the theory of complex variables - university of east angliah008/teaching/2c1y/complex... · 2012....
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The Theory of Complex Variables
Alan Camina, Graham Everest and Tom Ward
Autumn Semester 2000
Aims and Objectives
Aims
The aims of this part of the course are to:-
• revise the definition of a complex number
• introduce the geometry of the complex plane
• investigate the behaviour of functions from C → C with particular interest in the ideas ofcontinuity, differentiability and integration.
Objectives
The objectives are to :-
• understand the idea of open and closed sets and regions. To be able to describe geometricfigures using simple and parametric equations
• to know the definitions of continuity and differentiability and the Cauchy-Riemann equations.
• to understand the definition of an integral along a contour.
• to know and be able to apply the “fundamental theorem” of calculus and Cauchy’s Theorem.
• to know Taylor’s and Laurent’s theorems and to be able to calculate the appropriate series fora large class of functions
• to know the definition of singularity and pole and be able to classify them for a given function.
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Chapter 1
Introduction
These are lecture notes for the complex variable section of MTH-2A31. Hopefully theywill be available on the UEA intranet. The suggested book for the course is “Introductionto Complex Analysis” by H.A. Priestley (Oxford). It is not essential to have this book.
There are other books, Stewart and Tall, Churchill et al, Spiegel. They are all quite goodand cover most of the material in this course. Hopefully the notes will suffice and I willtry to make them available.
In this we wish to investigate the structure of complex numbers and show how to develop a theorywhich enables us to do for functions of complex numbers what calculus does for the real numbers.The first section is largely revision then we discuss the notions of continuity and differentiability. Itwill be interesting to note that these concepts are much stronger than for the real case. It will beshown that any complex function which is differentiable gives rise to two functions which are relatedand both satisfy the important partial differential equation
∂2f(x, y)∂2x
+∂2f(x, y)∂2y
= 0.
This is Laplace’s equation which is fundamental to much Applied Mathematics and Physics.
The final sections develop the theory of integration for complex functions. This depends on areduction to the real case but leads to many amazing theorems. Some of which seem to have nothingto do with complex numbers. For example∫ +∞
−∞
11 + x2 + x4
textdx =π√3.
Another perhaps surprising result isTheorem Let f(x) be a polynomial then it has at least one root in C.
There are many other applications but many of these may not be seen until the second semester.It is important to see the two units as one whole course.
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Chapter 2
The Complex Plane
This is to some extent revision.
Definition 2.1 The set of numbers {a + bi : a, b ∈ R; i2 = −1} are called Complex Numbers anddenoted by C. a is called the real part and b is called the imaginary part. A complex number iscalled real if b = 0 and purely imaginary if a = 0.
It is usual to denote complex numbers as z, z1, z2, w to distinguish them from the real numbers.If z = a + ib we write a = Re z and b = Im z. We add and multiply them in the obvious way andget the following rules
Definition 2.2
(a+ bi) + (c+ di) = (a+ c) + (b+ di)(a+ bi)× (c+ di) = (ac− bd) + (ad+ bd)i.
All the basic rules about multiplication and addition come from applying these definitions and usingthe rules we know about the real numbers.
Lemma 2.3 Let z1, z2, z3 be complex numbers then
(z1 + z2) + z3 = z1 + (z2 + z3)(z1z2)z3 = z1(z2z3)
z1(z2 + z3) = z1z2 + z1z3
z1 + z2 = z2 + z1
z1z2 = z2z1
Proof: The proofs are very routine exercises. I will prove the third one. Let zr = ar+bri, r = 1, 2, 3.Then the left hand side is
(a1 + b1i)((a2 + a3) + (b2 + b3))i = (a1(a2 + a3)− b1(b2 + b3))+ (a1(b2 + b3) + b1(a2 + a3))i= (a1a2 + a1a3 − b1b2 − b1b3)+ (a1b2 + a1b3 + b1a2 + b1a3)i.
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Then the right hand side is
(a1 + b1i)(a2 + b2i) + (a1 + b1i)(a2 + b3i) = (a1a2 − b1b2)− (a1b2 + a2b1)i+ (a1a3 − b1b3)− (a1b3 + a1b3)i= (a1a2 + a1a3 − b1b2 − b1b3)+ (a1b2 + a1b3 + b1a2 + b1a3)i.
As you can see this is straightforward but tedious.2
The next lemma tells us how to find negatives, trivial and inverses.
Lemma 2.4 Let z = a+ bi ∈ C with a, b ∈ R then −z = −a− bi and
z−1 = 1/z =a
a2 + b2− b
a2 + b2i
whenever z 6= 0.
Both are calculations. The only thing to note is that 0 = 0 + 0i and that if z 6= 0 then at least oneof a = Re z or b = Im z is non-zero so a2 + b2 6= 0.
From this we can see that any equation of the form uz + v = w has a unique solution for z ifu 6= 0, here we assume that all entries are complex. We have not proved uniqueness but it is quitesimple to prove. All the algebraic rules that work for Q and R also work in C and all these arecalled Fields. Another way we can interpret the statement above is to say that any linear equationuz + w = 0 has one solution in C if u 6= 0, where all the constants lie in C. The same statementis true if we replace C with R. Note however that in R the quadratic equation x2 + 1 = 0 has nosolution but that in C there are two solutions i, −i. What is remarkable is that for any polynomial
unzn + un−1z
n−1 + · · ·+ u0 = 0
has n solutions in C if un 6= 0 if we count the solutions carefully.
I want to return to Lemma 2.4 to look at the form of the equation for the inverse:-
z−1 =1z
=a
a2 + b2− b
a2 + b2i.
From this two particular definitions arise which are important in the development of complex theo-ries.
Definition 2.5 Let z = a+ bi where z ∈ C and a, b ∈ R then
1. The modulus of z is√a2 + b2 and is denoted by |z| and
2. The complex conjugate of z is a− bi and is denoted by z.
The two concepts are related.
Lemma 2.6 Let z be a complex number. Then
1. |z|2 = zz
2. 2Re z = z + z and 2Im z = z − z
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Proof: |z|2 = (Re z)2 + (Im z)2 = (Re z + iIm z)(Re z − iIm z) = zz. The second part is as easy.
We can now rewrite the rule for inverses as z−1 = z/|z| which looks much neater. These bothhave a neat geometric interpretation so we introduce the Argand diagram.1
z
θ
|z|
Figure 2.1: The Argand Diagram
In this we represent each complex number as a point on the plane R2 where the x-axis is the realcoordinate and the y-axis is the imaginary coordinate. When we do this we can see that |z| is thedistance of z from the origin and z is the reflection of z in the real axis. Thus for any two complexnumbers z1, z2 there is the distance between them |z1 − z2|. However beware, this is not quite likethe absolute value in the real numbers. Note that |1| = | − 1| = |i| = | − i| = | cos θ + i sin θ| for anyangle θ.
We now have some important properties of modulus and complex conjugate
Lemma 2.7 Let z, z1 and z2 be any three complex numbers. Then we have the following
1. z1 + z2 = z1 + z2 and z1z2 = z1 z2.
2. |z| = |z| and zz = |z|2
3. |z1z2| = |z1||z2|
4. |Re z| ≤ |z| and |Im z| ≤ |z|
5. |z1 + z2| ≤ |z1|+ |z2|1Jean Robert Argand was born on 18 July 1768 in Geneva, (now Switzerland) and died on 13 Aug 1822 in Paris,
France He was an accountant who was only an amateur mathematician. He is famed for his geometrical interpretationof the complex numbers where i is interpreted as a rotation through 90◦. The concept of the modulus of a complexnumber is due to him. He also gave a beautiful proof (with some gaps) of the fundamental theorem of algebra in 1814.
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6. ||z1| − |z2|| ≤ |z1 − z2|
Proof: We will prove a selection. First (1). Let z1 = a1 + b1i and z2 = a2 + b2i. Then zt = at − btiand so the right hand side of (1) is
a1a2 − b1b2 − (a1b2 + a2b1)i = a1a2 − b1b2 + (a1b2 + a2b1)i.
Which is just the left hand side. So we have proved (1).
Note that |z|2 = (Re z)2 + (Im z)2. Note that | | is always positive this gives (4).
For number (5) we first observe that |z|2 = zz so
|z1 + z2|2 = (z1 + z2)(z1 + z2) = z1z1 + z2z2 + (z1z2 + z1z2)= |z1|2 + |z2|2 + (z1z2 + z1z2
= |z1|2 + |z2|2 + 2Re (z1z2)≤ |z1|2 + |z2|2 + 2|(z1z2)|, by(4)≤ (|z1|+ |z2|)2.
For the final part assume first that |z1| > |z2| and so ||z1− z2|| = |z1− z2|. Now write u+w = z1
and w = z2 and then
|u+ w| ≤ |u|+ |w|
and so
|u+ w| − |w| ≤ |u|||z1| − |z2|| ≤ |z1 − z2|
If |z2| > |z1| put u+ w = z1 and v = z2 and repeat the argument. 2
This all looks very clear if we draw the picture in the Argand diagram. See below:
Once we have introduced the idea of the Argand diagram we can see how to do geometry withthe complex numbers. We will often refer to the Complex Plane in recognition of this fact. It is easyto describe a straight line and a circle
Example 2.1 Let u and v be two points in the complex plane. Then the set of points
{z : |z − u| = |z − v|}
is the straight line through the midpoint of the line joining u and v perpendicular to that line. Seefigure
Example 2.2 Here we show that it is very easy to define a circle. Choose a point a and a radiusr, note r ∈ R and r > 0 then the set of points with
{z : |z − a| = r}
is the circle we require.
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w
u
w+u
Figure 2.2: The Inequalities
Note that with slight variations we get the solid disc
{z : |z − a| ≤ r}
or the open discD(a; r) = {z : |z − a| < r}
or even the punctured disc{z : 0 6= |z − a| < a}
So far we think of the complex plane with respect to Cartesian coordinates. Thus we writez = a + bi : a, b ∈ R but we can also use polar coordinates z = r(cos θ + i sin θ) where r = |z|and θ is called the argument and is the angle between the between z and the real axis measured inan anti-clockwise direction. NOTE that this definition does not work for z = 0! Also r is uniquehowever θ is not since we can add any integer multiple of 2π without affecting the value of cos orsin. It might seem strange but at this point we make the following definition.
Definition 2.8 Let θ ∈ R then
exp iθ = eiθ = cos θ + i sin θ.
Given this we can write z = reiθ = |z| exp(iθ). Using the laws for compound angles we get
Lemma 2.9 Let θ ∈ R and φ ∈ R then
ei(θ+φ) = eiθeiφ.
Proof: We just use the definition to get
ei(θ+φ) = cos(θ + φ) + i sin(θ + φ)= cos θ cosφ− sin θ sinφ+ i(sin θ cosφ+ cos θ sinφ)= (cos θ + i sin θ)(cosφ+ i sinφ)= eiθeiφ
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Corollary 2.10 (de Moivre’s Theorem) Let θ ∈ R then
exp(iθ)n = exp(inθ)
This follows from the above by induction. We can also see that for two complex numbers z1 = r1eiθ1
and z2 = r2eiθ2 the product is given by z1z2 = r1r2e
i(θ1+θ2). So the modulus is just the product butthe arguments add. Notice here that you have to be quite careful about the argument and recallthat it is not uniquely defined.
This enables us to define a circle using parameters quite neatly.
Example 2.3 Let a be the centre of the circle and let the radius be r. Then the circle is given by
{z : z = a+ reiθ, 0 ≤ θ < 2π}
Note this works as |reiθ| = 1 for all θ ∈ R. Here is another example of curve being given byparameters.
Example 2.4 Let a and b be two points in the complex plane. Then the line joining them is givenby
{z : z = (1− t)a+ tb, t ∈ R}
This the whole line, if consider
{z : z = (1− t)a+ tb, 0 ≤ t ≤ 1}
then this is the line between a and b. If we consider t > 0 we get the line to the right of b and if wechoose t < 0 we get the line to the left of a.
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Chapter 3
Functions, Continuity andDifferentiability
3.1 Introduction to Functions
We are mainly interested in functions from C→ C. One question might be why are we so concernedwith functions. Because that is how we interpret the physical world around us. Once we can countor measure something we try to see connections and we represent those connections in the form offunctions. Most of the physical world we see as being both continuous and smooth which is why theemphasis on such functions. There are interesting philosophical points about whether this representsreality but so far we have made dramatic progress by making these assumptions.
Let f be any such function, we will often write this as w = f(z) where z and w are both complexvariables. It is this class of functions which we will be concerned with. However this gives rise totwo new functions both R2 → R given by
u(x, y) = Re (f(z))
andv(x, y) = Im (f(z))
where z = z + iy. This can be rewritten as f(z) = u(x, y) + iv(x, y).
Example 3.1 1. Consider w = z2 then u(x, y) = x2 − y2 and v(x, y) = 2xy
2. Let f(z) = 1/z, z 6= 0 then u(x, y) = xx2+y2 and v(x, y) = −y
x2+y2 .
3. If w = |z| then u(x, y) =√x2 + y2 and v(x, y) = 0.
Notice that we do not insist that the functions are necessarily defined on the whole of the complexplane nor do we insist that the image is the whole plane. It is also quite hard to visualize sincea mapping C → C is a mapping in real terms from R
2 → R2. Two represent this we need a four
dimensional model.
Given two functions u and v both R2 → R we can define a new function f(z) = u(x, y) + iv(x, y)where z = x+ iy. where z = z + iy. This can be rewritten as f(z) = u(x, y) + iv(x, y).
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Example 3.2 1. Consider u(x, y) = x2 + y2 and v(x, y) = 2xy then we get f(z) = |z| +iRe (z)Im (z)
2. A more interesting example is given by u(x, y) = ex cos y and v(x, y) = ex sin y.
We write this function as ez or exp z. When z is real it coincides with the usual definitionsince θ = 0 or π. If z is purely imaginary so that θ = ±π we get the definition we had earlier
3.2 Continuity
For this subsection we are going to consider a function f : S → C where S is a subset of C. We willhave to be careful about our set S and that will force us to introduce the notion of an open set. Thedesire for continuity is not to have the function jumping around or tearing the space. If we think ofthe complex plane as a rubber sheet and the function as deforming the sheet then when we say thatthe function is continuous we are trying to say that the deformation does not tear the sheet. (Notethat a similar argument works for the real line).
The problem we face is that if S is too odd then we cannot make sense of this idea. So we needto look at special subsets.
Definition 3.1 A subset S of C is said to be open if for any point z ∈ S there is a r > 0 withD(z; r) ⊂ S.
bu
a
Figure 3.1: An Annulus
We ought to see if there are any such sets.
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Example 3.3 (Some Open Sets) 1. The first must be D(a; s) for some a ∈ C and s > 0.Choose any z ∈ D(a; s) and let |z − a| = t. Note that t < s so choose r = min((s− t)/2, t/2)then D(z; r) ⊂ D(a; s)
2. Let S = {z : Re z > 0}. This time we can just take r = Re z/2
3. Choose 0 < a < b and let u ∈ C. Define S = {z : a < |z − u| < b}. This is an annulus. Letz ∈ S and note that if |z − u| = t then a < t < b. Put r = min((t− a)/2, (b− t)/2.
4. Let S = {z : |z| > 1}.
Definition 3.2 A subset S of C is said to be closed if C \ S is open.
This might seem an odd definition but we will show that it makes sense in another way. Note that{z : |z| ≤ 1} is closed. It is the complement to the last example above. Also a a proper non-emptysubset can be neither open nor closed for example the set {z : a < |z − u| ≤ b}.
Notice that the set {z : |z| ≥ 0} is not open. We can now define a continuous function much aswe do for real functions.
Definition 3.3 1 Let S be an open subset of C and let f be a function S → C.
1. Given a point u ∈ S we say that f has a limit at u if there exists w ∈ C such that for everyε > 0 there exists an r > 0 such that D(u; r) ⊂ S and z ∈ D(u; r) then |f(z) − w| < ε. Wewrite this as limz → u = w
2. We say that f is continuous at u if limz→u = f(u)
3. If f is continuous at each point z ∈ S we say that f is continuous on S.
Note that if we write a > 0 we are implying that a ∈ R because such a statement doesn’t make sensefor a ∈ C. Also I have reversed the order but the statement is the same. You could just as well say:-
Let S be an open subset of C and let f be a function S → C. Given a point u ∈ S we say that fis continuous at u if for every ε > 0 there exists an r > 0 such that |f(z)− f(u)| < ε if |z − u| < r.We could also have defined the limit of a function f at a point u to be a w such that if for everyε > 0 there exists an r > 0 such that |f(z)− w| < ε if |z − u| < r. Then the function is continuousif it has a limit and the limit is f(u).
Example 3.4 1. f(z) = |z|.
2. f(z) = z2. This follows as |z2 − u2| = (|z − u|)(|z + u|). So we can choose r < ε/2(|u|).
In fact we have the following
Lemma 3.4 Let f : S → C. The f is continuous if and only if Re f and Im f are continuous mapsfrom R
2 → R.
Proof: Let f(x + iy) = u(x, y) + iv(x, y) where u(x, y) and v(x, y) are real, that is Re f = u(x, y)and Im f = v(x, y). Now assume that f is continuous at z0 = x0 + iy0 ∈ S. Given ε > 0 there existsr > 0 such that |f(z)− f(z0)| < ε whenever |z − z0| < r. But now we use Lemma 2.7 to see that
|(Re (f(z))− Re (f(z0))| ≤ |f(z)− f(z0)| < ε1If we do not demand that S is open we replace the condition that z ∈ D(u; r) by z ∈ D(u; r) ∩ S.
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and|(Im (f(z))− Im (f(z0))| ≤ |f(z)− f(z0)| < ε.
So we see that u(x, y) and v(x, y) are both continuous at (x0, y0).
Now assume that both are continuous. Then
|f(z)− f(z0)| =√
(u(x, y)− u(x0, y0))2 + (v(x, y)− v(x0, y0))2.
Now given ε > 0 there exists r > 0 so that |u(x, y)−u(x0, y0)| < ε/√
2 and |v(x, y)−v(x0, y0)| < ε/√
2whenever |(x, y)− (x0, y0)| < r. This gives the other half.2
This lemma makes it easy to see when a function is continuous by looking at the real andimaginary parts separately. We can now introduce a simple generalisation of a real idea.
Definition 3.5 Let z1, z2, z3, . . . , be an infinite sequence of complex numbers. It is said that thesequence converges to w if for all ε > 0 there exists N ∈ N so that |zn − w| < ε for all n > N . Ifthere is such a w the sequence is said to be convergent. We write limn→∞(zn) = w.
Given a set S ⊆ C and a convergent sequence z1, z2, z3, . . . , with zn ∈ S, ∀n ∈ N where the set{zn | n ∈ N} is infinite, we say that limn→∞ zn is a limit point of S.
An alternative definition which might be easier to follow is that a point z is a limit point of aset S if and only if every punctured disc about z contains a point of S.
We use the ideas from Lemma 3.4 to prove.
Lemma 3.6 Let z1, z2, z3, . . . , be an infinite sequence of complex numbers. Then the sequence hasa limit if and only if the two sequences of real and imaginary parts are convergent.
We now get a nice result-
Proposition 3.7 A set S ⊆ C is closed if and only if every limit point of S is in S
Proof: Let z1, z2, z3, . . . , be an infinite sequence of elements of S with limn→∞ zn = w
First assume that S is closed. We have to show that w ∈ S. We know S′ = C \ S is open. Soassume that w 6∈ S, that is w ∈ S′. So there is ε > 0 with D(w; ε) ⊂ S′. But there is N ∈ N so that|zn − w| < ε if n > N . So zN+1 ∈ D(w, ε) ⊂ S′. But zN+1 ∈ S which gives a contradiction.
Now we assume that every limit point of S is an element of S. So we need to prove that S isclosed which means that we have to show that S′ = C \ S is open. Now assume that z ∈ S′ andthere is no open disc with centre z contained in S′. So for each ε > 0 the disc D(z; ε) 6⊂ S′. For eachn > 0, n ∈ N we can find zn ∈ D(z; 1/n). But then limn→∞ zn = z. This means that z is a limitpoint of S and so s ∈ S which gives a contradiction.2
We leave as an exercise the proof of:-
Proposition 3.8 Let f be a function S → C where S is open. Then f is continuous on S if andonly if
f( limn→∞
zn) = limn→∞
f(zn)
for all convergent sequences in S.
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3.3 Differentiability
We can now make a very similar definition as in the real case.
Definition 3.9 Let f : S → C be a function from an open set S to the complex plane. We say thatf is differentiable at the point z0 if
limz→z0
f(z)− f(z0)z − z0
exists. We write the limit as f ′(z0). If the function is differentiable at every point of S we say thatf is differentiable on S, analytic on S or the term we will use is holomorphic on S.
The key thing to note here is that we can approach z0 any way we like, from any direction andthis has a huge impact.
Example 3.5 Let f(x+ iy) = x+ y+ (x− y)i. Let z0 = 0. In any disc around 0 we will get pointson the real axis, say z = x but then
f(z)− f(z0)z − z0
=x+ ix
x= 1 + i.
In any disc around 0 we will get points on the imgainary axis but then
f(z)− f(z0)z − z0
=y − iyiy
= −1− ii
= −1− i.
So the function cannot be differentiable despite the real and imaginary bits both being well-behaved.
This idea leads us to the following important theorem.
Theorem 3.10 (Cauchy-Riemann) Let f : S → C be a function from an open set S to thecomplex plane. Let f(x+ iy) = u(x, y) + iv(x, y). If f is differentiable at z = z0 then
∂u
∂x=
∂v
∂y(3.1)
and
∂u
∂y= −∂v
∂x. (3.2)
The converse is true if and only if the equations hold and the partial derivatives are continuous insome disc about z0
Proof: We use the idea above and consider what happens when we approach z0 = x0 + iy0 hori-zontally and vertically. So we see what happens when z = x+ iy0
f ′(z0) = limz→z0
f(z)− f(z0)x− x0
= limx→x0
u(x, y0)− u(x0, y0) + i(v(x, y0)− v(x0, y0))x− x0
=∂u
∂x+ i
∂v
∂x
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f ′(z0) = limz→z0
f(z)− f(z0)iy − iy0
= limy→y0
u(x0, y)− u(x0, y0) + i(v(x0, y)− v(x0, y0))i(y − y0)
=−i∂u∂y
+∂v
∂y
Now comparing the two equations we see that
∂u
∂x=
∂v
∂y(3.3)
and
∂u
∂y= −∂v
∂x. (3.4)
To do the reverse implication is quite subtle so I will outline the proof that is I will write downthe equations and wave my hands. Let z0 = x0 + iy0 z = h+ z0 where h = a+ ib then consider
f(z0 + h)− f(z0)h
=u(x0 + a, y0 + b)− u(x0, y0) + i(v(x0 + a, y0 + b)− v(x0, y0))
h
=a
h
(u(x0 + a, y0 + b)− u(x0, y0 + b) + i(v(x0 + a, y0 + b)− v(x0, y0 + b))
a
)+
b
h
(u(x0, y0 + b)− u(x0, y0) + i(v(x0, y0 + b)− v(x0, y0))
b
)
Now wave your hands and this uses both the existence of the partial derivatives and their continuityto get:-
limh→0
f(z0 + h)− f(z0)h
=a
h
(∂u
∂x+ i
∂v
∂x
)+b
h
(∂u
∂y+ i
∂v
∂y
)Now using the Cauchy-Riemann Equations
=a
hf ′(z0) +
b
hif ′(z0)
= f ′(z0).
Sincef ′(z0) =
∂u
∂x+ i
∂v
∂x
andif ′(z0) =
∂u
∂y+ i
∂v
∂y.
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Cauchy2 and Riemann 3 are two of the greatest figures of 19th mathematics.
Example 3.6 Let f(z) = z. The f is not differentiable anywhere. Note that u(x, y) = x and2Augustin Louis Cauchy (born 21 Aug 1789 in Paris, died 23 May 1857 in Sceaux) pioneered the study of analysis
and the theory of permutation groups. He also researched in convergence and divergence of infinite series, differentialequations, determinants, probability and mathematical physics. Numerous terms in mathematics bear Cauchy’sname:- the Cauchy integral theorem in the theory of complex functions, the Cauchy-Kovalevskaya existence theoremfor the solution of partial differential equations, the Cauchy-Riemann equations and Cauchy sequences. He produced789 mathematics papers, an incredible achievement. This achievement is summed up in Belhoste’s biography ofCauchy as follows:-
... such an enormous scientific creativity is nothing less than staggering, for it presents research onall the then-known areas of mathematics ... in spite of its vastness and rich multifaceted character,Cauchy’s scientific works possess a definite unifying theme, a secret wholeness. ... Cauchy’s creativegenius found broad expression not only in his work on the foundations of real and complex analysis, areasto which his name is inextricably linked, but also in many other fields. Specifically, in this connection,we should mention his major contributions to the development of mathematical physics and to theoreticalmechanics... we mention ... his two theories of elasticity and his investigations on the theory of light,research which required that he develop whole new mathematical techniques such as Fourier transforms,diagonalisation of matrices, and the calculus of residues.
His collected works, Oeuvres compltes d’Augustin Cauchy (1882-1970), were published in 27 volumes.Political events drove Cauchy from France, and in 1831 he went to Turin. He taught there from 1832, and Luigi
Menebrea said that his courses:
were very confused, skipping suddenly from one idea to another, from one formula to the next, with noattempt to give a connection between them. His presentations were obscure clouds, illuminated fromtime to time by flashes of pure genius. ... of the thirty who enrolled with me, I was the only one to seeit through.
3Georg Friedrich Bernhard Riemann ( born on 17 Sept 1826 in Breselenz, Hanover and died on 20 July 1866 inSelasca, Italy) He showed a particular interest in mathematics and the director of the Gymnasium allowed Bernhardto study mathematics texts from his own library. On one occasion he lent Bernhard Legendre’s book on the theoryof numbers and Bernhard read the 900 page book in six days.
Stern, one of his teachers at university, certainly did realise that he had a remarkable student and later describedRiemann at this time saying that he:-
... already sang like a canary.
Riemann’s work always was based on intuitive reasoning which fell a little below the rigour required to make theconclusions watertight. However, the brilliant ideas which his works contain are so much clearer because his work isnot overly filled with lengthy computations. It was during his time at the University of Berlin that Riemann workedout his general theory of complex variables that formed the basis of some of his most important work.
Riemann’s thesis studied the theory of complex variables and, in particular, what we now call Riemann surfaces.It therefore introduced topological methods into complex function theory. The work builds on Cauchy’s foundationsof the theory of complex variables built up over many years and also on Puiseux’s ideas of branch points. However,Riemann’s thesis is a strikingly original piece of work which examined geometric properties of analytic functions,conformal mappings and the connectivity of surfaces.
Riemann was appointed to the chair of mathematics at Gottingen in 1859. A few days later he was elected to theBerlin Academy of Sciences. Riemann sent a report “On the number of primes less than a given magnitude” a workwhich changed the direction of mathematical research in a most significant way. In it Riemann examined the zetafunction
ζ(s) = Σ(1/ns) = π(1− p− s)− 1
Here the sum is over all natural numbers n while the product is over all prime numbers. Riemann considered a verydifferent question to the one Euler had considered, for he looked at the zeta function as a complex function ratherthan a real one. Except for a few trivial exceptions, the roots of z(s) all lie between 0 and 1. In the paper he statedthat the zeta function had infinitely many nontrivial roots and that it seemed probable that they all have real part1/2. This is the famous Riemann hypothesis which remains today one of the most important of the unsolved problemsof mathematics.
In the autumn of 1862 Riemann caught a heavy cold which turned to tuberculosis. He had never had good healthall his life and in fact his serious heath problems probably go back much further than this cold he caught. Riemanntried to fight the illness by going to the warmer climate of Italy. Having spent from August 1864 to October 1865 innorthern Italy, Riemann returned to G0ttingen for the winter of 1865-66, then returned to Selasca on the shores ofLake Maggiore on 16 June 1866.
15
v(x, y) = −y so we get∂u
∂x= 1 6= ∂v
∂y
anywhere.
Now we rarely use such methods for proving differentiability preferring to build the function fromsimpler examples. So we have a catch-all proposition, most of which has proofs identical to the realcase. But before that we introduce some notation. Let S ⊆ C.
H(S) = {f : f is holomorphic on S}
We use this most often when S = D(a; r) for some a ∈ C and r > 0
Proposition 3.11 Let S be an open set Let f, g ∈ H(S) be two functions .
Addition Then f ± g ∈ H(S) and (f + g)′ = f ′ + g′.
Product Then fg ∈ H(S) and (fg)′ = f ′g + fg′.
Division Let a ∈ S and assume that there exists r > 0 with f(z) 6= 0 for z ∈ D(a, r) then 1/f isdifferentiable at z = a and (1/f)′ = −f ′/f2.
Chain Rule Let h be holomorphic on f(S) then h ◦ f is holomorphic on Sand (h ◦ f)′ = (h′ ◦ f)f ′.
The proofs of these are the same as for the real line.
From this we get that all polynomials are holomorphic on C and that all rational functions p(x)q(x)
where p and q are polynomials are differentiable everywhere other than the roots of q(x).end of lecture 5
Example 3.7 Let a1, a2, . . . am be roots of the polynomial f(x) = c0 +c1x+c2x2 + · · ·+cn−1xn−1 +
xn. The we can define the matrix with m rows and n columns as the matrix of these linear equationsin the “unknowns”, c0, c1, . . . , cn−1. This is a VanderMonde determinant and so we see that m ≤ n.
We now get an interesting application of the Cauchy-Riemann equations.
Proposition 3.12 Let f ∈ H(D(0;R)) for some R > 0. Then
1. if f ′(z) = 0, ∀z ∈ D(0, R) then f is constant.
2. if |f(z)| is constant for z ∈ D(0, R) then f is constant.
Proof: For the first part we note that by the Cauchy-Riemann equations we have that
∂v
∂y− i∂u
∂y= 0
and∂u
∂x+ i
∂v
∂x= 0.
So we have that all the partial derivatives are zero on D(0, R). Now choose two points z1, z2 ∈ D(0, R)and let z1 = a+ bi and z2 = c+ di. At least one of a+ di or c+ bi are in D(0, R). Assume it is the
16
first and let w = a+di. Note that both u(x, y) and v(x, y) vary in the second variable along the linejoining z1 and w. So they can be thought of as functions of one real variable y whose derivatives are0. For the real case it follows from the mean value theorem that u(z1) = u(w) and that v(z1) = v(w).
Now consider the line joining w and z1 and note that y is constant along this line, so similarlyu(z2) = u(w) and that v(z2) = v(w). Putting this all together gives f(z1) = f(z2).
To do the second part we note that if |f(z)| is a constant then u2 + v2 is constant. So takingpartial derivatives we get
u∂u
∂x+ v
∂v
∂x= 0
andu∂u
∂y+ v
∂v
∂y= 0.
Now∂v
∂y
∂u
∂x− ∂u
∂y
∂v
∂x= 0.
Now apply the C-R equations to get
(∂u
∂x)2 + (
∂v
∂x)2 = 0
and(∂u
∂y)2 + (
∂v
∂y)2 = 0.
Hence f ′(z) = 0 for all z ∈ D(0;R) and the result follows by the first part.2
Example 3.8 Assume that we have a function f(z) holomorphic on C and satisfying f ′(z) = f(z)with f(0) = 1. Let a ∈ C and consider g(z) = f(z+a)/f(z). Then g′(z) = (f(z+a)f(z)−f(z)f(z+a))/f(z)2 = 0. So g(z) is a constant and when z = 0 we have g(0) = f(a). Thus f(z+a) = f(a)f(z).
We can introduce a large class of functions which are holomorphic on large subsets of C. Thatis power series but we had better define convergence for series.
Definition 3.13 Let z1, z2, . . . be an infinite sequence of complex numbers. Then we say that theinfinite series
∑∞1 zn is convergent of the sequence sn =
∑n1 zm is convergent.
Again we have a simple reduction to the real case.
Lemma 3.14 Let z1, z2, . . . be an infinite sequence of complex numbers. Then the series∑zn is
convergent if and only if both∑
Re zn and∑
Im zn are convergent.
Remark 3.1 Some basic facts which come from the real case.(i) If
∑|zn| is convergent we say that the series is absolutely convergent.
Definition 3.15 Let a0 + a1z + a2z2 + · · · be a power series. We define the radius of convergence
R to be the supremum, maximum of the set of real numbers r such that
∞∑0
|anrn|
is convergent. Note we have to allow R to be ∞.
17
Lemma 3.16 If R is the radius of convergence of a power series a0 + a1z + a2z2 + · · · then the
series converges for z, |z| < R.
Proof: Let z be such that |z| = r < R. Now by the definition of R there exists r < w < R with∑|anwn| convergent. But now we can use the comparison test.2
Now power series are beautifully behaved in D(0;R) where R is the radius of convergence. Inparticular they are holomorphic, which we will now prove in two steps.
Lemma 3.17 Let R is the radius of convergence of the power series∑anz
n then the series∑∞
1 nanzn−1
converges for z, |z| < R.
Proof: The simplest way to see this to see that
limn→∞
| (n+ 1)an+1zn
nanzn−1| = lim
n→∞|an+1
an||z|.
Thus this limit is greater than 1 for |z| < R. So the result holds.2
Before completing the proof that any power series is holomorphic within the radius of convergencewe prove the following lemma.
Lemma 3.18 Let∑an be an absolutely convergent series. For any ε > 0 there exists N ∈ N so
that
|∞∑N+1
an|ε.
Proof: Let sn =∑n
0 ar and s =∑∞
0 ar. For any ε there exists N ∈ N so that |sn − s| < ε/2 and|sm − s| < ε/2for m > n > N . So |sm − sn| < ε. Let m tend to infinity and the lemma follows forthis value of N .2
Theorem 3.19 Let R be the radius of convergence of a power series a0 + a1z + a2z2 + · · · . If we
let f(z) =∑anz
n for |z| < R then f ′(z) =∑nanz
n−1.
Proof: We already know that both the series converge so all we have to do is take the appropriatelimit.
The first thing to note is that it is true that
∞∑0
anzn −
∞∑0
anzn0 =
∞∑0
an(zn − zn0 ).
The proof can be seen by noting that it is true for any finite sum and then using the previous lemmato approximate to the infinite sum as accurately as we like.
What we now have to do is to calculate
limz→z0
( ∞∑n=1
anzn − zn0z − z0
− nanzn−10
)
and show that this is zero!
18
Consider ε > 0 and split the sum into two parts by choosing some N ∈ N and let
s1 =N∑n=1
anzn − zn0z − z0
− nanzn−10
and let
s2 =∞∑
n=N+1
anzn − zn0z − z0
−∞∑
n=N+1
nanzn−10 .
Now we can rewrite s2 as
s2 =∞∑
n=N+1
an
n−1∑1
zrzn−1−r0 −
∞∑n=N+1
nanzn−10 .
Now we can find r > 0 with |z| < r < R and |z0| < r < R so that the series∑nanr
n−1 is absolutelyconvergent. So we have
|s2| ≤∞∑
n=N+1
|an|nrn−1 +∞∑
n=N+1
n|an|rn−1.
But now we see that both bits of s2 can be made less than ε/3 by the previous lemma if wechoose N correctly.
The first sum is just the expression for the derivative of a polynomial so we can find d so that|s1| < ε/3 if |z − z0| < d. Thus we get the result.
Example 3.9 1.exp z = 1 + z + z2/2 + z3/3! · · · , R =∞
and exp′(z) = exp(z).
2.cos z = 1− z2/2 + z4/4!− · · · , R =∞
andsin z = z − z3/3! + z5/5! · · · , R =∞
sin(z)′ = cos z and cos(z)′ = − sin(z).
3.1
1− z= 1 + z + z2 + · · · .
We now have the following simple corollary
Corollary 3.20 Let R is the radius of convergence of a power series a0 + a1z + a2z2 + · · · . Define
the function f(z) =∑anz
n for |z| < R. Then f(z) is infinitely differentiable on D(0;R).
Remark 3.2 We have done all this based on discs about the origin. However we can change thefunction by translating through a. That is if we have a function and a power series f(w) =
∑anw
n
for |w| < R then if we put w = z − a we get a function f(w) = f(w − a) =∑an(z − a)n for
|z− a| < R = D(a;R). We could have done all or definitions for this class of series but it just makesthe proofs have even more notation.
19
Example 3.10 Consider the series
1− (z − 1) + (z − 1)2 − · · ·
Then this gives the function 1/z on D(1; 1)
To end this section we need a few comments and results which we will need later.
Definition 3.21 Let S ⊂ C. We say that S is bounded if there exists R > 0 such that |z| ≤ R forall z ∈ S or S ⊆ D(0;R).
Proposition 3.22 (Bolzano-Weierstrass) Let S be a bounded subset of C and assume that S isinfinite. Then S has a limit point.
Proof: We reduce to the real case by considering the set T = {x : x = Re z, z ∈ S}. Then T isbounded and let the sup of T be ζ. We assume that T is infinite, it is not hard to adapt the proof tothe situation when T is finite. Then for each n ∈ N there exists xn with |xn − ζ| < 1/n. Note thatlimxn = ζ. Let zn ∈ S be so that Re zn = xn. Let yn = Im zn and note that the set {yn : n ∈ N} isbounded. Let the sup of this set be η. Then for each m ∈ N there is a yn(m) with |yn(m)−η| < 1/m.Now we see that limm→∞ yn(m) = η. Thus we get that limm→∞(xn(m) + iyn(m)) = ζ + iη. Thisproves the result.2
This gives an immediate corollary.
Corollary 3.23 Any bounded sequence has a convergent subsequence.
Proof: Either the sequence has infinitely many elements so we can apply the above result or thesequence contains infinitely many identical terms. 2
Theorem 3.24 (Cauchy’s Condition) Let z1, z2, . . . be a sequence. Then the sequence convergesif and only if for all ε > 0 there exists N ∈ N so that |zm − zm| < ε for all N < n < m.
If the sequence has a limit, say w we just choose N so that |zn − w| < ε/2
So assume that we can find N so that |zm − zm| < ε for all N < n < m. In particular if weput ε = 1 we find M with |zm − zm| < 1 for all M < n < m. But now we see that the sequenceis bounded by sup{|z1|, z2|, . . . |zM |, |zM+1|+ 1}. So this has a convergent subsequence zn1 , zn2 , . . .with limit say w. Choose any ε > 0 and choose N so that both |znr −w| < ε/2 and |zm− zn|ε/2 forall m > n > n and all nr > N and the result follows.2
Corollary 3.25 The series a0 + a1 + a2 + · · · converges if and only if for all ε > 0 there existsN ∈ N so that |
∑mn ar| < ε for all N < n < m.
The important of this result is that there is no need to know what the limit is to deduce the existenceof a limit. This will turn out to be useful later.
Definition 3.26 Let S ⊂ C. S is said to be compact if it is both closed and bounded
Example 3.11 Let R > 0. Then the circle of radius R about a point z0 is compact.
Theorem 3.27 Let f : S → C be a continuous function from the compact set S. Then f is boundedon S and attains its bound on S.
20
Proof: We start by showing that the bound exists. Assume false and for each n ∈ N let zn besuch that |zn| > n. Note that as S is bounded the sequence z1, x2, . . . has a convergent subsequencew1, w2, . . . with a limit say w. Then w ∈ S as S is closed and so f(w) is defined but as f(w) =lim f(wn) as f is continuous we get a contradiction.
Now let M be the sup of |f(s)| : s ∈ S. Now for each n ∈ N choose zn so that |f(zn)| > M−1/n.Now choose a convergent subsequence and look at the limit.2
3.4 Roots and Logs–Funny Functions
Consider the function zn. We see that if we write this in polar coordinates we get (reiθ)n = rneinθ.Thus the modulus gets powered and the argument gets multiplied. It is now easy to find the rootsof unity for any given n ∈ N, for if rn = 1 then r = 1 but if nθ = 0 we get θ = 0 except that therepresentation of a point in the complex plane given by the polar coordinates is not unique. Wecan have nθ = 2mπ for any m ∈ Z. So we get n distinct roots of unity, {z0, z1, . . . , zn−1 wherezr = exp(2rπi)/n for 0 ≤ r < n. We note that these values are all distinct and that there are nvalues. Now we can see the problem with defining the root of a complex number. It is not uniquelydefined. Since for a mathematician a function has to have a unique value for each point of thedomain we have to consider objects like z1/n rather differently. This is related to the problem ofdefining the argument of a complex number. Given z we can define the modulus as +
√x2 + y2 and
this is well-defined there is no problem BUT the argument of z can have infinitely many values so
Arg z = {θ + 2rπi | r ∈ Z}.
This is really at the root of the problem. Whilst it is obvious what zn means for any n ∈ Q it isnot clear what za means for real a or even complex a. The neat answer for real numbers is
xa = ea log(x).
This is fine as log is well defined, well at least for x > 0 and since on the real numbers we cannotexpect to define x1/2 for x < 0 this is OK. So we wish to extend this to the complex plane:-
xa = ea log(x).
How are we to define log for complex values. One way might be to consider the inverse function toexp. Let us recall the definition of exp,
exp z = ex(cos(y) + i sin(y)).
Unfortunately this is not 1− 1, exp(z+ 2nπi) = exp z for all n ∈ Z, so we do not get a unique valuefor log. We define
Logz = log |z|+ iArgz.
Note that Log is a set of values, that is the set of w so that expw = z.
21
Chapter 4
Contours and Conformal Mappings
In this section we want to consider to sets of ideas which have some connection. The first is howto define curves and what are nice curves. The second are a class of maps which behave well withrespect to tangents but first we need some more definitions.
Definition 4.1 Let [a, b] be an interval in R. A curve γ is a continuous function γ : [a, b] → C.We say that γ(a) is the initial point and that γ(b) is the final or terminal point of γ. Notice thatthe curve is oriented. Further a curve is said to be(a) closed if γ(a) = γ(b) and(b) simple if for all a ≤ t1 < t2 ≤ b we have γ(t1) 6= γ(t2) implies t1 = a and t2 = b.
Thus a curve is a continuous line in C. Put in some examples.
Remark 4.1 1. Note that the curve is a function, if we want to refer to the set of points then wewrite γ∗. For example the curves given by γ1(t) = e2πit | t ∈ [0, 1] and γ2(t) = e4πit | t ∈ [0, 1]are different curves but have γ∗1 = γ∗2 .
2. We can go backwards, (−γ)(t) = γ((b+a− t). This just goes from the final point to the initialpoint.
3. If we have γ1 from [a1, b1] → C and γ2 from [a2, b2] → C we can define the join of themγ∧1 γ2 = γ from [a1, b1 + b2 − a2] by
γ(t) ={γ1(t) if t ∈ [a1, b1]γ2(t) if t ∈ [b1, b1 + b2 − a2]
4. If γ is differentiable for all t ∈ [a, b] we say that the curve is smooth.
The join will only give a continuous curve if γ1(b1) = γ2(a2). Thus we can do this for curves whichbegin and end at the same point. A similar problem occurs for smooth curves.
Definition 4.2 A path is a continuous curve which is the join of finitely many smooth curves.
A particular example of this is the following. This is given by γ2(t) = reiπt :, t ∈ [0, 1] andγ1(t) = t :, t ∈ [−1, 1]
22
r
a
2
γ1
γ
Figure 4.1: A Contour
Given a smooth curve we can ask what is the tangent at a point t0 ∈ [a, b]. It is obviouslyArgγ′(t0). Now if we have two curves crossing, let us say γ1(t) : t ∈ [a, b] and γ2(s) : s ∈ [c, d] whichmeet at z0 = γ1(t0) = γ2(s0). Then the angle between the curves is given by Argγ′1(t0)-Argγ′2(s0)
Definition 4.3 A contour is a continuous curve which is the join of finitely many smooth curveseach of which is either the chord of a circle or a straight line segment.
4.1 Conformal Maps
Definition 4.4 Let S ⊆ C. A conformal mapping is differentiable function f defined on S so thatf ′(z) 6= 0 for all z ∈ S. We say that a function f is conformal at a point z0 if there is an open discabout z0 for which f is conformal.
Proposition 4.5 Let f be conformal at a point z0. Let γ1 and γ2 be two smooth curves that meetat z0. Then the angle between the curves is the same as the angle between the images at f(z0).
Proof: Let γ1(t) = γ2(s) = z0. Then if θ is the angle between the curves
θ = Arg(γ′1(t))−Arg(γ′2(s))
Now what is the angle between the images, using the Chain Rule.
Arg(f(γ1(t))γ′1(t)))−Arg(f(γ2(s))γ′2(s)).
However this gives
Arg(f(γ1(t)) + Arg(γ′1(t))−Arg(f(γ2(s))−Arg(γ′2(s)).
And this is just θ as f(γ1(t)) = f(γ2(s))
What we would like to do is to examine confromal mappings from C to C. However many of hefunctions which behave nicely most of the time have isolated points of bad behaviour. So extendthe definition of the complex plane to C = {C∪ {∞}}. This breaks all or rules but we have to havesome idea of how to deal with ∞. I will just list them, they make sense just.
23
1. z ±∞ = ±∞ = ±∞+ z
2. z/∞ = 0,
3. z ×∞ =∞× z =∞, whenever z 6= 0
4. z/0 =∞, whenever z 6= 0 and
5. ∞+∞ =∞×∞ =∞ =∞
So we can now do arithmetic in C, though we have lost some nice algebraic properties in thisextension. The reason for this is that the extended plane is the natural place to do things. We willsee more of this later. We now have an important class of maps.
Definition 4.6 Let a, b, c and d be in C and assume that ad− bc 6= 0 then the map given by
w = f(z) =az + b
cz + d
is called a Mobius or bilinear or even fractional linear map from C to C.
Proposition 4.7 Given a, b, c and d be in C and assume that ad− bc 6= 0 then the map given by
w = f(z) =az + b
cz + d
is conformal and a bijection.
Proof: The proof is a series of computations.
f ′(z) =a(cz + d)− c(az + b)
(cz + d)2
=ad− bc
(cz + d)2
6= 0
Given z0 ∈ tcp then f(z) = z0 where z = (b − z0d)/(cz0 − a). Note that if cz0 − a = 0 thenz =∞ which is correct! So f is onto. We just have to work out what happens if f(z1) = f(z2) andthat is straightforward If
az1 + b
cz1 + d=az2 + b
cz2 + d
then(az1 + b)(cz2 + d) = (az2 + b)(cz1 + d)
Then z1(ad− bc) = z2(ad− bc) and the result follows.
Remark 4.2 (i) We can map any theree distinct points z1, z2, z3 to any other three points w1, w2, w3
by the map:-(w − w1)(w2 − w3)(w − w3)(w2 − w1)
=(z − z1)(z2 − z3)(z − z3)(z2 − w1)
(ii) It can be seen from the above that any Mobius transformation which fixes three points fixes allpoints.(iii) the Mobius transformations form a group under composition of mappings.
24
Chapter 5
The Integral
In the first part of the course you saw paths in C. In this section we describe how to integratecomplex–valued functions along paths. The assumption is that we can integrate real-valued func-tions, that is functions of the form f : [a, b]→ R. A simple extension enable us to integrate functionsf : [a, b]→ C as follows:-
Definition 5.1 Let f : [a, b]→ C be given. Then∫ b
a
f(t) dt =∫ b
a
Re (f(t)) dt+ i
∫ b
a
Im (f(t)) dt.
It is easy to give a simple example.
Example 5.1 Let f : [0, 1]→ C be given by f(t) = eiπnt, n 6= 0. Then∫ 1
0
eiπnt dt =∫ 1
0
cos(πnt) dt+ i
∫ 1
0
sin(πnt dt
= 0
when n is even. For odd values of n the integral is non–zero (see Exercise sheet).
So in this case we reduce the problem to that of integrating real functions. Now consider afunction f : C→ C. We need an extra ingredient and that is a path γ.
Definition 5.2 Let γ : [a, b] → C be a smooth cu‘rve and let f : C → C be a continuous functionγ∗ → C. Then ∫
γf(z) dz =
∫ b
a
f ((γ(t)) γ′(t) dt.
As with all definitions perhaps one should start by trying to do an example or two. But beforethat let me introduce three useful pieces of notation.
• In many situations the path γ and the function f are more important than the details of howone integrates in elementary terms. We will therefore often write
∫γf for
∫γf(z)dz.
25
• If a, b ∈ C then we denote by [a, b] = {z : z = a+(b−a)t, t ∈ [0, 1]}. This is just a complicatedbut precise way of saying the straight line joining a to b.
• Let r > 0 ∈ R and a ∈ C then γ(a; r) = {z : z = a+ reit, t ∈ [0, 2π]}. This is just the circle ofradius r about the point a.
Example 5.2 Let us calculate ∫[0,1+i]
z dz.
In this case
γ(t) = 0 + (1 + i)tγ′(t) = 1 + i
and so∫
[0,1+i]
z dz =∫ 1
0
(1− i)t(1 + i) dt
= 1.
Example 5.3 ∫γ(0;1)
zn dz ={
0 if n 6= −12πi if n = −1
In this case
γ(t) = eit
γ′(t) = ieit and so∫γ(0;1)
zn dz =∫ 2π
0
eintieit dt
=∫ 2π
0
iei(n+1)t dt
= 0 if n 6= −1= 2πi if n = −1.
One of the interesting facts about this is that integrating around a closed curve does not alwaysgive zero. You should ask what is different about n = −1.
We now consider what happens if we have a curve which is not smooth.
Definition 5.3 Let γ1, γ2 . . . , γn be smooth curves so that the end point of γj is the initial point ofγj+1 for j = 1, . . . , n− 1. Let γ be the union of these curves so that
γ∗ =n⋃1
γ∗j .
Let f be a continuous function defined on γ∗. Then∫γ
=n∑j=1
∫γjf(z) dz.
We call such a γ a path.
26
Definition 5.4 A simple path γ : [a, b] → C is such that γ(t1) = γ(t2) implies that t1 = a andt2 = b. The path is called a simple closed path when γ(a) = γ(b).
Definition 5.5 A contour is defined to be a closed path which is made up of a finite series of smoothcurves which are either circular or straight.
In the next lemma we sum up some of the basic results.
Lemma 5.6 Let γ be a path and let f1, f2, f : γ∗ → C be continuous functions. Then
1. If γ = γ1 + γ2 then ∫γf =
∫γ1
f +∫γ2
f.
2. If f = f1 + f2 then ∫γf =
∫γf1 +
∫γf2.
3. If c ∈ C then ∫γcf = c
∫γf.
4. ∫−γ
f = −∫γf
Proof:
1. Just a restatement of the definition.
2. Suppose that γ has parameter interval [a, b]. It is sufficient to verify this in the case when γ issmooth. Then∫
γ(f1(z) + f2(z))dz
=∫ b
a
(f1(γ(t)) + f2(γ(t)))γ′(t) dt
=∫ b
a
f1((γ(t))γ′(t) dt+∫ b
a
f2((γ(t))γ′(t) dt
=∫γf1 dz +
∫γf1 dz.
3. See Exercises.
4. We again assume that we have that γ is smooth. Note that −γ is given by −γ(t) = γ(a+b− t)and so −γ′(t) = −γ′(a+ b− t).Putting this all together gives∫
−γf(z) dz =
∫ b
a
f (γ(a+ b− t)) (−γ′(a+ b− t)) dt.
27
The integral on the right hand side is essentially a real integral so we substitute s = a+ b− tto get ∫
−γf(z) dz =
∫ a
s=b
f (γ(s)) (−γ′(s))− ds
= −∫ b
a
f(γ(s))γ′(s)ds
= −∫γf(z) dz
2 We now come to a key early result which plays a less important role in this theory than it doesin the case of real variables.
Theorem 5.7 (Fundamental Theorem of Calculus) Suppose that γ : [a, b]→ C is a path andF : G→ C is a function defined on some open set G with γ∗ ⊂ G. Further assume that F ′(z) existsand is continuous ∀z ∈ G. Then ∫
γF ′(z) dz = F (γ(b))− F (γ(a)).
Note that if γ is closed then∫γ F′(z) dz = 0.
Proof: We may assume that γ is smooth. Then we have∫γF ′(z) dz =
∫ b
a
F ′(γ(t))γ′(t)dt.
Now consider the function F ◦ γ(t). This is a function from R to C and this has a derivative(F ◦γ)′(t) = F ′(γ(t))γ′(t). (This can be shown by either using the chain rule or from first principles).Then ∫
γF ′(z) dz =
∫ b
a
(Re (F ◦ γ)′(t) dt+∫ b
a
(Im (F ◦ γ)′(t) dt
= Re (F (γ(b)))− Re (F (γ(a)) + i (Im (F (γ(b)))− Im (F (γ(a))) .= F (γ(b))− F (γ(a)).
2 Notice that the justification depends on the equivalent theorem for the real case.
Example 5.4 Let γ be the curve shown in Figure 5.1. Consider the integral∫γ
cos z dz.
Note that γ = γ1 + γ2 where
γ1 : [0, 1]→ γ1(t) = 2t− 1
γ2 : [0, π]→ γ2(t) = eit.
The crude way is just to substitute in the formula and get:-∫γ
cos z dz =∫ 1
0
2 cos(2t− 1) dt+∫ π
0
cos(eit)ieit dt.
28
�
�
�����
Figure 5.1: The contour γ
Now if you are really good perhaps you can do this but to be honest I can’t do it. But now we canuse the Fundamental Theorem. Just remember that sin′ z = cos z. So we get∫
γcos z dz = 0.
Example 5.5 Let n 6= −1 and f(z) = zn, z 6= 0. In these circumstances we can use the Fun-damental theorem because we have a well-behaved function F (z) = zn+1
n+1 as long as the contourdoesn’t include the origin. Notice that we may have to take as our region G something likeG = {z : r < |z| < R} for two real numbers 0 < r < R.
������������ ��������
�������������! �"$#%#�&�'�(*)+',�.-0/1
2�3�4�5�687$9�:!;�<$=%=�>�?A@CB+?ED�7�F0GIHJLKIMON.KPJRQTSIUOVXWZY�[�\]U Q_^�`ba*cTdfe�K�\�g�\
h
i�j�k�l�m�nobp!q+r�s%s�t�u�v*w+u�x.y
Figure 5.2: Choosing one branch of the many–valued function log
We now come to another powerful result which again depends on properties of the real integral.
29
Theorem 5.8 (Estimation Theorem) Let γ : [a, b] → C be a curve and let f be a continuousfunction defined on γ∗. Then
|∫γf(z) dz| ≤
∫ b
a
|f(γ(t))γ′(t)| dt.
Note that the theorem does not say that
|∫γf(z) dz| =
∫γ|f(z)| dz.
Try to find an example where this fails.Proof: This is in the book, result 3.9.2
Corollary 5.9 Let γ : [a, b]→ C be a path and let f be a continuous function defined on γ∗. Assumethat 0 < M ∈ R is such that |f(z)| ≤M,∀z ∈ γ∗. Then
|∫γf(z) dz| ≤M
∫ b
a
|γ′(t)| dt.
This is a straightforward deduction from the Estimation Theorem. Note that the integral∫ b
a
|γ′(t)| dt
is just the length of the curve. This is a very useful result as you will see later.
Example 5.6 Let
f(z) =1
z − 1and let γ = γ(0;R), R > 1. The key point here is to estimate the value of z − 1 on this circle. Weuse the inequality |a+ b| ≥ ||a| − |b|| to get |z − 1| ≥ R− 1 and so
|f(z)| ≤ 1R− 1
for z ∈ γ∗. Thus
|∫γf(z) dz| ≤ 2πR
R− 1.
This very quickly gives us an important criteria for deciding when it is possible to exchange aninfinite sum and an integral.
Theorem 5.10 Let γ : [a, b] → C be a path. Let U, u0, u1, u2 . . . be continuous functions definedon γ∗. Assume that
1.∞∑k=0
uk(z) = U(z) ∀z ∈ γ∗.
2. ∃Mk for k = 0, 1, 2 . . . such that∑MK converges and |uk(z)| ≤Mk for k = 0, 1, 2, . . . for all
z ∈ γ∗.
30
Then∞∑0
∫γuk(z) dz =
∫γ
∞∑0
uk(z) dz =∫γU(z) dz.
Proof: The main ingredient in the proof is the Estimation Theorem. However we first point outthat because |uk(z)| ≤ Mk and
∑Mk is convergent then
∑|Mk| is convergent. Let l > 0 be the
length of the contour. Then we have
|∫γuk(z) dz| ≤Mkl.
Som∑n
|∫γuk(z) dz| ≤
m∑n
Mkl
for all n < m ∈ N. We can now use Cauchy’s criterion for convergence.
In its sequence form it says that a sequence u0, u1, . . . is convergent if and only if ∀ε > 0 ∃N ∈N such that |un − um| < ε ∀, N < n < m. In its series form it says that a series
∑an is convergent
if and only if ∀ε > 0 ∃N ∈ N such that
|m∑k=n
ak| < ε, ∀N < n < m.
Let ε > 0 be given. Then ∃N ∈ N such that
m∑k=n
Mk <ε
l, ∀N < n < m.
So we getm∑n
|∫γuk(z) dz| ≤
m∑n
Mkl < ε ∀N < n < m.
This proves the convergence of the series. A little more thought is needed to show the last line: bytaking m→∞ above we see that∣∣∣∣∣
∞∑0
∫uk −
∫ N∑0
uk
∣∣∣∣∣ =∣∣∣∣∑N + 1∞
∫uk
∣∣∣∣ ≤ ε,showing that the sum of the integrals converges to the integral of the sum. 2
Example 5.7 Consider the following
∞∑0
zn =1
1− z.
Let w ∈ C be so that |w| = r < 1. Let γ = [0, w]. Then |zn| ≤ rn forall z ∈ γ∗. But∑rn is
convergent so we can apply the previous result and the Fundamental theorem gives
∞∑1
wn
n=∫
[0,w]
11− z
dz.
31
Chapter 6
Topology
The theory of complex variables has two sets of difficulties to contend with. The first is that ofanalysis, ie the problems of epsilons and deltas and infinities. However because it all lies in theplane there are also geometrical or topological difficulties. Perhaps one of the simplest reasonsis that there are infinitely many different directions. In some ways that is why continuity anddifferentiation is a stronger condition than for real variables.
Example 6.1 Draw the following sets
1. {z : |z| < 1} ∪ {|z| > 1}
2. {z : |z| ≤ 1} ∪ {|z| > 1}
3. {z : |z| < 1} ∪ {|z − 2| < 1}
4. {z : |z| < 1} ∪ {|z − 1| < 1}
5. {z : |z| ≤ 1} ∪ {|z − 1| ≤ 1}
In order to organise our thoughts more clearly we wish to define the the types of subsets whichare well behaved. As always there are more than one way to make our definitions so we choose oneand either show equivalence or leave it vague as to the connection.
Definition 6.1 Let G ⊆ C. Then we say that G is path connected if for any two points z and wboth in G there is a path γ : [a, b]→ C such that γ∗ ⊂ G with γ(a) = z and γ(b) = w.
Example 6.2 Consider the subset of R2 given by {(x, y) : x > 0, y = sin(1/x)} ∪ {(x, y) : x = 0}.In more advanced books on topology it is shown that this set is a connected subset of C but is notpath–connected.
We will just say that a subset is connected, dropping the reference to path, on the understandingthat none of our sets will be so pathological.
Definition 6.2 A region is an open connected subset of C.
A key result that we need is quite complicated but if we think straight is not too bad.
32
Theorem 6.3 (Covering Theorem) Let G be an open subset of C and let γ : [a, b] → C be acurve such that γ∗ ⊂ G. Then there exists a subdivision of [a, b], a = t0 < t1 < · · · < tn = b andopen discs D0, D1, . . . Dn−1 such that
1. Dj ⊂ G, ∀j, and
2. γ(t) ∈ Dj ∀t ∈ [tj , tj+1], 0 ≤ j < n.
Proof: Before we begin the proof proper a little piece of notation. Let x ∈ [a, b]. Then define γx tobe the curve obtained by restricting γ to [a, x]. The idea of the proof is to show that for all x ∈ [a, b]the theorem is satisfied for γx. Define C = {x : x ∈ [a, b] the theorem holds for γx}. We now showthat C = [a, b].
Step 1. The first thing to do is to show that a ∈ C.Recall that γ(a) ∈ G but that G is open. Thus there is a disc with centre a contained in G. So nobig deal.
Step 2. There is a real c which is an upper bound for C.Clearly C is bounded above by b. Also by Step 1, C is non-empty so C has a least upper bound sayc.
Step 3. c ∈ C.Note that c ∈ G so ∃δ > 0 so that D(γ(c); δ) ⊆ G. Also we should note that c ≤ b but that if c = bwe have completed the proof, so assume that c < b. But γ is continuous. So we have ε > 0 suchthat |γ(t)− γ(c)| < δ, ∀t with |t− c| < ε. Now we know that c is the least upper bound of C so thatthere is a d ∈ C with a ≤ d ≤ c and |d− c| < ε. Let γd be covered by discs D0, D1, . . . Dm. Now wecan add a new disc Dm+1 = D(γ(c), δ). But now we have covered γc and so c ∈ C.
Step 4 c = b.Assume that c < b. By Step 3 we have that existence of a particular ε and δ. But as above choosesome d with c < d < b and |d− c| < ε. This is a contradiction which completes the proof.2
Corollary 6.4 Let G ⊆ C be a region and f : G→ C be a function differentiable at all points of G.Suppose that f ′(z) = 0, z ∈ G. Then f is constant on G.
Proof: We have shown that the result is true if G is an open disc in the first part of the course.Let u, v ∈ G and let γ be a path that connects them in G. By the Covering Theorem we can coverγ∗ with open discs, any adjacent pair overlap. So we get f(u) = f(v). 2
Note that we need connectedness. We now introduce a further class of sets which are morerestricted but have some very good properties.
Definition 6.5 A set G ⊆ C is called convex if any two points in G can be joined by a straight linesegment in G.
There are a number of results which look sort of obvious which when you try to prove them gettrickier. The first is always called Jordan’s Curve Theorem1 but it was not Jordan who first provedit: he realised that it was important.
1Marie Ennemond Camille Jordan (born 5 Jan 1838 in Lyon, died 22 Jan 1922 in Milan) was highly regarded byhis contemporaries for work in algebra and group theory.
Jordan studied mathematics at the Ecole Polytechnique and from 1873 taught there and at the College de France.He introduced important topological concepts in 1866. He was aware of Riemann’s work in topology but did not
know of the work by Mobius. Jordan introduced the notion of homotopy of paths, looking at deformation of pathsone into the other. He defined a homotopy group of a surface without explicitly using group terminology.
33
Theorem 6.6 (Jordan’s Curve Theorem) Let γ be a simple closed path, ie no self intersectionsexcept for the end points. Then there is an inside and an outside. That is there are two open setsI, O with C \ γ∗ = I ∪O. I is bounded and is the inside and O is unbounded and the outside.
To appreciate this theorem, it is important to understand just how complicated a closed simplepath can be, and that if we allow non–simple closed paths the result can fail in a spectacular way.
Figure 6.1: A closed simple curve
An easier result than Theorem 6.6 is the following.
Theorem 6.7 Any polygon can be cut up into triangles.
The only tricky case is when the polygon is not convex. The next result is not stated too precisely.
Theorem 6.8 Let γ be a path. Then it can be approximated as closely as wished by a polygonalpath.
Jordan was particularly interested in the theory of finite groups. His introduction to the group concept in geometry(1869) was motivated by studies of crystal structure. In this work he considers classification of groups of Euclideanmotions.
His “Traite des substitutions et des equations algebraique” in 1870 gave a comprehensive study of Galois theory.For this work he was awarded the Poncelet Prize of the Academie des Science. The work contains the “Jordan normalform” for matrices, not over the complex numbers but over a finite field. He appears not to have known of earlierresults of this type by Weierstrass. It also brings permutation groups to a central role in mathematics.
Jordan is best remembered today for his proof that a simply closed curve divides a plane into exactly two regions.He originated the concept of functions of bounded variation and is known especially for his definition of the lengthof a curve. This appears in his “Cours d’analyse de l’Ecole Polytechnique” (3 volumes) (1882), the Jordan curvetheorem appearing in the 3rd edition (1909-1915).
He also generalised the criteria for convergence of Fourier series.Two of Jordan’s students, Sophus Lie and Felix Klein, drew upon his studies to produce their own theories of
continuous and discontinuous groups.
34
Chapter 7
Cauchy’s Theorem
We begin by stating Cauchy’s theorem in some generality. Then we will prove it in a special caseand then describe it in full generality. The difficulties in the general case are due to the topologicalproblems rather than the analytic ones. Hopefully I will be able to illustrate these ideas as wedevelop this material. During this section we are going to have one underlying hypothesis:-
Hypothesis 7.1 Let G be some region of C and let γ be a closed path so that γ∗ ⊂ G. Let f be afunction which is differentiable on some open subset of G which contains γ∗ and the inside of γ.
The key topological difficulty is
What is the inside?
The main theorem should be:-
Theorem 7.2 (Cauchy’s Theorem for a path) Let G, f, γ satisfy Hypothesis 7.1. Then∫γf(z) dz = 0.
Cauchy originally proved it, I think, by an appeal to Stokes Theorem. However that demands extraconditions and still needs the topology although this is nearly always ignored in most analyses.
I will begin by proving it for the special case when the path is a triangle. This special casecontains nearly all the analytical complexity.
Theorem 7.3 (Cauchy’s Theorem for a triangle) Let G, f, γ satisfy Hypothesis 7.1. Furtherassume that γ∗ is a triangle. Then ∫
γf(z) dz = 0.
Proof: Let the triangle γ0 have vertices p, q and r and denote any such triangular path by [p, q, r].Let
I =∫γf(z) dz.
35
p a q
bc
r
Figure 7.1: The initial triangle [p, q, r]
We are trying to show that I = 0. In the picture below let a, b and c be the midpoints a of [p, q], bof [q, r] and c of [q, r] as shown in Figure 7.1.
We can then consider the four triangular paths to obtain
I =∫
[a,q,b]
f(z) dz +∫
[b,r,c]
f(z) dz +∫
[c,p,a]
f(z) dz +∫
[a,b,c]
f(z) dz
LetI1 =
∫[a,q,b]
f(z) dz, I2 =∫
[b,r,c]
f(z) dz, I3 =∫
[c,p,a]
f(z) dz
andI4 =
∫[a,b,c]
f(z) dz.
A simple application of the triangle inequality gives
|I| ≤4∑j=1
|Ij |
Thus for at least one j we have that |Ij | ≥ |I|/4. Let this triangle give rise to the contour γ1. Notethat
|∫γ1
f(z) dz| ≥ 14|∫γ0
f(z) dz|. (7.1)
Step 1:Repeat this to construct a whole series of triangular paths γ0, γ1, . . . γn, . . . such that
|∫γnf(z) dz| ≥ 1
4|∫γn−1
f(z) dz|,
|∫γnf(z) dz| ≥ 1
4n|∫γ0
f(z) dz|
36
and I(γ∗n−1) ⊃ I(γ∗n), where I(δ) denotes the closure of the interior of a path δ.
Step 2:Claim that
⋂I(γn) 6= ∅. To prove this consider for each triangle I(γn) the centrum1 zn. Let the
length γ0 be l. Then the length of γn is l2n . Note that any two points u, v ∈ I(γn) satisfy
|u− v| ≤ l
2n(7.2)
From this we can see that|zn+1 − zn| ≤
l
2n.
Thus the sequence is a Cauchy sequence and has a limit, say w.
Step 3:We use the differentiability of f at w to estimate the integrals for large enough n. Recalling thedefinition of differentiability we see that for ε > 0 there exists r > 0 such that for all z ∈ D′(w; r)
|f(z)− f(w)z − w
− f ′(w)| < ε.
Rewritten as|f(z)− f(w)− (z − w)f ′(w)| < ε|(z − w)| ∀z ∈ D′(w; r).
So now we can choose N so that γN ⊂ D(w; r) and from 7.2 |u− v| ≤ l2n . Now comes the hard
bit of analysis. ∫γN
f(z) dz =∫γN
(f(z)− f(w)− (z − w)f ′(w)) dz
+∫γN
(f(w) + (z − w)f ′(w)) dz.
But
f(w) + (z − w)f ′(w) =d
dz
(f(w)z +
(z − w)2
2f ′(w)
)By the Fundamental Theorem∫
γNf(z) dz =
∫γN
(f(z)− f(w)− (z − w)f ′(w)) dz and so
|∫γN
f(z) dz| = |∫γN
(f(z)− f(w)− (z − w)f ′(w)) , dz|
From the Estimation Theorem
|∫γN
f(z) dz| ≤ length(γN ) supz∈γ∗N
|(f(z)− f(w)− (z − w)f ′(w)|
≤ l
2Nε supz∈γ∗N
|z − w|
≤ l2
4Nε
Hence from earlier
|I| ≤ 4N |∫γN
f(z) dz|
≤ l2ε.1The intersection of the three lines joining a corner with the mid–point of the opposite side.
37
Thus we get the required result by letting ε get smaller. 2
From this, which is probably the hardest proof you have seen, we can get some quite interestingresults.
Theorem 7.4 Let G be a convex region. Let f : G → C be continuous and assume that for alltriangular contours γ ∫
γf(z) dz = 0.
Choose a point u ∈ G. Define a function F : G→ C by
F (z) =∫
[u,z]
f(w) dw.
Then F ′(z) = f(z) for all z ∈ G.
Proof: Because G is convex we have that [u, z] ⊂ G, ∀z ∈ G. Thus the definition makes senseand we are in a hopeful situation. Now we need to use the hypothesis of the Theorem. The idea isto consider
F (z + h)− F (z)h
for h 6= 0 so that (z + h) ∈ G. By the hypotheses of the Theorem
0 =∫
[u,z]
+∫
[z,z+h]
+∫
[z+h,u]
(f(w) dw) .
Rewriting shows that
F (z + h)− F (z) =∫
[z,z+h]
f(w) dw
and hence thatF (z + h)− F (z)
h=
1h
∫[z,z+h]
f(w) dw.
To prove the theorem we need to show:-
limh→0
(F (z + h)− F (z)
h
)= f(z).
Once again rewriting we need to show
limh→0
1h
∫[z,z+h]
f(w) dw = f(z)
ielimh→0
1h
∫[z,z+h]
f(w)− f(z) dw = 0.
Choose an ε > 0. At this point we use the continuity of f to find a D′(z; r), r > 0 so that|f(z + h)− f(z)| < ε, ∀h ∈ D′(0; r). So by the estimation theorem we get
| 1h
∫[z,z+h]
f(w)− f(z) dw| ≤ 1|h||h|ε.
The result follows immediately. 2
We get two immediate corollaries from this.
38
Corollary 7.5 (The indefinite integral theorem) Let G be a convex region and let f ∈ H(G).Let u ∈ G and define
F (z) =∫
[u,z]
f(w) dw.
Then F : G→ C is an antiderivative of f .
Corollary 7.6 (Cauchy’s theorem for a convex region) Let G, f and γ satisfy Hypothesis 7.1with G convex. Then ∫
γf(w) dw = 0.
Proof: This follows from the previous corollary and the Fundamental Theorem of Calculus. 2
Let us recall that a contour is a path made up of a finite set of straight line segments or ofsegments of circles. I will prove a special case of Theorem 6.8 stated earlier.
Lemma 7.7 Let G, f and γ satisfy Hypothesis 7.1 and let γ be a contour. Then there is a polygonalpath δ : [0, 1]→ G such that ∫
γf(z) dz =
∫δ
f(z) dz.
Proof: It is at this point that we use the covering theorem. So there exist overlapping discsD0, D1, . . . , Dn with Di ⊂ G and γ∗ ⊂ ∪Di. Inside each disc pick some points on γ∗ and join themwith straight line, making certain that there is at least one point in each Di∩Di+1, i = 0, . . . , n−1.Since each disc is a convex set we can apply Cauchy’s theorem for convex sets to get that the integralis the same along each straight segment as along the contour. Adding all the bits gives the result.2
We can now use this to obtain
Theorem 7.8 (Cauchy’s Theorem for contours) Let G, f and γ satisfy Hypothesis 7.1 andlet γ be a contour.Then ∫
γf(w) dw = 0.
Proof: By the previous lemma we can assume that γ is polygonal. By Theorem 6.7 we can splitany polygon into triangles ∆i. But then∫
γf(w) dw =
∑∫∆i
f(w) dw.
But then ∫∆i
f(w) dw = 0
for all i by Cauchy’s theorem for triangles and the theorem follows. 2
This is the version of the theorem we use, although we rarely have more than two circular andthree or four straight segments. The proof of the main result is similar but we do have to checksome of the more subtle bits rather more carefully than we have done.
Definition 7.9 A path γ : [a, b] → C is said to positively oriented if for some u in the interior ofγ, γ(t) traces out a counter-clockwise path about u as t goes from a to b.
39
This leads to a rather useful result
Theorem 7.10 (Deformation Theorem) Let γ be a positively oriented contour such thatD(u; r) ⊂I(γ). Further assume that f is holomorphic inside and on γ∗ except possibly at u. Then∫
γf(z) dz =
∫γ(u;r)
f(z) dz.
Proof: The idea is to connect the circle to the contour with a two lines as shown in Figure7.2. Pick two points w1, w2 both on γ(u; r)∗ and let z1, z2 be on γ∗ such that [w1, z1] ∪ [w2, z2] ⊂(I(γ) \D(u; r).
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�
�� ��������
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��� �!
Figure 7.2: Cutting the contour γ
Now we can split γ up into
• γ1 from z1 to z2 and
• γ2 from z2 to z1.
This is done so that both are still oriented the correct way. Now do the same to γ(u : r)
• δ1 from w1 to w2 and
• δ2 from w2 to w1.
Now we have two contoursα1 = γ1 + [z2, w2] + (−δ1) + [w1, z1]
andα2 = γ2 + [z1, w1] + (−δ2) + [w2, z2].
It is clear that f is well behaved on and inside both α1 and α2. So by Cauchy’s theorem for contourswe have ∫
αi
f(z) dz = 0, i = 1, 2.
40
But
0 =∫α1
f(z) dz +∫α2
f(z) dz
=∫γ1
f(z) dz +∫γ2
f(z) dz −∫δ1
f(z) dz −∫δ2
f(z) dz
=∫γf(z) dz −
∫γ(u;r)
f(z) dz
and so∫γf(z) dz =
∫γ(u;r)
f(z) dz.
2
This really does make life a lot simpler: we have done the integral despite the fact f could bebadly behaved at u. Together with Cauchy’s Theorem, what this means is that the value of anyintegral around a closed path is determined only by the points where the function is not differentiable,together with information about whether those points lie inside or outside the path.
Example 7.1 Let γ be a positively oriented contour with 0 /∈ γ∗. Then∫γ
1zdz =
{2πi if 0 ∈ I(γ∗),0 if 0 ∈ O(γ∗).
Example 7.2 Let γ be any contour with 0 /∈ γ∗, 0 ∈ I(γ). Then∫γ
ez
zdz =
∫γ(0;1)
ez
zdz.
41
Chapter 8
Some consequences of Cauchy’sTheorem
Theorem 8.1 (Cauchy’s Integral Theorem) Let γ be a positively oriented contour and let fbe holomorphic inside and on γ∗. Let w 6∈ γ∗. Then
12πi
∫γ
f(z)z − w
dz ={
0 if w ∈ O(γ)f(w) if w ∈ I(γ)
Proof: We can find D(w; r) ⊂ I(γ). Then by Theorem 7.10
12πi
∫γ
f(z)z − w
dz =1
2πi
∫γ(w;r)
f(z)z − w
dz.
If w /∈ I(γ) it is clear that f(z)z−w is holomorphic both on and inside γ. Thus the integral is 0.
We have fairly easily that ∫γ(w;r)
1z − w
dz = 2πi
and so ∫γ(w;r)
f(w)z − w
dz = 2πif(w).
This gives us
| 12πi
∫γ(w;r)
f(z)z − w
dz − f(w)| =1
2π|∫γ(w;r)
f(z)− f(w)z − w
dz|
=1
2π|∫γ(w;r)
(f(z)− f(w)
z − w− f ′(w)
)dz|
≤ 12π
2πr supz∈γ(w;r)∗
|f(z)− f(w)z − w
− f ′(w)|
Since f is holomorphic the last expression above tends to zero as r tends to zero. The expression isindependent of r, so we have completed the proof. 2
All sorts of results follow very neatly from this.
42
Example 8.1 1. Let f be holomorphic in D(0; r) for any r > 1. Then∫γ(0;1)
f(z)z−0 dz = 2πif(0).
2. ∫γ(0;1)
ez
zdz = 2πi.
3. ∫γ(0:1)
z3 − 1z
dz = −2πi.
4. ∫γ(0:1)
sin zz
dz = 0.
Example 8.2 An interesting set of examples is based on partial fractions. We note that
1z2 + 4
=1
(z − 2i)(z + 2i)=
14i
(1
z − 2i− 1z + 2i
).
Thus ∫γ
cos(z)z2 + 4
dz =14i
(∫γ
cos(z)z − 2i
dz −∫γ
cos(z)z + 2i
)dz
=π
2(cos(2i)− cos(−2i))
= 0
Note we have to be slightly careful about γ.
The next result is due to Liouville1.1Joseph Liouville (born 24 March 1809 in Saint-Omer, died 8 Sept 1882 in Paris) is best known for his work on
the existence of a transcendental number in 1844 when he constructed an infinite class of such numbers.Liouville’s mathematical work was extremely wide ranging, from mathematical physics to astronomy to pure math-
ematics. One of the first topics he studied, which developed from his early work on electromagnetism, was a newtopic, now called the fractional calculus. He defined differential operators of arbitrary order (this order is usually aninteger but in the theory developed by Liouville in papers between 1832 and 1837, it could be a rational, an irrationalor most generally of all a complex number).
Liouville investigated criteria for integrals of algebraic functions to be algebraic during the period 1832-33. Havingestablished this in four papers, Liouville went on to investigate the general problem of integration of algebraic functionsin finite terms. His work at first was independent of that of Abel, but later he learnt of Abel’s work and includedseveral ideas into his own work.
Another important area which Liouville is remembered for today is that of transcendental numbers. Liouville’sinterest in this stemmed from reading a correspondence between Goldbach and Daniel Bernoulli. Liouville certainlyaimed to prove that e is transcendental but he did not succeed. However, his contributions were great and led him toprove the existence of a transcendental number in 1844 when he constructed an infinite class of such numbers usingcontinued fractions. In 1851 he published results on transcendental numbers removing the dependence on continuedfractions. In particular he gave an example of a transcendental number, the number now named the Liouvilliannumber
0.1100010000000000000000010000...
where there is a 1 in place n! for each n ≥ 1 and 0 elsewhere.His work on boundary value problems on differential equations is remembered because of what is called today
Sturm-Liouville theory which is used in solving integral equations. This theory, which has major importance inmathematical physics, was developed between 1829 and 1837. Sturm and Liouville examined general linear secondorder differential equations and examined properties of their eigenvalues, the behaviour of the eigenfunctions and theseries expansion of arbitrary functions in terms of these eigenfunctions.
43
Theorem 8.2 (Liouville’s Theorem) Let f be holomorphic and bounded in C. Then f is con-stant.
Proof: The assumptions imply that there exists M > 0 such that |f(w)| < M, ∀w ∈ C.Choosetwo points a, b ∈ C and let r > 2 max{|a|, |b|} and let γ = γ(0; r). Then |z − a| > r/2 and|z − b| > r/2 ∀z ∈ γ∗. So we have
f(a)− f(b) =1
2πi
∫γf(z)
(1
z − a+
1z − b
)dz
and so
|f(a)− f(b)| ≤ 12π
2πrM2|a− b|
r2
2
≤ 4M |a− b|r
So we can see that as r tends to infinity we get f(a) = f(b). 2
Corollary 8.3 (Fundamental Theorem of Algebra) Every non-constant polynomial over C hasa root in C.
Proof: Assume that the result is false. Then there is a non–constant polynomial f(z) suchthat f(z) 6= 0, ∀z ∈ C. Note that f is holomorphic on the whole plane. Since f has no zeros thefunction g(z) = 1/f(z) is also holomorphic on the plane. We can apply Liouville’s theorem to g aslong as we can show that g is bounded. The only way that g could be unbounded would be if therewas a sequence z1, z2, . . . with limn→∞ f(zn) = 0. If the sequence is bounded then it contains aconvergent subsequence with limit w say. Then, as f is continuous, f(w) = 0 which is false.
If the sequence is unbounded then we have |zn| → ∞ with f(zn) → 0. Write f = anzn +
an−1zn−1 + . . . a0 where an 6= 0. But then
|f(z)| ≥ |an||z|n −n−1∑
0
|aj ||zj |.
So we can see that |f(z)| → ∞ as n → ∞. Thus we can apply Louiville’s theorem to get that g isconstant and hence f is constant. 2
There is a slightly different proof of this in Priestly using the notion of compactness.
One of the difficult things in real variables is that the existence of all possible differentials andthe convergence of the Taylor series 2 does not guarantee that the Taylor series will converge to the
Liouville contributed to differential geometry studying conformal transformations. He proved a major theoremconcerning the measure preserving property of Hamiltonian dynamics (the existence of the “Liouville measure” inergodic theory and dynamical systems). The result is of fundamental importance in statistical mechanics and measuretheory.
In 1842 Liouville began to read Galois’ unpublished papers. In September of 1843 he announced to the Academy thathe had found deep results in Galois’ work and promised to publish Galois’ papers together with his own commentary.Liouville was therefore a major influence in bringing Galois’ work to general notice when he published this work in1846 in his Journal. However he had waited three years before publishing the papers and, rather strangely, he neverpublished his commentary although he certainly wrote a commentary which filled in the gaps in some of the proofs.Liouville also lectured on Galois’ work and Serret, possibly together with Bertrand and Hermite, attended the course.
In number theory Liouville wrote around 200 papers, working on quadratic reciprocity and many other topics. Hewrote over 400 papers in total.
2Brook Taylor (born 18 Aug 1685 in Edmonton, Middlesex, died 29 Dec 1731 in London).
44
function. A tedious calculation will show that the Taylor series of the function
f(x) ={e1/x2
when x 6= 00 when x = 0
is given by0 + 0x+ 0x2 + . . . ,
which certainly converges everywhere but does not converge to f for any value of x 6= 0.
The situation is very much better for complex function. The incredible fact is that being ableto differentiate once implies the existence of a Taylor’s series which itself implies the existence of alldifferentials.
Proposition 8.4 Let f be holomorphic on the open disc D(u;R). Then there exist constants cn ∈C, n = 0, 1, . . . such that for any z ∈ D(u;R) we have
f(z) =∞∑n=0
cn(z − u)n.
Further we have
cn =1
2πi
∫γ(u;r)
f(w)(w − u)n+1
dw,
where 0 < r < R.
Proof: Choose z ∈ D(u;R) and choose r with |z − u| < r < R. Use the integral formulaTheorem 8.1 to get
2πif(z) =∫γ(u;r)
f(w)w − z
dw
=∫γ(u;r)
f(w)(w − u)− (z − u)
dw
=∫γ(u;r)
f(w)w − u
(1
1− z−uw−u
)dw
The point of this is that v = z−uw−u has modulus less than 1 for w ∈ γ∗(u; r) and so we get
11− v
= 1 + v + v2 + . . .+ vn +vn+1
1− vIn 1708 Taylor produced a solution to the problem of the centre of oscillation which, since it went unpublished until
1714, resulted in a priority dispute with Johann Bernoulli.Taylor’s “Methodus incrementorum directa et inversa” (1715) added to mathematics a new branch now called the
“calculus of finite differences” and he invented integration by parts. It also contained the celebrated formula knownas Taylor’s expansion, the importance of which remained unrecognised until 1772 when Lagrange proclaimed it thebasic principle of the differential calculus.
Taylor also devised the basic principles of perspective in “Linear Perspective” (1715). Together with new principlesof linear perspective the first general treatment of the vanishing points are given.
Taylor gives an account of an experiment to discover the law of magnetic attraction (1715) and an improved methodfor approximating the roots of an equation by giving a new method for computing logarithms (1717).
Taylor was elected a Fellow of the Royal Society in 1712 and was appointed in that year to the committee foradjudicating the claims of Newton and of Leibniz to have invented the calculus.
45
So we get
2πif(z) =n∑k=0
∫γ(u;r)
f(w)(z − u)k
(w − u)k+1dw +An
where
An =∫γ(u;r)
f(w)(z − u)n+1
(w − u)n+1(w − z)dw
and so
2πif(z) =n∑k=0
(∫γ(u;r)
f(w)(w − u)k+1
dw
)(z − u)k +An
2
Our objective is now to show that limn→∞An = 0, which will complete the first part of thetheorem. Recall that f is continuous on the closed bounded set γ∗(u; r) so that there exists M > 0with |f(w)| ≤M, ∀w ∈ γ∗(u; r). We now use the estimation theorem to obtain
|An| ≤ 2πrM supw∈γ∗(u;r)
| z − uw − u
|n+1 1|w − z|
= 2πrM
(sup
w∈γ∗(u;r)
1|w − z|
)|v|n+1
Since |v| < 1 we get the result that we want. Note that the choice of r depended on z but by theDeformation Theorem 7.10 this doesn’t matter so
cn =∫γ(u;r)
f(w)(w − u)n+1
dw
is well defined and not dependent on r.
Theorem 8.5 (Taylor’s Theorem) Let f ∈ H(D(u;R)) then
1. There exists constants c0, c1, . . . ∈ C such that
f(z) =∞∑n=0
cn(z − u)n,∀z ∈ D(u; r).
2.
cn =1
2πi
∫γ(u;r)
f(w)(w − u)n+1
dw,
for any r with 0 < r < R.
3.
f (n)(u) = n!cn
=n!
2πi
∫γ(u;r)
f(w)(w − u)n+1
dw
Proof: The first two follow from the previous Proposition. The final result follows from the resultabout the differentiation of convergent series. 2
46
Corollary 8.6 Let G be an open set and f ∈ H(G). Then f ′ ∈ H(G).
Proof: Let u ∈ G and choose R > 0 such that D(u;R) ⊂ G. Then by Taylor’s Theorem 8.5 ∃cjsuch that
f(z) =∞∑n=0
cn(z − u)n
for z ∈ D(u;R). But then
f ′(z) =∞∑n=0
ncn(z − u)n−1
for z ∈ D(u;R). This is again a convergent power series and so in particular f ′ is differentiable atu.2
We now come to another theorem of Cauchy.
Theorem 8.7 (Cauchy’s Formula for derivatives) Let γ be a positively oriented contour andf be a function which is holomorphic inside and on γ. Let u 6∈ γ∗. Then
n!2πi
∫γ
f(w)(w − u)n+1
dw ={
0 if u ∈ O(γ)f (n)(u) if u ∈ I(γ)
Proof: If u ∈ O(γ) the result follows from Cauchy’s Theorem for a contour.
Now suppose u ∈ I(γ). Then there exists R > 0 such that D(u;R) ⊆ I(γ). If 0 < r < R thenusing Taylor’s Theorem 8.5 give
f (n)(u) =n!
2πi
∫γ(u;r)
f(w)(w − u)n+1
dw.
By the Deformation Theorem 7.10 we get
f (n)(u) =n!
2πi
∫γ
f(w)(w − u)n+1
dw.
2
Example 8.3 Let γ = γ(0; 1). Then ∫γ
ez
z2dz = 2πi.
Example 8.4 Let us evaluate
I =∫γ
ez
(z − 1)(z + 1)3dz
where we choose different contours γ.
1. Let γ = γ(0; 1/2). Then the integral is zero.
2. Let γ = γ(1; 1/2). So we write the integral as∫γ
ez
(z−1)(z+1)3
z − 1dz
Thus we have I = eπi/4.
47
3. Let γ = γ(−1; 1/2). Then we write the whole thing as∫γ
ez
(z−1)
(z + 1)3dz
Now put g(z) = ez
(z−1) and so by the previous formula we get that I = 2πi2! f
(2)(−1). This is astraightforward but tricky set of calculations.
f (1)(z) =(z − 1)ez − ez
(z − 1)2
=zez − 2ez
(z − 1)2
f (2)(z) =((ez + zez)(z − 1)2 − 2(zez − 2ez)
(z − 1)3
f (2)(−1) =116
(4(e−1 − e−1 − 2e−1) + 4(−e−1 − 2e−1))
= − 54e.
So ∫γ
ez
(z − 1)(z + 1)3dz = −5πi
4e.
Definition 8.8 A zero of a function f(z) is a point a such that f(a) = 0.
Assume that f is holomorphic in some disc D(a; r) and that f is not identically zero. Taylor’stheorem says we can write:
f(z) =∞∑n=m
cn(z − a)n,
where m ≥ 1, cm 6= 0 and z ∈ D(a; r).Then m is said to be the order of the zero at z = a. Note thatf has a zero of order m at a if and only if
f(a) = f ′(a) = ... = fm−1(a) = 0, fm(a) 6= 0.
Definition 8.9 For any function f let Z(f) = {z : f(z) = 0}
Basically we want to show that Z is not too big if f is holomorphic on a region.
Theorem 8.10 Let G be a region and let f ∈ H(G). Suppose that there exists w ∈ G which is alimit point of Z(f). Then f(z) = 0 ∀z ∈ G.
Proof: We begin by proving that there is a disc on which f is identically zero.
Let u be any limit point of Z(f) and let D ⊆ G be an open disc which contains u. Since f iscontinuous f(u) = 0. Assume that f is not zero in D and assume that u is a zero of order m. Picka disc D(u; r) so that D(u; r) ⊆ D. Then, by using Taylor’s Theorem, we get
f(z) = (z − u)m∞∑k=m
ck(z − u)k−m. where cm 6= 0
48
for z ∈ D(u; r). Put
g(z) =∞∑k=m
ck(z − u)k−m.
Then g is holomorphic and g(u) 6= 0. So ∃ δ > 0 so that for 0 ≤ |z − u| < δ we have g(z) 6= 0. Thisis a contradiction to u being a limit point of Z(f). So we get that f is zero on D(u; r).
Next we let E be the set of limit points of Z(f). Let a ∈ E then by continuity f(a) = 0 and soE ⊂ Z(f). By the previous paragraph we have shown that E is open. Let a ∈ G \ E. Then thereis a disc D′(a; s) on which f is never zero. But then D(a, s) ⊂ G \ E. Now both E and G \ E areopen which contradicts the fact that G is connected. 2
This gives very quickly the following beautiful result.
Corollary 8.11 (The Identity Theorem) Let G be a region and let f, g ∈ H(G). Suppose thereexists a sequence of distinct points z1, z2, . . . which converges to w ∈ G and f(zj) = g(zj)∀j ∈ N.Then f and g are identical on G.
Proof: Apply the previous result to the function f − g.2
Just two more theorems to end this section.
Theorem 8.12 (The local Maximum-Modulus Theorem) Let a ∈ C and let R > 0, R ∈ R.Suppose f ∈ H(D(a;R)) and |f(z)| ≤ |f(a)| ∀z ∈ D(a;R) then f is constant on D(a;R).
Proof: Let 0 < r < R. By Cauchy’s Integral Theorem 8.1 we get
f(a) =1
2πi
∫γ(a:r)
f(w)w − a
dw
=1
2πi
∫ 2π
t=0
f(a+ reit)reit
ireit dt
=1
2π
∫ 2π
t=0
f(a+ reit) dt.
Thus we have by using the hypothesis that
|f(a)| ≤ 12π
∫ 2π
t=0
|f(a+ reit)| dt
≤ 12π
∫ 2π
t=0
|f(a)| dt = |f(a)|
and so
12π
∫ 2π
t=0
|f(a+ reit)| dt = |f(a)|.
Thus ∫ 2π
t=0
|f(a+ reit)| − |f(a)| dt = 0.
But the value of the integrand on the left is always less than 0. The only possibility is that it isidentically zero. But now for any z ∈ D(a;R) there is an r so that z = a + reit and the theoremfollows from Proposition 3.12. 2
49
Theorem 8.13 (The Maximum-Modulus Theorem) Let G be a bounded region and let Gdenote G together its limit points. Suppose f : G→ C is continuous and differentiable at all z ∈ G.Then ∃ a ∈ G \G such that |f(z)| ≤ |f(a)| ∀z ∈ G.
Proof: We know that G is closed so we know ∃ a ∈ G such that |f(z)| ≤ |f(a)| ∀z ∈ G. Nowassume that a ∈ G. Then we can find a disc D so that a ∈ D ⊂ G. But then by the previousresult we have that f is constant on D. But then by the Identity Theorem f is constant on G. Asf : G→ C is continuous we have that f is constant on G and the result follows. If a /∈ G the resultfollows anyway. 2
50
Chapter 9
Laurent Series
At the end of the last section a very powerful description of holomorphic functions had been devel-oped. An essential tool in this was Cauchy’s Theorem and it is clear that saying that a complexfunction is holomorphic in some region is much more powerful than in the real case. However a lot ofvery useful ’objects’ fail to be holomorphic functions for two distinct reasons. The first is that theyare functions but are not holomorphic, the second reason is that they are not in general functions.The second group contains many inverse “objects” some of which we have already seen i.e log, sin−1
etc. In this section we will concentrate on the first category.
Our first objective is to see what we can do about Taylor series. Note that any function definedby a Taylor series in a disc is holomorphic in that disc. If we examine the function f(z) = 1/zwe cannot expect to find a Taylor series expansion about the point z = 0. But we do have a sortof series expansion if we allow ourselves negative exponents for z. Similarly we could look at thefunction
f(z) =2
(z − 2)2− 1z − 2
about the point z = 2. Clearly such a description describes the function for |z − 2| 6= 0 completely.
Thus we define a Laurent Series1 about a point z = a by a series of the form
∞∑−∞
cn(z − a)n.
The next theorem (6.1 in Priestleys book) gives conditions under which a function can have sucha series expression.
Theorem 9.1 Let A = {z : R < |z − a| < S}, (0 ≤ R < S ≤ ∞) and let f ∈ H(A). Then, forz ∈ A,
f(z) =∞∑
n=−∞cn(z − a)n
1Pierre Alphonse Laurent (born 18 July 1813 in Paris, died 2 Sept 1854 in Paris) was in the engineering corps andspent six years directing operations for the enlargement of the port of Le Havre.
He submitted a work for the Grand Prize of 1842, unfortunately after the final date for submission. Cauchy reportedon his work, which gives the Laurent series for a complex function, saying that it should be approved but it was not.
After Laurent’s death his widow arranged for two more of his memoirs to be presented to the Academy. One wasnever published, the second appeared in 1863.
51
wherecn =
12πi
∫γf(w)(w − a)−n−1 dw,
with γ = γ(a; r), (R < r < S).
Proof: We will assume that a = 0, this saves lots of notation. We also observe that the region A isan annulus. Thus we consider two contours γ, δ as shown in Figure 9.1, where R < P < |z| < Q < S.
S
QR
P
Figure 9.1:
The idea of the proof is two add the two contours together and see that the result is preciselythe difference of two circular contours. However first we note that
f(z) =1
2πi
∫γ
f(w)w − z
dw (9.1)
and
0 = 1/2πi∫δ
f(w)w − z
dw. (9.2)
Now add these together we see that
f(z) =1
2πi
(∫γ(0;Q)
f(w)w − z
dw −∫γ(0;P )
f(w)w − z
dw
).
52
Note that whenever w ∈ γ(0;Q)∗ |w/z| > 1. So we can use the binomial expansion to get that
f(w)w − z
=f(w)w
∞∑n=0
(z/w)n.
However when w ∈ γ(0;P )∗, |z/w| > 1 and so
f(w)w − z
=f(w)z
∞∑n=0
(w/z)n.
Now since on both contours f(z) is bounded we see that both the series are uniformly convergent.So by an earlier theorem we can reverse the order of integration and summation. Hence we obtainthe following:
2πif(z) =∞∑n=0
(∫γ(0;Q)
f(w)wn+1
dw
)zn
+∞∑m=0
(∫γ(0;P )
f(w)wm dw
)z−m−1
The result now follows by replacing the two contours by the given contour using the deformationtheorem and by putting n = −m− 1 in the last series. 2
This is a very satisfactory theorem in the sense that it gives a very explicit form to a functionwhich is holomorphic in an annulus. The bad behaviour is very clearly seen. The one difficulty withthe theorem as compared to Taylor’s series is that it is not easy to calculate the coefficients. Theyare not so clearly related to the differential coefficients. However it turns out that any method offinding them works. This is made precise in the following theorem.
Theorem 9.2 Let f ∈ H(A) where A = {z : R < |z − a| < S}, (0 ≤ R < S ≤ ∞), and suppose
f(z) =∞∑
n=−∞bn(z − a)n.
Then bn = cn for all n, where cn is as in Theorem 9.1
Proof: We may again assume that a = 0. Choose any r such that R < r < S. Then, by 9.1 wehave
2πicn =∫γ(0;r)
f(w)w−n−1 dw =∫γ(0;r)
∞∑k=−∞
bkwk−n−1 dw.
We can now split the sum into two halves and use uniform convergence (3.13) all over again
2πicn =
( ∞∑k=0
∫γ(0;r)
bkwk−n−1 dw
)+
( ∞∑m=1
∫γ(0;r)
b−mw−m−1−n dw
)and so the only term that appears is that with power of w equal −1. Thus
2πicn = 2πibn
as claimed. 2
Given this theorem we can see that any method of calculating the coefficients is good enough.Lets try an example to see how good we are.
53
Example 9.1 Consider the function f(z) = z/(z − 1) about the point z = 1. Clearly f(z) isholomorphic in an annulus 0 < |z− 1| < R for any R > 0. The trick here is to write z as (z− 1) + 1and so
f(z) =z − 1 + 1z − 1
;
now we can see that f(z) = 1 + 1z−1 . This is clearly the Laurent series for the function about z = 1.
Example 9.2 Let
f(z) =1
z2 − 1and consider expansions about z = 1, −1.
Example 9.3 Consider f(z) = csc(z), this is well behaved except at points z = kπ. So an appro-priate choice of annulus around the origin will work. Note csc(z) = 1/ sin(z) and the Taylor seriesfor sin(z) is
z − z3/3! + z5/5!− . . .
So we can writesin z = z
(1− z2/3! + h(z)
),
where |h(z)| ≤ Kz4 for some K > 0, we will use the ‘O’ notation, and write h(z) = O(z4). But now
csc z = z−1(1− z2/3! +O(z4)
)−1= z−1
(1 + z2/3! +O(z4
)This gives that all the terms cm = 0 if m < −1, it also gives the first two non-zero terms c−1 = 1and c1 = −1/6. We can also see that c0 = 0.
Very often all we need is to find the first few non-zero terms to get the properties of a function weare interested in. As a general rule it is possible to calculate the Laurent series for a function of theform 1/f(z) about a point z = a whenever f is holomorphic in some disc about a and f(a) = 0. Todo this we write down the Taylor series and then formally invert it. To show how to do this assumea = 0. Then the Taylor series looks like
zd∞∑n=0
anzn.
Let g(z) =∑bmz
m and assume g(z)f(z) ≡ zd, ignoring the problems of convergence. Using theusual rules for multiplying series we get the following equations:
a0b0 = 1,a0b1 + a1b0 = 0,
a0b2 + a1b1 + a2b0 = 0,and the general equation looks like:
k∑i=0
aibk−i = 0, if k > 0
Given that we know all the a’s then we have recursion formulae to calculate the b’s. The requiredLaurent expansion is given by taking z−dg(z).
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Example 9.4 Let f(z) =∑nzn, note that this is holomorphic for all 0 < |z| < 1. Choose an
annulus in this region with centre z = 0. The equations give b0 = 1, b1 = −2, b2 = 1, b3 = 0 andbn = 0 ∀n > 2. Thus the Laurent expansion is
1z− 2 + z =
(1− z)2
z.
We now make some definitions which play a key role in the development. Let f(z) : C → C be afunction.
Definition 9.3 1. We say that f is regular at a point a if f is holomorphic in a disc about a.
2. A point a is said to be a singularity of f if a is a limit point of regular points but is not itselfa regular point.We can be a little more precise,
(a) if there is a punctured disc about a in which f is holomorphic then a is called an isolatedsingularity;
(b) otherwise we say that a is a non-isolated essential singularity.
We also give a special name to the left-hand side of the expansion.
Definition 9.4 Given a Laurent series expansion of a function f(z) about the point z = a theprincipal part is
−1∑∞
an(z − a)n
.
To give some simple examples we see that the function 1/z has an isolated singularity at 0. Howevercsc(1/z) has an non-isolated essential singularity at 0, since sin(1/z) = 0 whenever z = 1/πk for aninteger k.
The next stage is to show how we can classify isolated singularities and their relation to Laurentseries. Let
f(z) =∞∑
n=−∞cn(z − a)n,
be a Laurent expansion about the point z = a.
Definition 9.5 Then the point a is said to be:
1. a removable singularity if cn = 0 ∀n < 0;
2. a pole of order m (m ≥ 1) if c−m 6= 0 and cn = 0∀n < −m;
3. an isolated essential singularity if there exist arbitrarily large n such that c−n 6= 0.
The reason for the term removable singularity is that the function can be redefined at thesingularity so that it becomes holomorphic at that point.
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Example 9.5 Consider the function sin(z)/z which has a singularity at z = 0, but if we define anew function
f(z) ={
sin(z)z if z 6= 0,
1 if z = 0
then we see that f(z) is holomorphic everywhere. This can be seen from the Laurent series forsin(z)/z, which is 1− z2/3! + . . . .
Our earlier examples show that csc(z) has a simple pole (of order 1) at z = 0, and in Example 9.1we had a simple pole at z = 1. Poles are specially well behaved singularities and the next few resultsenable us to make this more explicit.
Theorem 9.6 1. Let f ∈ H(D(a; r)). Then f has a zero of order m ≥ 1 at z = a if and only if
limz→a
(z − a)−mf(z) = C,
where C is some non-zero constant.
2. Let f ∈ H(D′(a; r)). Then f has a pole of order m ≥ 1 at z = a if and only if
limz→a
(z − a)mf(z) = D,
where D is some non-zero constant.
Proof: I will prove 1), since 2) is in the book. Suppose a is a zero of order m. Then by theremarks above:
f(z) = (z − a)m( ∞∑n=0
cn(z − a)n),
where c0 6= 0. Then clearlyf(z)(z − a)−m = c0 + zh(z),
where h(z) is holomorphic in D(a; r). Since c0 6= 0 one half follows.
Now assume the limit condition holds. Again using the Taylor series it is clear that the coefficientsof (z − a)d have to be 0 if d < m, and so the result follows. 2
Corollary 9.7 Let f be holomorphic in some disc D(a; r). Then f has a zero of order m at z = aif and only if 1/f has a pole of order m at z = a.
Proof: This is straightforward.2
Example 9.6 Let f(z) = z sin(z). Note that f(z) has zeroes at z = nπ , for n ∈ Z. Nowf ′(z) = sin(z) + z cos(z) and f (2)(z) = 2 cos(z) − z sin(z). Thus f ′(nπ) 6= 0 if n 6= 0 and so 1/f(z)has simple poles at z = nπ, n 6= 0. But f ′(0) = 0 however f (2)(0) 6= 0 and so z = 0 is a double poleof 1/f(z).
Example 9.7 We already know the answer for the function 1/(z2− 1). However just to check notethat if g(z) = z2 − 1, g′(1) = 2 and g′(−1) = −2, so we see that f(z) has simple poles at z = ±1.
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We now consider what happens for the three different types of singularities. Firstly we examine thecase that z = a is a removable singularity. Since f(z)→ c0 as z → a we can redefine f(a) = c0 andthen f(z) is now holomorphic at z = a, (strictly we should invent a new function but one rarelybothers). If f(z) has a pole at z = a then 1/f(z) has a zero at z = a so clearly
limz→a|f(z)| =∞
Because ∀k > 0 ∃δ > 0 such that |1/f(z)| < 1/k, ∀z with |z−a| > δ and so |f(z)| > k ∀0 < |z−a| <δ.
The most interesting case is that of the essential singularity. Unfortunately the results in thiscase take us beyond the scope of this course. However we can make some comments. One thingthat does not happen is that f(z) has a limit as z → a. The only possible limit could be ∞, (in theextended complex plane) but then 1/f(z)→ 0 as z → a and we would see that we could show thatf(z) had a pole at z = a. Now consider the function exp(1/z) at z = 0. This has a Laurent series:-
1 +1z
+1
2z2+
16z3
+ . . . |z| > 0.
Choose any ε > 0 and let w ∈ C, w 6= 0. I claim that we can find z such that exp(1/z) = w and0 < |z| < ε. So
1/z = Log(|w|+ i(arg(w) + 2πk) where k ∈ Z.
Now we choose k very large and that makes z as small as we wish. That is in any disc about a wecan make exp( 1
z ) take any non-zero value we choose. This holds for any function f(z) which has anessential singularity at a (this is a form of Picard’s Theorem2).
We know want to look at functions which are defined on the whole extended complex plane C.The key idea is to generalise the ideas of disc and behaviour of functions to the point ∞. Thus wedefine a disc D(∞; r) to be the set {z | |z| > r} ∪∞ and we examine the properties of the functionf(1/z) at z = 0 to see the behaviour of f(z) at z =∞. So for example f(z) = 1/z is nicely behavedat∞ as is sin(1/z) that is both functions are holomorphic on some disc about z =∞. Now we maketo definitions.
Definition 9.8 Let G ⊆ C be an open set. A function f : G → C is said to be meromorphic if itsonly singularities on G are poles
Lemma 9.9 Let S be an infinite closed subset of C. Then S has a limit point in S.2Charles Emile Picard (born 24 July 1856 in Paris, died 11 Dec 1941 in Paris). There are two theorems in complex
analysis that carry his name. Picard’s Little Theorem says that if f ∈ H(C) and there are two distinct complexnumbers α and β which are not in that range of f , then f is constant. The so–called Big Picard Theorem is a localversion: if f has an isolated singularity at a point z0 and if f omits two values in some neighbourhood of z0, then z0is a removable singularity or a pole of f .
He was appointed lecturer at the University of Paris in 1878 and professor at Toulouse in 1879. In 1898 he wasappointed professor at the Sorbonne in Paris. He made his most important contributions in the field of analysisand analytic geometry. He used methods of successive approximation to show the existence of solutions of ordinarydifferential equations.
Building on work by Abel and Riemann, Picard’s study of the integrals attached to algebraic surfaces and relatedtopological questions developed into an important part of algebraic geometry. On this topic he published, with GeorgesSimart, “Theorie des fonctions algebriques de deux variables independantes” (2 volumes) (1897, 1906). Picard alsoapplied analysis to the study of elasticity, heat and electricity.
He married Hermite’s daughter. Picard’s daughter and two sons were all killed in World War I. His grandsons werewounded and captured in World War II.
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Proof: It is appropriate to make the definition of limit point clear. A point w is a limit point of asubset S if every puncture disc about w contains a point of S. This is the same as saying that thereis a sequence whose limit is w but that the sequence is not eventually constant.
If the set is bounded the result was proved earlier, 3.22. However if the set is unbounded thatmeans that for each n ∈ N there is some zn ∈ S with |zn| > n. Now the sequence has zn has limit∞.
Remark 9.1 We can now see that meromorphic functions can only have a finite number of poles.Let S be the set of poles. If this were infinite then the set would contain a limit point by the abovelemma. But this point would not be an isolated singularity and so could not be a pole.
We now end the course with the following theorem.
Theorem 9.10 Let f be a function defined on C. Then(i) If f is holomorphic on C then f is constant and(ii) if f is meromorphic on C then f is a rational function, that is we can write f = p/q where p(z)and q(z) are polynomials.
Proof:(i) By Liouville’s Theorem 8.2 if we can show that f is bounded on C it will be constant. The functionis continuous so it is bounded on any compact set,3. Here is a compact set S1 = {z | |z| ≤ 1}, so thereexists M ∈ R so that |f(z)| ≤M for all z ∈ S1. Now consider g(w) = f(1/z) for S2 = {w | |w| ≤ 1}.Note that g(w) is a continuous function on S2 so there exists L ∈ R so that |g(w)| ≤ L for all w ∈ S2.But we see that f(z) is bounded for |z| ≥ 1 and for |z| ≤ 1 so f is bounded and we can apply 8.2.
(ii) By the remark above there are just finitely many poles. We will use induction on the numberof poles, say n. Note that if n = 0 the result follows from (i). Now assume that the result is true forfunctions meromorphic with n− 1 poles. Let z = a be a pole of order m. Let the principal part be
h(z) =1∑m
c−n(z − a)n
.
Note that h(z) = p(z)/(z − 1)m where p(z) = c−n + c−n+1(z − a) + · · · + c−1(z − a)m−1, which isa polynomial. Thus h(z) is a rational function. Now f(z) − h(z) is a meromorphic function whichhas a removable singularity at z = a. Remove it and then the function has only n− 1 poles so thetheorem is proved.
3if we had shown that C was compact, it would be the end of the story.
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