msc math complex variables

8/10/2019 MSc Math Complex Variables

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Mathematical PhysicsMSc courseDr. Ali Abdulateef Kareem


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In the early days of modern mathematics, people were puzzled by

equations like this one:

For this reason, mathematicians dubbed an imaginary number. We

abbreviate this by writing i in its place, that is:

Now have solutions in terms of complex numbers, i.e

2 1 0x

1

1i

x i

Which have no solution in the real number system

It is useful to note that

2

2 4 2

2

3 2

1 3

2 3

1, , 1

1 1 1, 1,

i i i i

ii i i i

ii

i

i i

i i

i


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Functions of a complex variable provide us some powerfuland widely useful tools in in theoretical physics.

Some important physical quantities are complex variables(the wave-function )

Evaluating definite integrals.

Integral transforms

, i kx t x t A e

representing a wave travelling in the positivexdirection,

recall that: cos sinix

e ix x


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The familiar numbers, such as 1, 1/3, 2, and which arerepresented by points on a line, will be referred to as real numbers.

A complex number is an expression of the form

z x iy

where xandyare real numbers. The number xis called the real part

of zand is writtenRex z

The numbery, despite the fact that it is also a real number, is called

the imaginary part of zand is written

Imy z

If x= 0, then z= iy is a pure imaginary number.

Two complex numbers are equal if and only if their real parts are

equal and their imaginary parts are equal.


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A complex number (The sum of real and an imaginary umber) can beplotted on a plane with two perpendicular coordinate axes

The horizontal x-axis, called the real axis

The vertical y-axis, called the imaginary axis.

Each complex number z = x + iy corresponds to the point P(x, y)in the xy-plane.

r is the absolute value or modulus of z, then, is just the distancefrom the point P(x, y) to the origin.

2 2z r x y

*

2 2

( )( )zz x iy x iy

x y

Note that :

x

yP(x, y)

z r

Re

Im

O

,z z

z x iy


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There is a further interest and significance if we make use the usualpolar coordinates in the xy-plane for representation the complex

numbers which gives

x

yP(x, y)

z r

Re

Im

O

cos , sinx r y r

siny r

cosx r

2 2

1tan , if 0, tan

r x y

y yx

x x

2 2 2 2sin , cos

y x

x y x y

z x iy cos sinz r i or r cis


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From the complex number we have2 4z i

4 24 16 20, sin , cos

20 20z r

2 4z i

2 4z i The complex conjugate of

orz z 2 4z i

Then 20, 20z z

Therefore, =z z

In other words in the diagram, the

complex conjugate is the mirror image

of z in the real axis


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1 2 1 2

1 2 1 2

1 1

2 2

2

. .

(a)

(b)

(c)

(d)

(e)

(f)

z z z z

z z z z

z z

z z

z z

z z

z z z


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Example:Find the polar representation 1z i

Solution:

cos sinz r i


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Addition and Subtraction of Complex Numbers

The complex numbers satisfy the commutative and associativelaws

(10 4i) (5 2i)

= (10 5) + [4 (2)]i

= 5 2i

(4 6i) + (3 + 7i)

= [4 + (3)] + [6 + 7]i

= 1 + i

1 2 1 1 2 2

1 2 1 2 1 2

z z x iy x iy

z z x x i y y

1 1 1 2 2 2If and are two complex numbersthen

z x iy z x iy

1 2 2 1

1 2 3 1 2 3

z z z z

z z z z z z

Examples


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Multiplication of Complex Numbers

The product of and is defined as

1 2 1 1 2 2

1 2 1 2 1 2 2 1

( )( )

= ( ) ( ).

z z x iy x iy

x x y y i x y x y

Division of Complex NumbersThe division of and is defined as1z

2 2 1 12 2 2

1 1 1 1 1 1 1

x iy x iyz x iy

z x iy x iy x iy

2

2 1 2 1 2 1 2 1

2 2 2

1 1

x x x iy iy x i y y

x i y

2 1 2 1 2 1 2 12 2

1 1

x x y y y x x y i

x y

2 1 2 1 2 1 2 1

2 2 2 2

1 1 1 1

x x y y y x x yi

x y x y

Denominatorconjugate

2z


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4(a)

2 3

i

i

4 2 3 5 14 5 14

2 3 2 3 13 13 13

i i ii

i i

1 2 2(b)

3 4 5

i i

i i

1 2 3 4 2 5

3 4 3 4 5 5

5 10 5 10 2

25 25 5

i i i i

i i i i

i i

H.W.30 193

2 1

i i

i

Express each of the following complex numbers in polar form.

( ) 2 2 3 , ( ) 5 5 , ( ) 6 2 , ( ) 3a i b i c i d i


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We know that

find

1 1 1 1 2 2 2 2andcos sin cos sinz r i z r i 1

1 2

2

andz

z z

z

1 1 11

2 2 2 2

cos sin

cos sin

r iz

z r i

2 2

2 2

cos sin

cos sin

i

i

1 2 1 2 1 2 1 212 2

2 2 2

cos cos sin sin sin cos cos sin

cos sin

ir

r

1 1 1 2 1 22 2

cos sinz r

iz r

1 2 1 1 1 2 2 2cos sin cos sinz z r i r i 1 2 1 2 1 2 1 2 1 2cos cos sin sin sin cos cos sinr r i

1 2 1 2 1 2 1 2cos sinz z r r i


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Find the product of4(cos50 sin 50 ) and 2(cos10 sin10 ).i i

4(cos50 isin50) 2(cos10 isin10 )

4 2 cos(50 10) isin(50 10)

8(cos60 isin60)

8 1

2 i

3

2

4 4i 3

Find the division of 16(cos70 sin 70 ) 16= cos(70 40 ) sin(70 40 )4(cos40 sin 40 ) 44cos30 sin30

3 14 2 3 2

2 2

i ii

i

i i


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We know 1 2 1 2 1 2 1 2cos sinz z r r i

If we have a set of n complex numbers, a generalization of the

above equation leads to:

1 2 1 2 1 2 1 2cos sinn n n nz z z r r r i

putting

1 2 1 2

1 and givesn n

r r r

cos sin cos sinnnz i n i n

1 1cos sin cos sin .nn n

z r i r n i n

DeMoivres

theorem

n

i nie e The results is equivalent to the statement

In general


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First, find trigonometric notation for (

1

i) using

Theorem 1 2 cos225 sin 225i i

5

5

5

1 2 cos 225 sin 225

using De Moiver's theorem we get

cos( ) sin( )

4 2 cos112

2 5 225 5 22

5 sin1125

2 24 2

5

2 2

4 4

i i

i

i

i

i

52 , 225

4r

cos sinz r i


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2 3 4

12! 3! 4!

x x x xe x Substitute ixforx.

2 3 4 5 61

2! 3! 4! 5! 6!

ix ix ix ix ix ixe ix

2 2 3 3 4 4 5 5 6 6

12! 3! 4! 5! 6!

ix i x i x i x i x i xe ix

2 3 4 5 6

12! 3! 4! 5! 6!

ix x ix x ix xe ix

2 4 6 3 5

12! 4! 6! 3! 5!

xi x x x x xe i x

Factor out the iterms.


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2 4 6 3 5

12! 4! 6! 3! 5!

ix x x x x xe i x

This is the series for

cosine.

This is the series

for sine.

cos sinz r i or r cis

cos sinix

e ix x Eulers Formula

Using EulerMs formula, the polar form of a complex number canbe rewritten as :

(cos sin )

i

z r i x y

z r e


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Show that

(a) (b)

cos 2

i ie e

sin

2

i ie e

i

Solution

we have cos sin and cos sini ie i e i

(a) Adding (1) and (2)

2cos or cos2

i ii i e ee e

(a) subtracting (2) and (1)

2 sin or sin2

i i

i i e ee e ii

H.W: Prove

(a) (b)33 1

sin sin sin 3 ,4 4

41 1 3

sin cos 4 cos 2 .8 2 8


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If , then to each value of wthere correspondsone value of z. Conversely, to a given there correspond

precisely n distinct values of w. Each of these values is called annth rootof z, and we write

( 1, 2, )nz w n 0z

nw z

Hence this symbol is multivalued, namely, n-valued. The n

values of can be obtained as follows.n z

andcos sin cos sinz r i w R i Then the equation becomes, by De Moivresformulanw z

cos sin nnw R i cos sin cos sinnR n i n z r i

We write zand win polar form


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i.e. cos sin cos sinnR n i n r i

Form the Equating of these equations, The modulus

n nR r R r

and the arguments

thu2

2 , s k

n kn n

where kis an integer. For we get ndistinct values

of w. Further integers of kwould give values already obtained. For

instance,

0, 1, 2, ..., 1k n

gives 2 / 2k n k n

Re

Im

4


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hence the wcorresponding to k = 0, etc. Consequently,

for , has the n distinct values

n z

0z

2 2cos sinn n

k kz r i

n n

where . These n values lie on a circle of radius

with center at the origin and constitute the vertices of

polygon of n sides.

0, 1, 2, ..., 1k n n r


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Find the square roots of

Trigonometric notation:

For k= 0, root is

For k= 1, root is

1 3i

1 3 2 cos60 sin 60i i

11

2260 360 60 360

2 cos60 sin 60 2 cos sin2 2 2 2

2 cos 30 180 sin 30 180

i k i k

k i k

2 cos30 isin30

2 cos210 isin210

2 2cos sinn n

k kz r i

n n


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Find all fourth roots of Write the roots inrectangular form.

Write in trigonometric form.

Here r= 16 and = 120. The fourth roots of this number haveabsolute value

8 8 3.i

8 8 3 16 cis 120i

4

16 2.120 360

30 904 4

kk

0 30 90 30

1 30 90 120

2 30 90 210

3 30 90

0

1

2

3 300

k

k

k

k

There are four fourth roots, let k= 0, 1, 2 and 3.


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Using these angles, the fourth roots are:

2 cis 30 , 2 cis 120 ,

2 cis 210 , 2 cis 300

written in rectangular form

3

1 3

3

1 3

i

i

i

i

The graphs of the roots areall on a circle that has center

at the origin and radius 2.

13


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Find each of the indicated roots and locate themgraphically.

1 41 3

1 , ( ) 2 3 2a i b i

Find the square roots of 15 8i


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When dealing with integrals that have complex numbers wetreat the imaginary number ias a constant

Example: The integral 1 i xe dx

Let 1

then 11

u i x

dudu i dx dx

i

Making the substitution yields

11

.1 1 1

i xuu e ee du

i i i

Removing the complex denominator by multiplying it by the conjugate

gives

111 1

1 1 2

i xi xe i i

ei i

the constant will be a complex number as well thus 1 2c c ic 1 1

1 2

1

2

i x i xi

e dx e c ic


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Some real-valued integrals can be solved using complex-valued

integrals.Integration such as and canbe solved using integration by parts.

Using the integral we can solve integration easily

cosxe x dx sinxe x dx 1 i x

e dx

rearrangement of the integral and its value 1 i xe dx 1 1 212

i xie c ic

cos sin

cos sin

x ix

x

x x

e e dx

e x i x dx

e x dx i e x dx

Firstly, rearranging and applying Euler's Formula gives 1 i xe dx


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Rearranging the value of the integral yields 1 1 21

2

i xie c ic

11 21

2 2i xi e c ic

1 2

1 2

1 2

1 2

1 2

1

1cos sin

2 2

1cos sin cos sin2 2

1 1cos sin cos sin

2 2

1 1 1 1cos sin cos sin

2 2 2 21 1

cos sin sin cos2 2

1 1cos sin

2

x

x x

x x

x x x x

x x

x

ie x i x c ic

ie x i x e x i x c ic

e x i x e i x x c ic

e x i e x i e x e x c ic

e x x c i e x x ic

e x x c i

2sin cos

2

xe x x c


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Since the integral and its value

rearranged, are equivalent, it means that the real parts areequal to each other, as well as the imaginary parts

cos sinx x

e x dx i e x dx

Therefore

1cos cos sin

21

sin sin cos2

x x

x x

e x dx e x x c

e x dx e x x c

cos sinx xe x dx i e x dx 1 2

1 1cos sin sin cos

2 2

x xe x x c i e x x c

msc math complex variables

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