0

[ f ( x )+g ( x ) ]=l+m

ii) limx→x0

[ f ( x ) . g ( x ) ]=lm

iii) limx→x0

[ f ( x )÷g ( x ) ]=l ÷m,m≠0

Corollary. Let f and g be defined on an interval ¿a ,b¿ except possibly at some points x0∈ ¿a ,b¿ and limx→x0

f (x)=l and limx→x 0

g(x )=m. Then:i. lim

x→x0[cf (x )]=cl, where c is a constant

ii. limx→x0

¿

iii. limx→x0

[ f 1(x)+ f 2(x)+… ]=l+m+…

iv. limx→x0

[ f 1(x)× f 2(x)×… ]=l×m×…

Examples1. Find the limit x7 as x→2.Solution:

limx→2

x7=lim ¿¿

¿ limx→2

x . limx→2

x . limx→2

x .… limx→ 2

x .

¿2×2×2×…×27׿

113

¿27

2. Find limx→1

(5 x2−x+2).Solution: lim

x→15x2− lim

x→1x+ lim

x→12=5 lim

x→1x2−¿ lim

x→1x+lim

x→12¿

¿5 limx→1

x limx→1

x−limx→1

x+2

¿5×1×1−1+2

¿6

3)Find limx→1

x2+12 x+1

.Solution: lim

x→1

x2+12 x+1

=limx→1

(x2+1)

limx→1

(2 x+1)=23

Theorem. Let limx→x0g ( x )= y0 and limy→ y0

f ( y )=l. Then limx→x0

f (g ( x ) )=l.Proof. lim

x→x0f (g ( x ) )=lim

x→x0f ( lim

x→x0g (x))

¿ limx→x0

f ( y0)

¿ l

Examples1) Find the lim

x→2(x+1)2.

Solution: limx→2

(x+1)2=( limx→2

(x+1))2

¿32

¿9

2) Find limx→−1

√4 x2+1

Solution:limx→−1

√5 x2+1=√5(−1)2+1

114

¿√5+1=√6

3) Show that limx→0

cos 1x does not exist.

Show. Let x= 12πn where n=1 ,2 ,3 ,… Then

limx→0

cos 1x=lim

n→∞cos2 πn becouse if n=∞ in 12πn , x=0 and vice-versa

limx→0

cos 1x=1…(1)

Again, le x= 1

2πn+ π2

. Then limx→0

cos 1x=lim

n→∞cos (2πn+ π

2)

¿ limn→∞

cos (2πn) limn→∞ ( π2 )

¿1× limn→∞

cos( π2 )¿0

Since(1) is not equal to (2), the limit as x→0 of cos 1x does not exist.Inf inite Limits and Limits at InfinityDefinition( limx→x0

f (x )=∞). Let f de defined on the interval ¿a ,b¿ except possibly at x0∈¿a ,b¿. Then f is said to tend to positive infinity +∞ as x tends to x0 if ∀M>0 (no matter how large), ∃ δ>0∋ f ( x )>M whenever 0<¿x−x0∨¿δ. limx→x 0

f (x )=l

Definition( limx→x0f (x )=−∞). Let f de defined on the interval ¿a ,b¿ except possibly at x0∈ ¿a ,b¿. Then f is said to tend to positive infinity −∞ as x tends to x0 if

∀M>0 (no matter how large), ∃ δ>0∋ f ( x )←M whenever 0<¿x−x0∨¿δ.Definition(lim

x→∞f (x )=l). Let f :R→R be a function. Then f is said to tend to l as x approaches +∞ if given ε>0 ,∃M>0 such that |f (x )−l|<ε ,∀ x>M .

115

Definition( limx→−∞

f (x )=l). Let f :R→R be a function. Then f is said to tend to l as x approaches −∞ if given ε>0 ,∃M>0 such that |f (x )−l|<ε ,∀ x←M .

Examples1) Let f :R→R be a function. Show that the limit fails at infinity and find limit at

∞ of f ( x )=1x .

Solution: limx→ 0−¿ 1

x=−∞¿

¿ and limx→ 0+¿ 1

x=∞¿

¿

The limit fails at 0.But at ∞, limx→∞−¿ 1

x=−0¿

¿ and limx→∞+¿ 1

x=+0¿

¿

Hence , limx→∞❑

1x=0

2) Let f :R→R be a real function defined by f ( x )=√ x23 x

. Find the limit at infinity of f (x).

Solution: We consider two cases here.Case I (When x<0). Let y=−x. Thenlimx→−∞

f (x )=√ x23x

. This implies limx→∞

g ( y )= limx→−∞

√ y2

3 y=13

Case II(When x>0):limx→∞

f (x )=limx→∞

x3 x

=limx→∞

13=13

Thus, limx→−∞

f (x )=13= lim

x→∞f (x ). The limit at ∞ of f ( x )=√ x2

3 x is 13 .

3) Let f :R→R be a real function defined by f ( x )=√ x2+2 x+32 x+1

. Find the limit of f (x) at ∞.

Proof: Two cases are also involve here. If x>0, then:116

√x2+2 x+32 x+1

=√ x2

x2+ 2 xx2

+ 3x2

2x√ x2

+ 1√x2

=√1+ 2x + 3x2

2+ 1x

And, limx→∞

√1+ 2x + 3x2

2+ 1x

=√1+0+02+0

=√12

=12

Case II (When x<0): Let y=−x. Then x=− y and g ( y )=√ y2−2 y+3−2 y+1

¿ √1−2y + 3y2

−2+ 1y

limx→−∞

f (x )=limy→∞

g ( y )=limy→∞

√1− 2y + 3y2

−2+ 1y

=−12

Continuity of FunctionsDefinition: Let I be an open interval and let f : I →R be a function. Then f is said to be continuous at x0∈ I if given ε>0 ,∃ δ>0 such that |f (x )−f (x0 )|<ε whenever ¿ x−x0∨¿ δ.Definition: Let f : I →R. Then f is said to be continuous on I if it is continuous at every point of I.Examples1) Let f :R→R be a real valued function. Then prove that f ( x )=c is continuous in R.Solution: We assume ¿ x−x0∨¿ δ and see what will be ¿ f ( x )−f (x0)∨¿.

|f (x )−f (x0 )|=|c−c|=|0|=0

Now, ε>0 implies 0<ε and we have117

|f (x )−f (x0 )|=0<ε which is what is required. f is continuous every where since x0 was taken arbitrary.2) Let f :R→R be a real valued function. Then prove that f ( x )=x is continuous in R.Solution: Let δ=ε and ¿ x−x0∨¿ δ. Then|f (x )−f (x0 )|=|x−x0|<δ which implies ¿ x−x0∨¿ ε. This is what is required.Hence, f is continuous every where since x0 was taken arbitrarily.3) Let f be a real valued function. Then prove that f ( x )=¿x∨¿ is continuous in R, where ¿ x∨¿ is absolute value of x.Solution: Assume ε=δ and let ¿ x−x0∨¿ δ. Then|f (x )−f (x0 )|=||x|−|x0||≤∨x−x0∨¿ by Triangular Inequality.Hence, |f (x )−f (x0 )|≤|x−x0|<δ which implies |f (x )−f (x0 )|<δ implies¿ f ( x )− f (x0)∨¿ε , as required.Therefore, f is continuous every where in R since x0 was arbitrarily taken.4) Let f be a real valued function. Then prove that f ( x )=x2 is continuous in R.Proof: It is early to assume here, but what we want is |f (x )−f (x0 )|<ε if |x−x0|<δ.Now, |f (x )−f (x0 )|=|x2−x0

2|=|(x−x0 ) (x+x0 )|≤∨x−x0∨¿ x+x0∨¿ from Triangular Inequality.¿∨x−x0∨¿ x+x0+x0−x0∨¿

¿∨x−x0∨¿ x−x0+2x0∨¿

¿∨x−x0∨¿

Assume |x−x0|<1 (we can assume is less than 1 since we can assume is less than even δ<1). Then |f (x )−f (x0 )|≤δ(1+2|x0|). Let δ=min (1, ε

1+2|x0|) and

|f (x )−f (x0 )|≤|x−x0|(1+2|x0|)≤δ (1+2|x0|)= ε1+2|x0|

(1+2|x0|)=ε.118

Hence, f is continuous since |f (x )−f (x0 )|<ε if ¿ x−x0<δ∨¿

5) Le f be real valued function. Then prove thatf ( x )={ 1 , x>0¿0 , x≤0 is not continuous at 0.

Proof. We assume |x−x0|=|x−0|=|x|<δ. Let ε=12 for any ∈¿ 0 , δ ¿ , δ>0.Then|f ( x )−f (0 )|=|1−0|=|1|=1 which is not less than 12. So, there is no such δ that |f (x )−f (x0 )|<ε=12 whenever |x−0|<δ .Theorem. Let f :¿ a ,b¿ and x0∈¿a ,b¿ (ie f is defined at x0). Then f is continuous at x0 if and only if

limx→x0

f (x )=f (x0)

Proof: We first proof the necessity ie the if as follows: Suppose f is continuous at x0. Then given ε>0 ,∃ δ>0 such that |f (x )−f (x0 )|<ε whenever |x−x0|<δ. In particular, |f (x )−f (x0 )|<ε when ever 0<|x−x0|<δ implies limx→x0

f (x )=f (x0).We can now proof the sufficiency (ie the only if) as follows:If limx→x0

f (x )=f (x0)

If limx→x0f (x )=f (x0), then given ε>0 ,∃ δ>0 such that 0<|x−x0|<δ implies |f (x )−f (x0 )|<ε.

But at x=x0 , limx→x0

f ( x❑ )=f (x0) and at x≠ x0, 0<|x−x0|<δ implies |f (x )−f (x0 )|<ε and |x−x0|<δ implies |f (x )−f (x0 )|<ε

That is, f is continuous at an arbitrary x0. f is continuous in R.Example

1) Let f :¿0,1 ¿ and x0∈¿0,1¿ (f is defined at x0). Then show that f is not continuous at 0 if f ( x )={sin xx , x ≠0

¿ 0 , x=0.

119

Show. limx→0

sin xx

=1, but f (0 )=0. Thus,limx→x0

f (x )≠ f (x0 )

Hence, f is not continuous at 0.2) Let f :¿0,1¿ and x0∈¿0,1¿ (f is defined at x0). Then show that f is continuous

at 1 if f ( x )={sin xx , x ≠0

¿1, x=0.

Show. limx→1

sin xx

=limx→11=1=f (0)

Hence, f is continuous at 1.Theorem. Let f and g be defined on I and that f and g are continuous at x0∈ I. Then:i. f +g is continuous at x0ii. fg is continuous at x0iii. f

g is continuous at x0, where g(x )≠0Examples1) If f ( x )=x2+3 and f ( x )=x2+1 are continuous, then f ( x )= x2+3

x2+1 is continuous.

That is,Theorem. Composition of continuous functions is continuous.Example1) Since f ( x )=x2+1 and g ( y )=sin y are continuous functions, then

fog ( x )=sin (x2+1 ) is also continuous.Definition. Let f :¿0,1¿ and x0∈¿0,1¿ (f is defined at x0). Then:i. f is said to be continuous on the right at x0 if lim

x→x0+¿f ( x )=f ( x0)¿

¿

ii. f is said to be continuous on the left at x0 if limx→x0

−¿ f ( x )=f (x0)¿¿.

120

Theorem. Let f :¿0,1 ¿ and x0∈¿0,1¿ (f is defined at x0). Then: f is said to be continuous at x0 if and only if f is continuous on the right at x0 and continuous on the left at x0.Examples1) Show that f :¿−1 ,5¿ defined by f ( x )={x−4 ,−1≤ x≤2

¿x2−6 ,2<x<5is continuous at 2.

Show. f (2 )=2−4=−2 , limx→2+¿ f (x )=−2 , lim

x →2−¿ f ( x )=22−6=−2¿¿ ¿

¿, ie f is continuous on the right at 2. Also, lim

x→ 2−¿ f ( x )=f (2)=−2 ¿¿

Thus, f is continuous on the left at 2. Hence, f is continuous at 2.2) Let f :¿0,1 ¿ and x0∈¿0,1¿ (f is defined at x0). Show that f ( x )={tan xx , x≠1

¿1 , x=1 is

continuous at 0.Show. lim

x→1

tan xx

=limx→1

sin xcos xx

=limx→1

1cos x

limx→ 1

sin xx

=1×1=1 and f (x0 )=f (1 )=1

Hence, limx→1

f ( x )=f (1) and f is continuous at 1.Definition (Discontinuity). Let f :¿0,1¿ and x0∈ ¿0,1¿ (f is defined at x0). Then f is discontinuous at x0 if it is not continuous at x0.Definition (Discontinuity of the 1st kind). Let f :¿0,1¿ and x0∈ ¿0,1¿ (f is defined at x0). Then f is said to have discontinuity of the first kind if lim

x→x0+¿f ( x)¿

¿ and lim

x→x0−¿ f (x)¿

¿ exist.Otherwise it is discontinuous of the 2nd kind.Definition. f is said to have a removable discontinuity at x0 if limx→x0

f (x) exist, but it is not equal to f (x0).Here, by assigning another f (x0), the discontinuity is removed.

121

Definition. f is said to have a jump or removable discontinuity if limx→x0

+¿f ( x)¿¿ ≠

limx→x0

−¿ f (x)¿¿

Theorem. If f is continuous, then limx→∞

f (xn=f ( limx→∞

xn)).Proof. Since limn→∞

xn=x0 and δ>0, we have |❑❑−❑❑|<δ implies |f (xn)−f (x0 )|<ε ,∀ n≥ N .Examples1) Show that lim

n→∞e1n=1.

Proof. We know that ex is continuous. Hence,limn→∞

exn=elimn→∞

xn from Theorem, and¿e

limn→∞

1n=e0=1

2) Show that limn→∞

sin( 3 π2 + 3n )=−1.

Proof. Since sin x is continuous on R, we have from Theorem, limn→∞

sin( 3 π2 + 3n )=sin( limn→∞

3π2

+ limn→∞

3n )

¿ sin( 3 π2 +0)¿sin 3π

2

¿−1

Theorem (Boundedness Theorem). Let f is continuous on the closed interval [a ,b]. Then f is bounded.Theorem (Intermediate Value Theorem). Let f be continuous on the closed interval [a ,b] and c is a number between f (a) and f (b). Then ∃ x0∈[a ,b ] ∋ f (x0 )=c.Theorem. The image of an interval is also an interval

122

Theorem (Extreme Value Theorem). Let f is continuous on the closed interval I=[a ,b ]. Then ∃ x0 , x1∈[a ,b ] such that f (x0 )≤ f ( x )≤ f (x1 ) ,∀ x∈ I .That is f attains its maximum and minimum values on I.Uniform ContinuityDefinition. Let f : I →R be a function.Then f is said to be uniformly continuous on I if ∀ ε>0 ,∃ δ>0∋|f (x1 )−f (x2)|<ε ,∀ x1 , x2∈ I whenever ¿ x1−x2∨¿δ .Examples1) Show that f : I →R be defined on [0,1 ] by f ( x )=x2 is uniformly continuous.Show. |f (x1 )−f (x2 )|=|x12−x2

2|=|x1+x2||x1−x2|≤2|x1−x2|,∀ x1 , x2∈ I. Assume δ= ε2 . Then

|f (x1 )−f (x2 )|<2. ε2 which implies |f (x1 )−f (x2 )|<ε whenever |x1−x2|<δ

2) Let f :¿→R be define by f ( x )=1x . Show that f is continuous, but not uniformly continuous on I.

Show. Let x1=1n and x2= 1n+1

, n≥1 , n∈Z+¿¿ and set ε=12. Then

|f (x1 )−f (x2 )|=|11n

−

11

n+1|=|n−(n+1 )|=|1|=1

Thus, we see that for any δ>0, |f (x1 )−f (x2 )|=1 is not less than 12. Hence, f is not uniformly continuous on I=¿.Theorem (Uniformly Continuity Theorem). If f is continuous on the closed interval [a ,b], then f is uniformly continuous.DifferentiationThe word differentiation comes from its root word difference. This is the slope or difference between two points.Definition. Let f be defined on an open interval and let x0∈ I. Then f is differentiable at x0 iff

123

limx→x0

f ( x )−f (x0)x−x0

exist and it is denoted by f '( x0), called derivative of f at x0. Let h=x−x0. Then x→ x0 implies x−x0→0 and x=h+x0. Equivalently,

f ' (x0 )=limh→0

f (x0+h )−f (x0)h

Theorem. If f is differentiable at x0∈ I, then f is differentiable on I.Examples1) Let f :R→R be defined by f ( x )=c. Show that f ' (x0 )=0.Show. lim

x→x0

f ( x )−f (x0)x−x0

=limx→x0

c−cx0−x

= limx→x0

0=0

That is, f ' (x0 )=0 ,∀ x0∈ R anf f ( x )=c is deffentiable on R.2) Let f :R→R be defined by f ( x )=x. Show that f ' (x0 )=1.Solution.

limx→x0

f ( x )−f (x0)x−x0

= limx→x0

x−x0x−x0

= limx→x0

1=1

That is, f ' (x0 )=1, ∀ x0∈R and f ( x )=x is differentiable on R.3) Let f :R→R be defined by f ( x )=3 x2−5 x+10. Show that f ' (x0 )=6 x0−5.Solution.f ' (x0 )=lim

h→0

f (x0+h )−f (x0)h

¿ limh→0

[3 (x0+h )2−5 (x0+h )+10 ]−[3 x02−5 x0+10 ]h

¿ limh→ 0

6 x0h+3h2−5h

h¿ lim

h→0[6 x0+3h−5]

¿6 x0−54) Let f :R→R be defined by f ( x )=sin x. Show that f ' (x0 )=cos x0.Solution.

124

f ' (x0 )=limh→0

f (x0+h )−f (x0)h

¿ limh→ 0

sin x0cosh+cos x0 sin h−sin x0h

¿ limh→ 0

sin x0 (cosh−1 )+cos x0sin hh

¿ limh→0

sin x0(cosh−1)h

+cos x0sin h

h

¿0× limh→0

sin x0+cos x0 .1=cos x0

Theorem. Let f : I (open interval )→R. If f is differentiable at x0∈ I, then f is continuous.Proof. Suppose f is differentiable at x0∈ I, we need to show that f is continuous at x0, ielimx→x0

f (x )=f (x0) or limx→x0f (x )−f (x0 )=0. We hence check

limx→x0

f (x )−f (x0 )= limx→x0

f (x )−f (x0 )x−x0

(x−x0)

¿ limx→x0

f ' ( x )(x−x0)

¿ limx→x 0

f ' ( x ) limx→x0

❑(x−x0)

¿ f ' (x0 )×0=0

This is what is required. Hence, f is continuous at x0.Note: The reverse is not the case. Continuity does no always implies differentiability.Proof. We prove this by providing a counter example that is continuous, but not differentiable. Let us consider f ( x )=¿x∨¿. Clearly, f is continuous at 0 whencelimx→x0

f (x )= limx→x0

|x|=|x0|=f (x0) , but f is not differentiable at 0.

125

limx→0+¿ f ( x )−f (0 )

x−0 = limx→0+ ¿ x−0

x−0= lim

x→0+¿ xx= limx→ 0+ ¿1=1¿

¿ ¿

¿¿

¿¿

¿ since f ( x )=|x|={ x , x>0−x , x<00 , x=0

limx→ 0−¿ f ( x )−f (0)

x−0 = limx→0−¿−x−0

x−0= lim

x→ 0+¿−xx

= limx→0+ ¿−1=−1¿

¿¿

¿¿

¿¿

¿

Thus, limx→ 0+¿ f ( x )−f (0 )

x−0 ≠ limx →0−¿ f ( x )− f (0 )

x−0¿

¿¿

¿.So, f '( x0) does not exis and f is not differentiable at 0.Theorem (Rules of Differentiation). Let f and g be functions on an open interval I, and f and g are differentiable at x0∈ I. Then f +g , f−g , fg and fg where g≠0 and their derivatives are as follows:i. ( f ± g )' ( x0 )= f '(x0)± g

' (x0)

iii. ( fg )' (x0 )=g (x0 ) f ' (x0 )+f (x0 ) g' (x0)

iv. ( fg )'

=f ' (x0 )g (x0 )−g' (x0 ) f (x0)

[ g( x0)]2

Theorem (Local Maximum). Let f be defined on [a ,b] and let f has a local maximum at a point x0∈[a ,b] and f '( x0) exists. Then f ' (x0 )=0.Note: A similar result holds for local minimum.Proof. If x0 is a local maximum, then f (x0+h≤ f (x0)) for h sufficiently small and f (x0+h )− f (x0)≤0. If h>0 , f (x0+h )−f (x0)

h≤0 and lim

h→ 0+¿ f (x0+h )−f (x0)h

≤ 0¿

¿ ie f ' (x0 )≤0…(1)

If h<0 , f (x0+h )−f (x0)h

≥0 and limh→ 0+¿ f (x0+h )−f (x0)

h≥ 0¿

¿ ie f ' (x0 )≥0…(2)

From (1) and (2), f ' (x0 )=0.Theorem (Rolle Theorem). Let f be continuous on [a ,b] and let f be differentiable on ¿a ,b ¿. If f (a )=f (b )=0, then ∃ c∈¿ a ,b¿ ∋ f ' (c )=0.Proof. Since f is continuous on [a ,b ] ,by the extreme value theorem, f attains its maximum and minimum values on [a ,b]. If both the maximum and the minimum 126

values are attained at the end points a and b , f (a )=f (b )=0 implies f ( x )=0 ,∀ x∈ [a ,b ] .Hence, f ' (c )=0∀ c∈¿a ,b¿.However, if f attains its maximum and minimum values at some points c∈ [a ,b ] , thenf has a local maximum or minimum at c and by the above theorem, f ' (c )=0.Theorem (Mean Value Theorem). Let f be continuous on [a ,b ] and differentiable on ¿a ,b¿. Then ∃ϑ∈¿a ,b¿

Proof. Construct a function F on [a ,b ]as that looks as follows:F ( x )=f ( x )−f (a )−

f (b )−f (a)b−a (x−a)

Then F (a )=F (b )=0

Also, F is continuous on [a ,b] and differentiable on (a ,b). Thus, from Rolle Theorem, ∃ c=ϑ ∋F ' ( c )=F (ϑ )=0 and F '(ϑ )=f ' (ϑ )−

f (b )− f (a )b−a

=0. This implies F '(ϑ )=f (b )−f (a )

b−a, ie f (b )−f (a )=f ' (ϑ )(b−a)

Definition. Let f be a function defined on an interval I. Then f is said to be:i. Increasing on I if ∀ x , y∈ I with x< y , f ( x )≤ f ( y)ii. decreasing on I if ∀ x , y∈ I with x< y , f ( x )≥ f ( y)iii. constant on I if ∀ x , y∈ I with x< y , f ( x )=f ( y)iv. Strictly Increasing on I if ∀ x , y∈ I with x< y , f ( x )< f ( y )v. Strictly decreasing on I if ∀ x , y∈ I with x< y , f ( x )> f ( y )

Corollary. Let f is continuous on [a ,b] and differentiable on (a ,b). Then :i. If f ' ( x )≥0 , ∀ x∈¿ a ,b¿; then f is increasing on ¿a ,b ¿ii. If f ' ( x )≤0 , ∀ x∈¿ a ,b¿; then f is decreasing on ¿a ,b ¿iii. If f ' ( x )=0 , ∀ x∈ ¿a ,b¿; then f is constant on ¿a ,b ¿iv. If f ' ( x )<0 ,∀ x∈¿a ,b¿; then f is strictly increasing on ¿a ,b¿v. If f ' ( x )<0 ,∀ x∈¿a ,b¿; then f is strictly decreasing on ¿a ,b ¿

Proof. Theorem(Generalized Mean Value Theorem). Let f and g are continuous on [a ,b] and both f and g are differentiable on (a ,b) and g' (x )≠0 ,∀ x∈(a ,b) and g (b )−g (a)≠0. Then:127

∃ϑ∈ (a ,b )∋ f (b )−f (a)b−a

= f ' (ϑ )g '(ϑ )

Note: This theorem reduces to the 1st mean value theorem if we put f ( x )=x or g ( x )=x.Theorem (L’ Hospital Rule). Let f and F are continuous functions on [a ,b] and f ' and F ' exist on (a ,b ) with F ' ≠0 and x0∈(a ,b) and limx→x0

f (x )=limx→x0

F ( x )=0∨∞ andlimx→x0

f (x)

limx→x0

F ( x )=00 (¿∞∞ ) and lim

x→x0

f ' (x )F' (x )

=L .Then: limx→x0

f (x )F(x )

=L

Examples1. Find lim

x→0

sin xx .

Solution: limx→0

sin xx

=00 which is undefined. But f ( x )=sin x and f ' ( x )=cos x

F ( x )=x and F ' ( x )=1

Using the L’Hospital Rule, limx→0

sin xx

=limx→ 0

f '(x )F' (x )

=limx→ 0

cos xx

=limx→0

11=1

2. Find limx→0

x2

sin x using L’Hospital Rule.

Solution: limx→0

x2

sin x=lim

x→0

2xcos x

=limx→02 x . 1

limx→0cos x

=0.1=0

3. Find limx→∞

lin xx .

Solution:limx→∞

ln xx

=limx→∞

ln x

limx→∞

x=

limx→∞

1

x1

=01=0

4. Find limx→∞

xex .

128

Solution: limx→∞

xex =

∞∞

Applying L’Hospital Rule, we have:limx→∞

x

limx→∞

ex=limx→∞

1

limx→∞

ex=1

limx→∞

e x=0

5. Find limx→0

1+x−e x

2x2.

Solution: limx→0

1+x−e x

2x2. By applying L’Hopspital Rule, we have:

limx→0

1−ex

4 x= 1−14(0)

=00Reapplying, we have:

limx→0

−ex

4=−14Theorem (Generalized L’Hospital Rule). Let f n and Fn exist on (a ,b )∧x0∈[a ,b]

and limx−. x0f n−1(x )

limx−. x0

Fn−1 =00∨∞∞ and limx−. x0

f n(x )

Fn(x )=L. Then limx−. x0

f n(x )

Fn(x )=L, where t n is the nth

derivative of t.Examples1. Find lim

x→1 ( 1lin x− 1x−1 ).

Solution: limx.→1 ( 1

lin x−1

x−1 )=limx→1 ( x−1−lin x( lin x )(x−1) )= 00

Applying the L’Hospital Rule, we have:limx→1 ( 1−1

x

lin x+( x−1) 1x

)=limx→1 x−1x lin x+x−1

=00

Reapplying the L’Hospital Rule, we have:limx→1

❑ 1lin x+1+1

= 10+1+1

=12

129

2. Find limx→ 0+¿( 1x )

sin x

. ¿

¿

Solution: We know that e lin y= y. Then e ln( 1x )sin x

=( 1x )sin x

But e ln( 1x )sin x

=esin xln( 1x )

Now, lim

x→ 0+¿ sin x ln( 1x)= limx→+0

ln 1xcosec x= lim

x→+0

1x

−cosec x cot x ¿

¿

¿ limx→ 0+¿ 1

x 1sin x .

cos xsin x

= limx →0+¿ sin2 x

x cosx= lim

x→0+¿ x ( sin xx )2 1cos x

¿ ¿¿

¿ ¿¿

¿

¿0×12× 11=0

Hence, limx→0+¿( 1x )

sin x

=e0=1 ¿

¿.Higher Order DerivativesIf f is differentiable at x0 , we call the derivative at x0, f '( x0). Also, if f '( x0). This continuous to generality and we call f (n )(n) the nth derivative of f (if it exists).Examples

1. If f ( x )=x2, find f v (x).Solution: f ( x )=x2 which implies f ' ( x )=2 x which implies f ' ' ( x )=2 which implies f ' ' ' ( x )=0

2. Find f iv(x ) if f ( x )=sin x. Show that f iv ( x )−f ( x )=0

Solution:f ( x )=sin x

f ' ( x )=cos x

f ' ' ( x )=−sin x

f ' ' ' ( x )=−cos x

130

f iv ( x )=sin x

Hence, f iv ( x )−f ( x )=0

Examples1. Find y ' of y=x x2.

Solution: Take log of both sides of y=x x2as follows:log y= log x x2 which implies log y=x2 log x

Differentiating both sides, we have:1ydydx

=2x log x+x

dydx

= y (2 x log x+x)

y '=xx 2(2 x log x+x)

2. Find y ' of y=(1+x2)5.Solution: dydx=dy

du. dudx where u=1+ x2

¿2 x .5(1+x2)5

¿10 x (1+ x2)4

3. Find y ' of y=sin3 (3 x2+2x−5 ) .

Solution: Let u=3x2+2 x−5. Then dudx=6 x+2 and v=sin u , dvdu=cosu

Saddle Points (or Points of Inflexion)A saddle point is said to occur at any point p on a curve at which the tangent at p crosses the curve at p. If on a curve of y=f (x ) and f ' ' ( x )<0, then we have maximum and if on a curve of y=f (x ) and f ' ' ( x )<0, then we have maximum and if a curve of y=f (x ) and f ' ' ( x )>0, then we have minimum and if f ' ' ( x )=0, we have the saddle point (or point of inflexion).For example, the saddle point of f ( x )=x3 is 0 ie f ' ' ( x )=0

131

Theorem (Taylor Theorem). Let a function f be defined on [a ,a+h] is such that: f n−1 is continuous on [a ,a+h] and f n exists on [a ,a+h]. Then

f (a+h )=f (a )+ h1 !

f ' (a )+ h2

2 !f ' ' (a )+ h3

3 !f ' ' ' (a )+…+ hn−1

(n−1 ) !f n−1 (a )+Rn

Where Rn is called a Taylor Remainder after nth term which some mathematicians presented its value asi. For p=1, Rn=

hn (1−θ )n−1

(n−1 )!f n(a+θh), the Cauchy Remainder

ii. For p=n, Rn=hn

(n )!f n(a+θh), the Lagrange Remainder

IntegrationYou are certainly familiar with the ordinary integral of calculus, ∫

a

b

f (x )dx. This integral is designed to make sense from limits of sums of the form ∑

i=1

n

f (x i)∆ x i. Sums of this form have many important interpretations in physical sciences and engineering. One such interpretations is that of area between the function f and the x−axis (convince your self ). In this chapter, we shall use this interpretation to introduce a definition of the integral, ∫

a

b

f (x )dx, which is familiar from calculus and lends itself best to an interpretation as area.Definition. A partition p of the closed bounded interval [a ,b] is a finite se of points p= {x0 , x1 ,…, xn } of [a ,b] labelled so that a=x0<x2<x1<…<xn=b.Riemann SumLet f : [a ,b ]→R be a function and a=x0 , x1 ,…, xn−1 , xn=b be a partition of [a ,b ]and f i is such that x i−1≤ f i≤x i. Then the quantity ∑

i=1

n

f (f i) (x i−x i−1) is called a Riemann Sum for f corresponding to this partition.IllustrationThen

∑i=1

n

f (f i) (x i−x i−1 )=f (f 1) (x1−a )+ f ( f 2 ) (x2−x1 )+ f ( f 3 )(b−x2)

132

¿ Area(A)

Definition. Let f : [a ,b ]→R. Then f is said to be Riemann Integrable iff there exist a number A such that ∀ ε>0 ,∃ δ>0 with partition a=x0 , x1 ,…, xn−1 , xn=b of width less than δ and |∑i=1

n

f (f i ) (x i−x i−1 )−A|<ε whenever f i∈[ xi−1 , xi]

The number A is called the Riemann Integral of f between a and b, and is denoted by integral A=∫

a

b

f (x )dx.Theorem. The Riemann Integral of a function between a and b is unique if it exists.Examples

1. Let f : [a ,b ]→R be defined by f ( x )=c. Find ∫a

b

f (x )dx.Solution: The graph isLet a=x0 , x1 ,…, xn−1 , xn=b be any partition of [a ,b]. Then

∑i=1

n

f (f i) (x i−x i−1 )=∑i=1

n

c (x i+1−x i)

¿c∑i=1

n

(x i+1−xi)

¿c (b−a)

We know the area A of a rectangle is l ×b=c×(b−a)

A=c (b−a)

Clearly,|∑i=1

n

f (f i ) (x i−x i−1 )−c(b−a)|=|0|=0<ε ,∀ δ

Therefore,133

∫a

b

f (x )dx=∫a

b

cdx=c (b−a) 2. Let f : [a ,b ]→R be defined by f ( x )={ 1 , x=rational

¿0 , x=irrational . Find ∫a

b

f (x )dx.Solution: Any sub interval of [a ,b] contains both rationals and irrational points. Let f i be rational in the partition of [a ,b]. Then, the corresponding Riemann Sum is

SR=∑i=1

n

f (f i) (x i−x i−1 )=∑i=1

n

1 (x i−x i−1 )=b−a

On the other hand, if f i is irrational, then the Riemann Sum isSR=∑

i=1

n

f (f i) (x i−x i−1 )=∑i=1

n

0 (xi−xi−1 )=0

For any choice of partition of width less than δ , a number no matter how small, there are Riemann Sums equal to b−a and 0. Therefore, f is not Riemann Integrable because the Riemann Integral if it exists, it must be unique.Theorem. Let f and g are integrable on [a ,b ] and c∈[a ,b ] is a constant. Then:

i. f +g is integrable on [a ,b ] and∫a

b

[ f ( x )+g ( x ) ] dx=¿∫a

b

f ( x )dx+¿∫a

b

g ( x ) dx¿¿

ii.cf is integrable and ∫a

b

cf (x)dx=c∫a

b

f (x)dx

ii. fg is integrable.Corollary. Let f and g are integrable on [a ,b]. Then f−g is integrable on [a ,b] and ∫a

b

[ f ( x )−g (x ) ] dx=¿∫a

b

f (x )dx−¿∫a

b

g ( x ) dx¿¿

Theorem. Let f is integrable on [a ,b ] and c∈[a ,b ]. Then ∫a

b

[ f ( x ) ]dx=¿∫a

c

f ( x )dx+¿∫c

b

f ( x )dx¿¿

Theorem. Let f is integrable on [a ,b] and f ( x )≥0 ,∀ x∈[a ,b]. Then ∫a

b

f (x )dx ≥0

134

Corollary. Let f ( x )≤g ( x ) ,∀ x∈[a ,b] and both are integrable on [a ,b]. Then integral ∫a

b

f ( x )dx ≤∫a

b

g(x )dx.Proof. If f ( x )≤g ( x ) ,∀ x∈ [a ,b ] ,then g ( x )−f ( x )≥0 ∀ x∈ [a ,b ]and ∫

a

b

[ g ( x )−f (x ) ] dx≥0from Theorem above, ie∫a

b

g ( x )−∫a

b

f (x )dx ≥0from Corollary aboveAnd, ∫

a

b

g ( x )dx≥∫a

b

f ( x )dx

Corollary. Let f is integrable on [a ,b]. Then|∫ab

f (x)dx|≤∫ab

|f ( x )|dx

Proof. We know that ∫a

b

f ( x )dx ≤∫a

b

¿ f (x)∨dx and−∫

a

b

f ( x )dx≤∫a

b

¿ f (x)∨dx

Thus, ¿∫a

b

f ( x )dx|≤∫a

b

¿ f (x)∨dx

Corollary. Let f is integrable function on [a ,b] and m and M are numbers such that m≤ f (x )≤M ,∀ x∈[a ,b ]. Then m (b−a )≤∫

a

b

f (x )dx≤ M (b−a).Proof. m≤f (x )≤M which implies ∫

a

b

mdx≤∫a

b

f (x)dx≤∫a

b

M dx

m∫a

b

dx≤∫a

b

f (x )dx≤M∫a

b

dx

m [ x]ab≤∫

a

b

f (x)dx≤ M [x ]ba

135

m (b−a )≤∫a

b

f (x ) dx≤ M (b−a )

Examples1. Show that 0≤∫

0

1 x73√1+x8

dx≤ 18 .

Solution: Let f ( x )= x73√1+x8

, x∈[0 ,1]

Since 0≤ x≤1,0≤1+ x and 0≤1+ x8 which implies 0≤ 3√1+x8 and 1≤ 3√1+x8 which implies 1

3√1+x8≤1 which implies x7

3√1+x8≤ x7 which implies 0≤ x7

3√1+x8≤x7

∫0

1

0 dx≤∫0

1 x73√1+x8

≤∫0

1

❑x7dx

0≤∫0

1 x73√1+x8

≤[ x88 ]0

1

0≤∫0

1 x73√1+x8

≤ 18

[18−08 ]

0≤∫0

1 x73√1+x8

≤ 18

Theorem. Let f is continuous on [a ,b ]. Then f is integrable on [a ,b ] .

Theorem (Mean Value Theorem for Integrals). Let f be a continuous function on [a ,b ] .Thenif ∃ a number τ∈ [a ,b ]∋∫

a

b

f ( x )dx=f (τ ) (b−a ) .

Proof. Let m and M be the minimum and maximum values of f on [a ,b]. Then m≤ f (x )≤M implies

m (b−a )≤∫a

b

f (x ) dx≤ M (b−a )

m≤ 1b−a∫a

b

f ( x ) dx ≤M

By the extreme value theorem, ∃ x0 , x1∈ [a ,b ]∋m= f (x0) and M= f (x1).136

Hence, f (x0 )≤c≤ f (x1) where c= 1b−a∫a

b

f (x )dx.iBy the intermediate value theorem, there exists a number τ∈ [a ,b ]∋ f (τ )=c, ie

τ∈ [a ,b ]∋∫a

b

f ( x )dx=f (τ ) (b−a ) .

Darboux IntegralDefinition (Upper and Lower Darboux Sums). Let f : [a ,b ]→R be a fixed bounded real valued function defined on the interval [a ,b] (recall that this implies in particular, that “¿ f (x)”) and “inf f (x)” exist for every x∈[a , b]. Let p be any partition of [a ,b] and let

M i ={f ( x ): x i−1≤ x ≤ x i, i=1, 2 ,… ,n } ¿mi=inf {f ( x ) : x i−1≤ x≤ x i , i=1 ,2,…,n }

The upper and lower Darboux Sums are defined respectively byU (f , P )=∑

i=1

n

Mi (x i−x i−2 )=∑i=1

n

M i∆ x i

L ( f , P )=∑i=1

n

mi (x i−x i−2 )=∑i=1

n

mi∆ xi

From the definition,m (b−a )≤L (P , f )≤U (P , f )≤M (b−a )

Now, each partition gives rise to a pair of sums, the upper and lower sums. By considering all partitions of [a ,b ] ,weget a set u of upper sums and set l of lower sums. The inequalitym (b−a )≤L (P , f )≤U (P , f )≤M (b−a ) clearly shows that the serts u and l are bounded and so each has a supremum and an infimum.Now, the infimum of u (the set of upper sums) is called the upper Darboux Integral:

∫a

b

f ( x )dx=Inf U

Inf {U (P , f ) :Pis a partition of [a ,b]} and the supremum of l (the set of all lower sums) is called the Darboux Lower Integral:137

∫a

b

f ( x )dx=l ¿¿

Or¿ {L (P , f ):Pis a partitionof [a ,b ]}∧∫

a

b

f (x )dx≤∫a

b

f (x )dx

Theorem (Darboux Condition of Integrability). f ( x ) is said ¿be Riemann Integrable if∫a

b

f (x )dx=∫a

b

f ( x )dx=∫a

b

f ( x )dx

Examples1) Let f : [a ,b ]→R be defined by f ( x )=c. Show that f is integrable on [a ,b ] ,where c is a constant.Solution: Since f ( x )=c, where c is a costant, m=M=c. Let p be a partition a=x0 , x1 ,…, xn=b. Then∑i=1

n

c∆ x i=c(b−a)which implies U (P , f )=L(P , f )

So ,∫a

b

f (x )dx=∫a

b

f ( x )dx=∫a

b

f ( x )dx=c (b−a)

2) Show that f : [a , b ]→R defined by f ( x )={ 0 , x=rational¿1 , x=irrationalis not integrable on

[a ,b].Solution:

U ( f , P )=∑i=1

n

Mi (x i−x i−2 )=∑i=1

n

1∆ x i=(b−a)

L (f , P )=∑i=1

n

mi (x i−x i−2 )=∑i=1

n

0∆ x i=0

Hence,∫a

b

f (x)dx=b−a≠0=∫a

b

f ( x )dx

138

∫a

b

f (x )dx does no exist. So, f is not Riemann Integrable.Theorem (Fundamenal Theorem of Calculus). Let f be continuous on [a ,b] and defined on [a ,b] byF ( x )=∫

a

x

f (t)dt . Then F is differentiable on ¿a ,b ¿ and F ' ( x )= f ( x ) ,∀ x∈¿a ,b¿.Proof. Since f is continuous on [a ,b], F (x) is defined on [a ,b ] ,∀ x. We show

limh→0

F (x0+h )−F (x0)h

=f (x0)

Now, f (x0+h )=∫a

x0+h

f (t )dt=∫a

x0

f ( t )dt+ ∫x0

x0+h

f ( t )dt

f (x0+h )=∫a

x0+h

f (t )dt= f (x0)+ ∫x0

x0+h

f ( t )dt

f (x0+h )−f (x0)=∫x0

x0+h

f (t )dt

Since f is continuous on [a ,b], f is continuous on [ x0 , x0+h]. By Mean Value Theorem, ∃ τ∈ [ x0 , x0+h ] such that ∫x0

x0+h

f (t )dt=f ( τ ) (h+ x0−x0 )=f (τ )h

Thus, f (x0+h )− f ( x0 )=h f (τ ) which implies F (x0+h )−F (x0)h

=f (τ )

limh→ 0

F (x0+h )−F (x0)h

=limh→0

h f (τ )h

=limh→0

❑ f (τ)

As h→0 , τ→ x0 andlimh→ 0

f (τ )= limτ→x0

f ( τ )= f (x0) since f is continuous on [a ,b ]∧F ' ( x )=f ( x ) , ∀ x∈ [a ,b ]. Then ∫a

b

f ( x )dx=F (b )−F (a )

139

Note: Any function F such that F '=f is called an antiderivative or primitive or indefinite integral of f .Corollary. ∫

a

b

f ( x )dx=−∫b

a

f ( x )dx

Examples1. Find F ' ( x ) if F ( x )=∫

0

x

sin t dt .

Solution: f ' ( x )=sin x

F ( x )=∫a

x

f (t)dt which implies F ' ( x )= f (x) F ' ( x )=sin x

2. Find F '(x ) if F ( x )=∫0

x2

sin t dt.Solution: If y=x2, then F ( y )=∫

0

y

sin t dt

dydx

=2 x and F ' ( y )=sin y=dFdy

F ' ( x )=dFdx

=dFdy

. dydx

=(sin y )2 x=2 x sin y=2x sin x2

3. Find F '(x ) if f ( x )=∫x2

x3

lin t dt.Solution: F ' ( x )=dF

dx=dF

dy. dydx . Let y=x3 and dy=3x2 and f ( y )=∫

y23

y

lint dt

F ' ( x )=( 1y )3 x2=3x2

y=3x

2

x3=3x

But F ( x )=∫x2

c

lin t dt+∫c

x3

lin t dt

F ( x )=−∫c

x3

lin t dt+∫c

x3

lin t dt

140

F11 ( x )=∫

c

x3

lin t dt∧F21 ( x )=∫

c

x3

lin t dt

Then F ' ( x )=F21 ( x )−F1

1(x)

F11 ( x )=dy

dx=(lin y )2 x=2 x lin x2 and F21=d F2

dy. dydx

=(lin y )3 x2=9x2lin x

F ' ( x )=9 x2lin x−4 x lin x

4. Show that the area between the curve of f ( x )=9−x2 and x−axis from 0 to 3 is 18.Solution: We want to show that ∫

0

3

(9−x2 )dx=18. Let partition [0 ,3 ] into n equal sub intervals of width 3n and let A be the area:

Sn=3n [9−(

3n)2]+ 3n [9−(

3n×2)

2]+…+3n [9−(

3n×n)

2]¿ 3n

[9n ]−3n×( 3n )

2

[12+22+…+n2]

¿27−27n2 [n (n+1 )(2n+1)

6 ]A=lim

x→∞Sn= lim

x→∞ [27−27n2 ( n (n+1 )(2n+1)6 )]

¿ limn→∞ [27−27n2 ( 2n

3+3n2+n6 )]

¿ limn→∞ [27−271 ( 2+ 3n + 1

n2

6 )]¿27−27( 26 )=27−273 =27−9

¿18

141

5. Find F ' ( x ) if F ( x )=∫x

x2 tlin t

dt.Solution: F ( x )=∫

x

0 tlin t

dt+∫0

x2 tlint

dt

F ' ( x )=−∫0

x tlint

+∫0

x2 tlin t

dt

F ' ( x )=−F11+F2

1 (x )= x3

lin x− xlin x

= xlin x

[ x2−1 ]

QUESTION PAPER OF COMPLEX ANALYSIS II1(a) Define a complex function.(b) Use Euler Formula and De Moivre Theorem to find the 8 roots of unity.(c) If f (z) is analytic inside a closed circular annulus C1:r ≤|z−z0|<ρ and w is in the open annulus C2:r<|z−z0|<ρ, show that:

limn→∞ [ 12πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw+ 12πi∫C2

❑

( w−z0z−z0 )

n f (w)z−w

dw]=0.2(a) Define a pole of order N and a zero of order N of a complex valued function.(b) i. Find the Laurentz Series Expansion of f ( z )= 1

z (z−1) for 0<|z|<1and for ¿ z∨¿1. ii. Write a Taylor Series Expansion of sin zz .(c) Let f(z) is single-valued and analytic inside and on a circle C except at the point z=a. If f ( z )= ∑

n=−∞

∞ 12πi∮C

❑ f (z)(z−a)n+1 dz and ∮ f (z )dz=2 πia−1 when n=1, show that

a−1=limz→a

1(m−1 ) !

dm−1

dzm−1 [(z−a)m f (z)] .

3(a) State and prove the maximum modulus theorem.(b) Show that∫

−R

R x2dx(x2+1)2 (x2+2 x+2 )

+∫Γ

❑ z2dz(z2+1)2 ( z2+2 z+2 )

+i∫−Q

Q y2dy( y2+1)2 ( y2+2 y+2 )

=7 π50 ,

where Q ,R→∞.(c) Find the nth derivative off (a )= 12πi∮C

❑ f (z )z−a

dz.142

4(a) State Morera Theorem and prove Argument Theorem.(b) Use Argument Theorem to prove Rouche Theorem.(c) Use Rouche Theorem to show that all the roots of z7−5 z3+12=0 lie between the circles |z|=1 and |z|=2.5(a) i. Define conformal and isogonal mappings. ii. Why is the bilinear transformation w=αz+β

γz+σ a combination of translation, rotation, stretching and inversion? Otherwise, show that ασ−βγ ≠0.(b) Find the fixed or invariant points of w=z9.(c) Show that ∫0

∞ cosmxx2+1

dx= π2e−m ,m>0.

6(a) Let the rectangular region R in the z-plane be bounded by x=0 , y=0 , x=2 , y=1. Determine the region R ' of the w-plane into which R is mapped under the transformationw=√2 e

πi4 z+(1−2i .)(b) Explain the concept of analytic continuation of the functions F1(z ) and F2( z) in the regions R1 andR2.(c) Use Riemann Sphere to explain why complex plane C does not require a new infinity besides the aleph naught of counting numbers, N , and the continuum hypothesis of R, the x-axis.

7(a) State and prove the Fundamental Theorem of Algebra.(b) Find a bilinear transformation that maps points z=0 ,−i ,−1 into w=i ,1 ,0 respectively.(c) Show that ∫

0

2π

[ 1a+bsinθ +

13−2cosθ+sinθ ]dθ=π ( 2+√a2+b2

√a2+b2 ) .MARKING SCHEME OF MTH 43121(a) A complex variable is a function from C to C, i.e. f :C→C.1(b) Euler Formula is given as e iθ=cosθ+i sinθ and De Moivre Theorem states that zn=r n(cosnθ+i sin nθ). Thus, z 1n=r

1n (cos θn+isin θ

n ). If θ=2πk where 2π is the circumference of the unit circle and k=0 ,1 ,2 ,…n−1 . e2πki=cos 2 πkn +isin 2πk

n since r1n=1

1n=1. Since we are looking for the 8-roots of unity, we divide 360 by 8 to get: 45.00, 90.00, 135.00, 180.00, 225.00, 270.00, 315.00, 360.00 and the roots are:

143

e45 .00 i=cos 45 .00+isin45 .00=√ 12 (1+i ) , e90 .0

0 i=cos90 .00+isin90 .00=i ,

e135.00 i=cos135 .00+isin135.00=√ 12 (−1+i), e180.0

0 i=cos180 .00+isin180 .00=−1

, e225.00i=cos225 .00+isin225 .00=√ 12 (−1−i ) , e270 .0

0 i=cos 270.00+isin270 .00=−i

, e315.00 i=cos315 .00+isin315.00=√ 12 (1−i) , e360 .0

0 i=cos360 .00+isin360 .00=1

1(c) Proof: Since w is on C1; we have: | z−z0w−z0|=γ<1, where γ is a constant. By

Boundedness Theorem [since f is analytic (continuous) in a closed circular annulus (closed interval)]: |f (w)|<M. And, |w−z|=|(w−z0 )−(z−z0)|≥r 1−|z−z0|. Hence, |Rn|=

12π |( z−z0

w−z0 )n f (w)w−z

d w|≤ 12π γ nMr1−|z−z0|

.2π r1=γn Mr1

r1−|z−z0|→0asn→∞.

That is, limn→∞

12 πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw=0.Also, since w is on C2; we have: |w−z0

z−z0 |=k<1. By Boundedness Theorem, |f (w)|≤ M for all w on C2 and |z−w|=|( z−z0 )−(w−z0 )|≥|z−z0|−r. Therefore, |Pn|=

12 π |∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw|≤ 12π

k nM|z−z0|−r

.2πr= kn M|z−z0|−r

→0 as n→∞.That is, lim

n→∞

12 πi∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw=0

Therefore, limn→∞

12 πi∫C2

❑

(w−z0z−z0 )

n f (w)z−w

dw+ limn→∞

12 πi∫C1

❑

( z−z0w−z0 )

n f (w)w−z

dw=0

2(a) Consider the Laurentz Series Expansion, f ( z )= ∑

n=−N

∞

cn ( z−z0 )n=c−N

( z−z0 )N+

c−N+1

( z−z0 )N−1 +…+c−1

z−z0+c0+c1 ( z−z0 )+…, where N is a

positive integer and c−N≠0. Then z0 is called a pole of order N.144

Consider the Laurentz Series Expansion, f ( z )=∑n=N

∞

an ( z−z0 )n. Then z0 is called a zero of f of order N.2(b)i. For 0<|z|<1, f (z )= 1

z (z−1)= 1z−1

−1z= −11−z

−1z=− (1+ z+z2+…)−1

z . For |z|>1 , 1

z−1=1

z ( 1

1−1z )=1z (1+1z + 1

z2+ 1z3

+…)= 1z+ 1z2+ 1z3 +…

ii. sin zz

=1− z2

3 !+ z4

5!− z6

7 !+…

2(c)Proof [a−1=lim

z→a

1(m−1 ) !

dm−1

d zm−1 {( z−a )m f (z )}]. Suppose f (z) has a pole a of order m. Then the Laurentz Series of f (z) is f ( z )=

a−m

(z−a)m+

a−m+1

(z−a)m−1+…+a−1

z−a+a0+a1 ( z−a )+a2(z−a)2+…(1)

Multiplying (1) by (z−a)m from both sides, we have:(z−a)m f (z )=a−m+a−m+1 ( z−a )+…+a−1(z−a)m−1+a0(z−a)m+…(2)Differentiating both sides of (2), m−1׿ , with respect to z❑; we have:dm−1

dzm−1 {( z−a )m f (z )}=(m−1 )! a−1+m (m−1 )…2a0 ( z−a )+…(3)Let z→a in (3). Then lim

z→a

dm−1

dzm−1 {( z−a )m f (z) }=(m−1 ) !a−1, i.e.a−1=lim

z→a

1(m−1 ) !

dm−1

d zm−1 {( z−a )m f (z )}

3(a)145

Maximum Modulus Theorem: Suppose f (z) is analytic inside and on a simple closed curve C. Then the maximum values of ¿ f (z)∨¿ occurs on C, unless when f (z) is a constant.Proof: By Cauchy Integral Formula,

f ( a )= 12πi∮C

❑ f (z )z−a

dz=¿ f (a )= 12π∫0

2 π f (a+r e iθ) ir e iθdθr e iθ

= 12π ∫0

2π

f (a+r e iθ)dθ

|f (a)|≤ 12π∫0

2π

|f (a+r e iθ)|dθ… (1)

Let us suppose that ¿ f (a)∨¿ is a maximum, so that ¿ f (a+r e iθ)∨≤∨f (a)∨¿ and|f (a+r e iθ )|<¿ f (a)∨¿ for one value of θ, then by continuity of f , it will hold for a finite arc, say θ1<θ<θ2. But |f (a+r eiθ )|<|f (a)| contradicts (1). Therefore, in any σ neighbourhood of a, |z−a|<σ ; f (z) must be a constant. If f (z) is not a constant, the maximum value of |f(z)| must occur on C.3(b)Solution: The poles of z2

( z2+1 )2 ( z2+2 z+2 ) enclosed by the contour C are z=i of order 2

and z=−1+i of order 1. Res (i )=lim

z→ i

ddz {( z−i )2 z2

( z+i )2 ( z−i )2 ( z2+2 z+2 ) }=9 i−12100

Res (−1+i )= limz→−1+ i

( z+1−i ) z2

( z2+1 )2 ( z+1−i ) ( z+1+i )=3−4 i

25

∮ z2dz( z2+1 )2 ( z2+2 z+2 )

=2πi{9 i−12100+ 3−4 i25 }= 7 π50

Hence, ∫−R

R x2dx(x2+1 )2 (x2+2 x+2 )

+∫Γ

❑ z2dz( z2+1 )2 ( z2+2 z+2 )

+i∫−Q

Q y2dy( y2+1 )2 ( y2+2 y+2 )

=7 π50 , where

R ,Q→∞.3(c)

146

Given that f (a )= 12πi∮C

❑ f (z )z−a

dz, its first derivative is f ' (a )= 12πi∮C

❑ f ( z)( z−a )2

dz, the second derivative is f ' ' (a )= 1

2πi∮C❑ f (z)

( z−a )3dz and the nth derivative is f n (a )= n!

2πi∮C❑ f (z )

(z−a )n+1 dz since ddz (z−a)−1=−(z−a)−2 , d

dz(−( z−a )−2 )=−1×−2 (z−a)−3=2! (z−a)−3

4(a) Morera Theorem: Let f (z) be a continuous function in a simply connected region R and suppose ∫

C

❑

f ( z )dz=0around every simple closed curve C in R. Then f (z) is analytic in R.4(b)Proof [Argument Theorem]: Let C1 and Γ1 be non-overlapping circles lying inside C and enclosing z=α and z=β respectively. Then: 12πi∮

f ' (z)f (z )

dz= 12πi∫C1

❑ f ' (z)f (z )

dz+¿ 12πi∫Γ 1

❑ f '(z )f (z)

dz ¿…(1)Since f (z) has a pole of order p at z=α❑, we have: f ( z )= F (z )

( z−α )p …(2). Taking

logarithmic differentiation of (2), we have: f ' (z )f (z)

= F '(z )F (z )

− pz−α

…(3), so that 12πi∫C1

❑ f ' (z)f (z)

dz= 12πi∫C1

❑ f ' (z)f (z)

dz−¿ p2πi∫C1

❑ 1z−α

dz=0−p=− p… (4)¿

Since f (z) has a zero of order n at z=β, we have: f ( z )= ( z−β )nG ( z )… (5)

By logarithmic differentiation of (5), we have: f ' (z )f (z)

= nz−β

+G' (z )G(z)

…(6)

So, 12πi∫Γ1❑ f ' (z)

f (z)dz= n

2πi∮Γ 1

❑ dzz−β

+ 12πi∮

G' (z)G (z)

dz=n…(7)

From (1), (4) and (7), we have the required result as:12πi∮

f ' (z)f (z )

dz= 12πi∫C1

❑ f ' (z)f (z )

dz+¿ 12πi∫Γ 1

❑ f '(z )f (z)

dz=n−p ¿

4(c)Proof [Rouche Theorem]: Let F ( z )= g(z )f ( z) , so that g ( z )=f ( z ) F(z ) or briefly g= fF.

Then if N1 and N 2 are the number of zeros inside C of f +g and f respectively, from Argument Theorem using the fact that these functions have no poles inside C, 147

N 1=12πi∮C

❑ f '+g 'f+g

dz,N 2=12πi∮C

❑ f 'fdz. Then:

N 1−N2=12 πi∮C

❑ f '+ f ' F+fF 'f + fF

dz− 12πi∮C

❑ f 'fdz= 1

2πi∮C❑ f ' (1+F)

f (1+F)dz− 1

2πi∮C❑ f '

fdz= 1

2πi∮C❑

{ f 'f + F '1+F }dz− 1

2 πi∮C❑ f '

fdz= 1

2πi∮C❑ F '1+F

dz= 12πi∮C

❑

F ' (1−F+F2−F3+…)dz

Using the fact that |F|<1 on C, the series is uniformly convergent on C because term by term integration yields the value zero. Thus, N1=N2.4(d)Solution: Consider the circle C1:∨z∨¿1. Let f ( z )=12 andg ( z )=z7−5 z3. On C, we have: |g ( z )|=|12−5 z3|≤|12|+|5 z3|≤60<27=¿ f (z)∨¿

By the Rouche Theorem, f ( z )+g ( z )=z7−5 z3+12 has the same number of zeros inside |z|=2 as f ( z )=z7, i.e. all the zeros are inside C2. Hence, all the roots lie inside |z|=2 but outside |z|=1, as required.5(a) i. Conformal Mapping: Suppose that under a transformation, point (x0 , y0 ) of the xy-plane is mapped into point (u0 , v0 ) of the uv-plane while curves C1 and C2 are mapped, respectively, into curves C1

' and C2' intersecting at (u0 , v0 ). Then, if the transformation is such that the angle at (x0 , y0 ) between C1

' and C2' both in magnitude and direction, the transformation or mapping is said to be conformal at

(x0 , y0 ). Isogonal: A mapping that preserves the magnitude of angle but not direction is called isogonal.ii. w=αz+β

γz+σ is a translation because it is of the form αzγz+σ

+ βγz+σ , it is a rotation

because α may equal e i θ0 , it is a stretch because α may equal az, it is an inversion of 1

γz+σ . Now, if w= z+ βz+ β

=1, it is not a bilinear transformation. It is so if α=γ and β=σ ⇒ α

γ=1orσβ=1. Only negation works, that is, αγ ≠1 ,and

σβ≠1⇒ ασ

γβ≠1⇒ ασ ≠ γβ⇒ασ−βγ ≠0.

148

5(b) The fixed or invariant points of w=z9 are:

z=0 , z=√ 12 (i+1 ) , z=i , z=√ 12 (−1+i ) ,

z=−1 , z=√ 12 (−1−i ) , z=−i , z=√ 12 (1−i )andz=1.5(c) Solution: The integrand has simple poles at z=± i ,but only z=i lies inside C.Res ( i )=lim

z→i {( z−i ) e imz

(z−i )( z+i)}= e−m

2i and ∫

−∞

∞ cosmxx2+1

dx=2 πi( e−m

2 i )=π e−m

That is, ∫0

∞ cosmxx2+1

dx= π2e−m.

6(a) Given w=√2 e14 z+(1−2 i ) ,u+ iv=(1+i ) ( x+iy )+1−2 i and u=x− y+1 , v=x+ y−2. The lines x=0 , y=0 , x=2 , y=1 are mapped respectively into

u+v=−1 , u−v=3 , u+v=3 ,u−v=1.

6(b) Analytic Continuation: Let F1(z ) be a function of z which is analytic in a region R1. Suppose that we an find a function F2( z) which is analytic in a region R2 and which is such that F1(z )=F2(z) in the region common to R1 and R2. Then we say that F2( z) is an analytic continuation of F1(z ). This means that there is a function F(z) analytic in the combined regions R1 and R2 such that F ( z )=F2(z) in R1 and F ( z )=F2(z) in R2. It suffices for R1 and R2 to have only a small arc in common such as LMN as follows:

149

6(c) Let P be the complex plane and consider a sphere Q tangent to P at z=0. The diameter NS is perpendicular to P and we call points N and S the north and south poles of Q. Corresponding to any point A on P, we can construct line NA intersecting Q at point A’. Thus, to each point of the complex plane P, there corresponds one and only one point of the sphere Q, and we can represent any complex number by a point on the sphere. For completeness, we say that the point N itself corresponds to the point at infinity of the plane. The set of all points of the complex plane including the point at infinity is called the entire complex plane, the entire z-plane, or the extended complex plane. The sphere is sometimes called the Riemann Sphere. When the diameter of the Riemann Sphere is chosen to be unity, the equator corresponds to the unit circle of the complex plane.

7(a) Fundamental Theorem of Algebra: Every polynomial equation p ( z )=a0+a1 z

1+a2 z2+…+an z

n=0, where the degree n≥1 and an≠0, has at least one root.Proof: Suppose on the contrary, p ( z )=0 has no root, then f ( z )= 1p(z ) analytic for all

z. Also, |f (z)|= 1|p(z)| is bounded (and in fact approaches zero) as |p ( z )|→∞. Then it

follows that f (z) and p(z ) must be constants. This is a contradiction and we thus conclude that p ( z )=0 must have at least one root.7(b)150

Solution: Haven known that w=αz+βγz+σ , we have: i=α (0 )+ β

γ (0 )+σ… (1 ) ,1=α (−i )+β

γ (−i )+σ…(2),and

0=α (−1 )+βγ (−1 )+σ

…(3). From (3), β=α. From (1) and (3), σ= βi=−iα . From (2), γ=iα .

Therefore, w= αz+αiαz−iα

=1i ( z+1z−1 )=−i( z+1z−1 ).

7(c)Solution: Let z=eiθ. Then sin θ= z−z−1

2i, cosθ= z+z−1

2, dz=izdθ, so that

∫0

2π dθ3−2cosθ+sinθ

=∮C

❑ 2dz(1−2i ) z2+6 iz−1−2 i

. The poles of 2dz(1−2 i ) z2+6 iz−1−2i

are 2−i and 2−i

5 , but only 2−i5 lies inside C.

Res( 2−i5 )= lim

z→ 2−i5

{z−2−i5 }{ 2dz

(1−2 i) z2+6 iz−1−2i }=lim

z→ 2−i5

2

2 (1−2 i ) z+6 i= 12 i

by L-Hospital Rule.Therefore, ∮

C

❑ 2dz(1−2 i ) z2+6 iz−1−2i

=2 πi( 12 i )=π. And, ∫0

2π dθa+bsinθ

=∮C

❑ 2dzb z2+2aiz−b

. The poles of 2

b z2+2aiz−b are ( – a+√a2−b2

b )iand( – a−√a2−b2

b ) i. only ( – a+√a2−b2

b )i lies inside C since |( – a+√a2−b2

b ) i|=|√a2−b2−ab

√a2−b2+a√a2−b2+a|<1. Whenever a>¿b∨¿,

Res(z1=( –a+√a2−b2

b )i)=limz→z1

2b z2+2aiz−b

=limz→z2

22bz+2ai

= 1b z1+ai

= 1√a2−b2i

by L’ Hospital Rule. Therefore, ∮

C

❑ 2dzb z2+2aiz−b

=2πi ( 1√a2−b2 i )= 2π

√a2−b2. And, ∫

0

2π 1a+bsinθ

= 2π√a2−b2

. Thus, ∫0

2π

( 1a+bsinθ+

13−2cosθ+sinθ )dθ=π ( 2+√a2+b2

√a2+b2 ).

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