Methods of Complex Functions Lecture Notes Bristol MATH20001maajh/methods/math20001_dec_20… · methods of complex functions lecture notes bristol math20001 5 Note that Im z and

Methods of Complex FunctionsLecture Notes Bristol MATH20001Karoline Wiesner, School of Mathematics, University of Bristol2015 (updated 10 November 2015)

Reading listThere are many textbooks suitable as reference for this course. I

recommend the following:Complex Analysis by Joseph Bak and Donald J. Newman. 3rd ed.Springer, 2010. This is available as e-book on the University library’sweb site, http://www.bristol.ac.uk/library/.

Basic Complex Analysis by Jerrold E. Marsden and Michael J.Hoffman, Freeman, 1999. 1 1 There are plenty of copies of Marsden

in the Queens library.For an introduction to functions of complex numbers and theirrelevance in physics I highly recommend The Road To Reality: AComplete Guide to the Laws of the Universe by Roger Penrose, inparticular Chapters 4, 5, 7, and 8.

Some course information:

• Drop-in sessions Tuesdays 3.30pm-4.30pm, office 3.1a, beginningweek 8.

• Homework given out every Friday, hand-in following Friday 10amin main maths building

• In week 7, 8, 10, 12: problem class Friday 11am.

• In weeks 9 and 11: small-group problem classes, time and place tobe announced.

• Maths Cafe with Dan Taylor Lewis: Friday 12pm - 1pm, Weekcommencing 9th November - 30th November (weeks 7-10) - SM3,Week commencing 7th December (week 11) - PC3, Week com-mencing 14th December (week 12) - SM3

This material provided exclusively for educational purposes and is to be down-loaded or copied for your private study only.

http://www.bristol.ac.uk/library/

methods of complex functions lecture notes bristol math20001 2


1 Complex numberssection 1 is a quick review of what youknow from Linear Algebra 1.How is it that −1 can have a square root? The square of a positive number

is always positive, and the square of a negative number is again positive.It seems impossible that we can find a number whose square is actuallynegative. Yet, this is a situation similar to when people were looking for asquare root of the number 2 which has no square root within the system ofrational numbers. In that case they resolved the situation by extending theirsystem of numbers from the rationals (Q) to a larger system, the system ofreals (R). We will do the same by extending the number system of reals byintroducing a single quantity, called ‘i′, which is to square to −1, and adjoinit to the system of reals, allowing combinations of i with real numbers toform expressions such as x + iy, where x and y are arbitrary real numbers.Any such combination is called a complex number. There are two square roots of −1: i and

−i.Let us start by introducing the system of complex numbers formally asordered pairs of real numbers.

Definition 1.1 (The complex numbers). The complex numbers are theset of ordered pairs of real numbers (x, y) ∈ R2 with addition andmultiplication of two complex numbers defined by

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) (1)

(x1, y1)(x2, y2) = (x1x2 − y1y2, y1x2 + x1y2). (2)

The associative and commutative laws for addition and multiplicationas well as the distributive law follow easily from the same properties of realnumbers. The additive identity, or zero, is given by (0, 0), and hence theadditive inverse of (x, y) is (−x,−y). The multiplicative identity is (1, 0). Tofind the multiplicative inverse of any nonzero (x1, y1) we set

(x1, y1)(x2, y2) = (1, 0), (3)

which has the solution

x2 =x1

x21 + y2

1, y2 =

−y1

x21 + y2

1. (4)

This shows that the set of complex numbers, denoted C, together withaddition and multiplication as defined in Definition 1.1 form a field. Two complex numbers are equal if and

only if their real and imaginary partsare equal.

We associate complex numbers of the form (x, 0) with the correspondingreal numbers x. It follows that (x1, 0) + (x2, 0) = (x1 + x2, 0) correspondsto addition of two real numbers, x1 + x2, and that (x1, 0)(x2, 0) = (x1x2, 0)corresponds to multiplication of two real numbers, x1x2.

We can now see that (0, 1) is a square root of −1 since

(0, 1)(0, 1) = (−1, 0) = −1 (5)

and henceforth (0, 1) will be denoted i.The following notations for complexnumbers are equivalent:

(x, y) ≡ (x, 0) + (0, y) ≡ x + iy , x, y ∈ R. (6)

The standard letter used for a complex number is z and we will usually usethe notation z = x + iy.


We can now see where the rules for addition and multiplication in Def-inition 1.1 come from. Adding two complex numbers z1 = x1 + iy1 andz2 = x2 + iy2 we get

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2) , (7)

where the RHS is again in the form of a complex number. Let us find theproduct of z1 and z2. Expanding the factors using the ordinary rules ofalgebra we get

(x1 + iy1)(x2 + iy2) = x1x2 + i2y1y2 + i(y1x2 + x1y2) (8)

= (x1x2 − y1y1) + i(y1x2 + x1y2) (9)

where we have applied the rule i2 = −1 and the final answer for the productof two complex numbers is again in the form of a complex number. In fact,there are two square roots of any nonzero complex number a + bi. To seethis, we solve (x + iy)2 = a + bi, by multiplying out the square and comparethe real and imaginary parts. That is, we set x2 − y2 = a and 2xy = b whichis equivalent to 4x4 − 4ax2 − b2 = 0 and y = b/2x. Solving first for x2, wefind the two solutions are given by

x = ±

√a +√

a2 + b2

2(10)

y = ±

√−a +

√a2 + b2

2sign(b) (11)

where sign(b) = 1 if b ≥ 0 and sign(b) = −1 if b < 0.

Example Following the same steps, find the following square roots.

1.√

2i

2.√−5− 12i

Solution. 1. The two square roots of 2i are 1 + i and −1− i.

2. The square roots of −5− 12i are 2− 3i and −2 + 3i.

It turns out that any quadratic equation with complex coefficients admitsa solution in the complex numbers. Indeed, we will see that any polynomialof order n has n roots in the complex numbers.

Definition 1.2. Let z ∈ C be a complex number, z = x + iy, x, y ∈ R. Wedefine

the real part of z: Re z = x;the imaginary part of z: Im z = y ;the conjugate of z: z̄ = x− iy;the modulus of z: |z| =

√x2 + y2;

the argument of z: tan(arg z) = y/x, x, y 6= 0 .


Note that Im z and Re z are real num-bers.Both z̄ and z∗ are used to denote theconjugate of z.

A special flavour of complex analysis arises because one may think of thecomplex numbers C both algebraically as a number system and geometri-cally as a vector space. It is essential therefore to have a good geometricalintuition for the complex plane.

To each complex number z = x + iy we associate the point (x, y) in theCartesian plane. Real numbers are thus associated with points on the x -axis,called the real axis while the purely imaginary numbers iy correspond topoints on the y-axis, designated as the imaginary axis. This plane is some-times called the Argand plane2. 2 Jean-Robert Argand (1768-1822) was a

Parisian bookkeeper. He wrote a pam-phlet in 1806 with the title “Essay onthe Geometrical Interpretation of Imagi-nary Quantities”. The mathematician A.Legendre (1752-1833) mentioned it in aletter to Francois Francais, a professorof mathematics. It was published aspart of an article in 1813 in the Annalesde Mathémathiques giving the basics ofcomplex numbers.

Let z = x + iy and |z| = r and arg z = θ. It follows that x = r cos θ,y = r sin θ and the so called polar form of z is

z = r(cos θ + i sin θ). (12)

r and θ are called the polar coordinates of z. This form is especially usefulfor multiplication. Let z1 = r1(cos θ1 + i sin θ1), z2 = r2(cos θ2 + i sin θ2). Then

z1z2 = r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2)). (13)

Thus, if z is the product of two complex numbers, |z| is the product of theirmoduli and Arg z is the sum of their arguments.3 It follows by induction 3 Similarly z1/z2 can be obtained by

dividing the moduli and subtracting thearguments:

z1

z2=

r1

r2(cos(θ1 − θ2) + i sin(θ1 − θ2)) .

that if z = r(cos θ + i sin θ) and n is any integer,

zn = rn(cos nθ + i sin nθ) . (14)

Eq. 14 is also called de Moivre’s Theorem, for r1 = r2 = 1.

Example Solve z3 = 1.

Solution. We write it in the polar form

r3(cos 3θ + i sin 3θ) = 1(cos 0 + i sin 0) ⇔ r = 1 , 3θ = 0(mod2π) .

Hence the three solutions are given by4 4 or in Cartesian coordinates

z1 = 1 , z2 = − 12+ i√

32

, z3 = − 12− i√

32

.z1 = cos 0 + i sin 0 , z2 = cos(2π

3) + i sin(

2π

3)

z3 = cos(−2π

3) + i sin(

−2π

3) ,

These three roots of z3 are the vertices of an equilateral triangle inscribedin the unit circle centred at the origin. Similarly the n-th roots of 1 are lo-cated at the vertices of the regular n-gon inscribed in the unit circle with onevertex at z = 1.

Sketch...


Addition also has a geometric interpretation: the sum of z1 and z2 corre-sponds to the vector sum z1 + z2 = (x1 + x2, y1 + y2).

Sketch...

The geometric interpretation of arg z is the angle which the vector (orig-inating from 0) to z makes with the positive x-axis. Thus arg z is definedmodulo 2π as that number θ for which

cos θ =Re z|z| , sin θ =

Im z|z| . (15)

From the geometric interpretation of complex numbers we can immedi-ately establish the so-called triangle inequality satisfied by complex numbersz1 and z2,

|z1 + z2| ≤ |z1|+ |z2|, (16)

and the further inequality

|u1| − |u2| ≤ |u1 − u2|, (17)

which follows immediately from putting z1 = u1 − u2 and z2 = u2.Since the argument is only defined up to modulo 2π we introduce the

principle argument Arg z.

Definition 1.3. The principle argument Arg z of a complex numberz ∈ C is defined as that unique value of arg z s.t. −π < arg z ≤π.

For example, all points on the negative real axis have Arg z = π. Anothercommon convention for the principle argument sets Arg z ∈ [0, 2π). Notethat any convention for the principle argument will introduce a discontinuityinto the arg function.

Using these definitions we can identify regions of the complex plane withsubsets of the complex numbers. Let’s look at some examples.


Examples 1. {z : Re z > 0} is represented geometrically by the righthalf-plane.

2. {z : z = z̄} is the real line.

3. {z : −θ < Arg z < θ} is an angular sector (wedge) of angle 2θ.

4. {z : |z + 1| < 1} is the disk of radius 1 centred at −1.

Sketch...

1.1 Sets in the complex plane

We will often need the notion of open disk and open set. There is another way of defining anopen set: A point z0 in a set S is calledinterior point if there is a neighbour-hood of z0 completely contained in S.If every point in a set S is an interiorpoint S is an open set.

There is also an alternative way ofdefining a closed set: A point z0 in aset S is called boundary point if everyneighbourhood of z0 contains at leastone point in S and at least one point notin S. S is closed if it contains all of itsboundary points.

Definition 1.4 (Open disk). A domain D(z0, R) of radius R > 0 centredon some point z0 and defined as D(z0, R) = {z : |z− z0| < R} is calledan open disk.

Definition 1.5 (Open set, closed set). A (possibly infinite) union or afinite intersection of open disks is called an open set.

A set S is called closed if its compliment C \ S is open.

In other words, a set S is called open if for any z ∈ S there exists an Rsuch that the open disk D(z, R) is also in S.

Definition 1.6 (Closed curve, simple closed curve, interior of a curve).


We call a curve a closed curve if its initial and terminal points coincide.C is a simple closed curve if no other points coincide, that is, the curvedoes not intersect with itself other than at the end points.

For a given closed curve C we call Int C the interior of the curve.

The interior of a closed curve is necessarily an open set.

Examples A triangle (4) is a simple closed curve, a figure eight (∞) is aclosed curve but not simple, a straight line from some point a to some pointb 6= a (–) is not a closed curve, and hence not a simple closed curve.

Very important in the following are the different topologies of sets.

Definition 1.7 (Connected set). A set S is said to be connected if anytwo points in S can be connected by a curve which is wholly inside S.

Definition 1.8 (Simply connected set). A connected set S is said to besimply connected if any closed curve in S can be shrunk continuouslyinside S to a point also inside S.

Examples

The complex plane is simply connected.

The complex plane minus the real axis is not simply connected since it is notconnected.

The annulus {z : 1 < |z| < 3} is connected but not simply connected.

The unit disk minus the positive real axis is simply connected.


2 Power series

One area where complex numbers are very useful is the convergence be-haviour of power series.

A power series is an infinite sum of the form

a0 + a1x + a2x2 + a3x3 + . . . . (18)

Because this sum involves an infinite number of terms, it may be the casethat the series diverges, which is to say that it does not settle down to aparticular finite value as we add up more and more of its terms. For anexample, consider the series

1 + x2 + x4 + x6 + x8 + . . . (19)

(where a0 = 1, a1 = 0, a2 = 1, a3 = 0 etc). If we put x = 1, then, adding theterms successively, we get

1, 1 + 1 = 2, 1 + 1 + 1 = 3, 1 + 1 + 1 + 1 = 4, etc (20)

and we see that the series has no chance of settling down to a particularfinite value, that is, it is divergent. On the other hand, if we put x = 1/2,then we get

1, 1 + 1/4 = 5/4, 1 + 1/4 + 1/16 = 21/16, etc (21)

and it turns out that these numbers become closer and closer to the limitingvalue 4/3, so the series is now convergent.

Figure 1: In the complex plane, thefunctions (1− z2)−1 and (1 + z2)−1

have the same circle of convergence,there being poles for the former atz = ±1 and poles for the latter atz = ±i, all having the same (unit)distance from the origin. (Figure from?.)

We can explicitly write down the answer to the sum of the entire series as

1 + x2 + x4 + x6 + x8 + · · · = (1− x2)−1 (22)

When we substitute x = 1, we find that this answer is 0−1, which is ‘infinity’,and this provides us with an understanding of why the series has to divergefor that value of x. When we substitute x = 1/2, the answer is 4/3, and theseries actually converges to this particular value.

To see how complex numbers fit into the picture, let us consider a func-tion just slightly different from (1 − x2)−1, namely (1 + x2)−1, and askwhether it has a sensible power series expansion. There is, indeed, a simple-looking power series for (1 + x2)−1, only slightly different from the one thatwe had before, namely

1− x2 + x4 − x6 + x8 − · · · = (1 + x2)−1, (23)

the difference being merely a change of sign in alternate terms. This seriesbehaves slightly different from the one we studied first. We get the followingbehaviour for x = 1 and x = 1/2:

x = 1 : 1, 0, 1, 0, 1, etc. (24)

x = 1/2 : 1, 3/4, 13/16, 51/64, etc. (25)

We see that convergence occurs only in the case x = 1/2, where the an-swer comes out correctly with the limiting value 5/4. Whereas if we put inx = 1 into the function (1 + x2)−1 we get the number 1/2 which is not theliit of the series which merely fluctuates between 0 and 1. To get a betterunderstanding of the limiting behaviour of series we move to the complex


plane and consider the complex values of thes functions rather than restrict-ing our attention to real ones. We simply write these extended functions as(1− z2)−1 and (1 + z2)−1, respectively. In the case of the first real function(1− x2)−1, we were able to recognise where the divergence trouble starts,because the function is singular (in the sense of becoming infinite) at the twoplaces x = 1 and x = −1; but, with (1 + x2)−1, we can see no singularityat these places and, indeed, no real singularities at all. However, in terms ofthe complex variable z, we see that these two functions are much more on apar with one another. We have noted the singularities of (1− z2)−1 at twopoints z = ±1, of unit distance from the origin along the real axis; but nowwe see that (1 + z2)−1 also has singularities, namely at the two places z = ±i,these being the two points of unit distance from the origin on the imaginaryaxis.5 But what do these complex singularities have to do with the question 5 In the particular cases (1− z2)−1 and

(1 + z2)−1 the singularities are of asimple type called poles, something wewill revisit in a later chapter.

of convergence or divergence of the corresponding power series?The most important infinite series with complex terms are power series,

in which an is of the form an = cnzn or, somewhat more general an =

cn(z− z0)n. A power series P(z) is of the form

P(z) :=∞

∑n=0

cn(z− z0)n, with cn, z0, z ∈ C. (26)

2.1 Convergence of power series

Without proving it, we state that if a power series converges for z = ξ itconverges absolutely for every value of z such that |z − z0| < |ξ − z0|.Equally, if a power series diverges for z = ν it diverges for every valueof z such that |z − z0| > |ν − z0|. In other words, one can always find acircle of radius R in the complex plane (where R can be 0 or ∞ or anythingin between) centred at z0 with the property that if the complex number zlies strictly inside of the circle then the series converges for that value of z,whereas if z lies strictly outside of the circle then the series diverges for thatvalue of z. Whether or not the series converges when z lies actually on thecircle of convergence does not have a general answer. R is called the radiusof convergence. The fixed parameter z0 is the point of expansion, the centreof the circle of convergence. Figure 2 illustrates this for the power series of(1± z2)−1 in the neighbourhood of z0 = 0. Here, the radius of convergence is1.

Figure 2: In the complex plane, thefunctions (1− z2)−1 and (1 + z2)−1

have the same circle of convergence fora power expansion around the origin.The poles for the former are z = ±1and poles for the latter at z = ±i.(Figure from ?.)

We have the following result for the radius of convergence:

Theorem 2.1. A power series P(z) (Equation 99) converges for anyz ∈ C inside a disk of radius

R =1

limn→∞ |cn|1/n , (27)

and diverges for any z ∈ C outside of this disk. Nothing in general canbe said about the case {z : |z− z0| = R}.

Examples Find the radius of convergence R. If R < ∞ what happens forpoints in the set {z : |z− z0| = R}?


1. ∑∞n=1 nzn

2. ∑∞n=1(z

n/n2)

Solution. 1. We have cn = n. Since n1/n → 1 as n → ∞, we find R = 1.Thus, ∑∞

n=1 nzn converges for |z| < 1 and diverges for |z| > 1. The seriesalso diverges for |z| = 1 since |n1n| = n→ ∞ as n→ ∞.

2. ∑∞n=1(z

n/n2) also has radius of convergence equal to 1. In this case,however, the series converges for all points z on the unit circle since∣∣∣∣ zn

n2

∣∣∣∣ = 1n2 for |z| = 1 .

and ∑∞1 1/n2 = π2/6.

2.2 Examples of power series

The power series of the elementary functions in real variable calculus canimmediately be extended to the complex numbers. In other words, we candefine our first elementary complex functions in terms of their power seriesexpansion. The most important one are listed here and can be assumed forthe rest of the course.

ez =∞

∑n=0

zn

n!, (28)

cos(z) =∞

∑n=0

(−1)n z2n

(2n)!, (29)

sin(z) =∞

∑n=0

(−1)n z2n+1

(2n + 1)!, (30)

cosh(z) =∞

∑n=0

z2n

(2n)!, (31)

sinh(z) =∞

∑n=0

z2n+1

(2n + 1)!. (32)

The above series converge for all z ∈ C, that is their radius of convergenceis ∞. The following relations immediately follow:

cos(z) + i sin(z) = eiz , (33)

cosh(z) = cos(iz) (34)

i sinh(z) = sin(iz) . (35)

As an example of a power series with a finite radius of convergence con-sider the series

log(1 + z) =∞

∑n=1

(−1)n+1 zn

n. (36)

This series has a radius of convergence R = 1, that is it converges for all|z| < 1.


2.3 Functions of a complex variable

We could continue to focus on functions of complex numbers which arerepresented as a power series. But instead we will start with a more generalpoint of view. A function of a complex number is, to start with, simply amapping from the point (x, y) ∈ R2 to a point (u, v) ∈ R2. This means thatu(x, y) and v(x, y) are two real-valued functions.

Here, we will limit the functions we consider to those for which u(x, y)and v(x, y) are two continuous functions and that they possess continuousderivatives w.r.t. x and y, in short, that ux, uy, vx, vy are also continuous.

More formally we have the following definition.

Definition 2.1. A complex function is a map f : C→ C,

z = x + iy 7→ w = u(x, y) + iv(x, y) (37)

which we may also regard as a map f : R2 → R2,

(x, y) 7→ (u(x, y), v(x, y)) . (38)

Example Let f (z) = z2. Find u(x, y) and v(x, y).

Solution. z2 = (x+ iy)2 = x2− y2 + 2ixy. I.e. u(x, y) = x2− y2, v(x, y) = 2xy.


3 Complex differentiability

The complex numbers have algebraic properties which are very similar tothose of the real numbers. This similarity means that we can define differ-entiability in the complex case in exactly the same way as we do in the realcase. Let f (x, y) = u(x, y) + iv(x, y) where u and v are real-valued func-tions. The partial derivatives fx ≡ ∂ f /∂x and fy ≡ ∂ f /∂y are defined byfx = ux + ivx and fy = uy + ivy respectively, where we assume the latterexist.

Definition 3.1. A complex function f : A → C, whith domain A ∈ C, issaid to be differentiable at z ∈ A if

limh→0

f (z + h)− f (z)h

exists. In this case, the limit is denoted f ′(z).

It is important to note that h can be a complex number. Hence, the limitmust exist irrespective of the manner in which h approaches 0 in the complexplane.

Definition 3.2 (Holomorphic function). A complex function is calledholomorphic at z0 ∈ C if it is differentiable for all z in an open diskD(z0, R).

You say analytic, I say holomorphic.In some text books on complex analysisthe terms holomorphic and analyticare used interchangeably, for a goodreason. A function is analytic if it islocally (i.e. in a neighbourhood) givenby a Taylor series and hence (infinitely)differentiable. However, in R, not everyfunction differentiable in a neigh-bourhood (i.e. holomorphic) can berepresented by a Taylor series. Hence,in R analytic implies holomorphic butnot vice versa. It is a major theoremof complex analysis that the two setscoincide in C.

Definition 3.3 (Entire function). A complex function which is differen-tiable on C (and hence holomorphic on C) is called entire.

Similar proofs to the ones in the real case produce exactly the same ele-mentary properties of differentiation of complex functions.

Theorem 3.1. Let f , g : A → C be two complex functions differentiableat z ∈ A, then

1. ( f + g)′(z) = f ′(z) + g′(z)

2. ( f · g)′(z) = f ′(z)g(z) + f (z)g′(z)

3. ( f /g)′(z) = [ f ′(z)g(z)− f (z)g′(z)]/g2(z), for g 6= 0.

4. (g ◦ f )′(z) = g′( f (z)) f ′(z).

3.1 Differentiation of power series

We call an expression of the form

Pk(z) = c0 + c1z + c2z2 + c3z3 + · · ·+ ckzk (39)


with (fixed) complex coefficients cn a polynomial of order k in z. A polyno-mial Pk(z) may be differentiated with respect to the independent variable z inexactly the same way as for real variables. We will say more about differen-tiability later on. In the first place, notice that the identity

zk1 − zk

z1 − z= zk−1

1 + zk−21 z + · · ·+ zk−1 (40)

holds. If we now let z1 tend to z, we get

ddz

zk = limz1→z

zk1 − zk

z1 − z= kzk−1 . (41)

In the same way we get, for a general polynomial of order k,

ddz

Pk(z) = limz1→z

Pk(z1)− Pk(z)z1 − z

=k

∑n=1

ncnzn−1 . (42)

A power series

P(z) =∞

∑n=0

cn(z− z0)n (43)

is a polynomial of infinite order. And as such, a power series can also bedifferentiated, as summarised in the following theorem.

Theorem 3.2. Let P(z) be a power series

P(z) =∞

∑n=0

cn(z− z0)n (44)

with radius of convergence R, i.e. P(z) converges within the domainA = {z : |z− z0| < R} ⊂ C. Then P(z) may be differentiated term-by-term inside A . That is, the limit

P′(z) = limz1→z

P(z1)− P(z)z1 − z

, z ∈ A, (45)

exists, and

P′(z) =∞

∑n=1

ncn(z− z0)n−1 (46)

is also convergent in A.

The theorem states that a power series can be differentiated arbitrarilymany times in the interior of its radius of convergence. The derivative P′(z)of a power series is, again, a power series with the same radius of conver-gence. 6 6 The radius of convergence of a power

series remains unchanged upon dif-ferentiation since lim |ncn|1/(n−1) =lim(|ncn|1/n)n/(n−1) = lim |cn|1/n.

By differentiating the power series in Equation 28–32 term-by-term, we


find

ddz

ez = ez , (47)

ddz

cos(z) = − sin(z) , (48)

ddz

sin(z) = cos(z) , (49)

ddz

cosh(z) = sinh(z) , (50)

ddz

sinh(z) = cosh(z) . (51)

By differentiating Equation 36 term-by-term we get

ddz

log(1 + z) =1

1 + zz 6= −1 . (52)

We will learn more about the complex logarithm later on.

3.2 The Cauchy-Riemann equations

For a complex function f to be differentiable the limit in Definition 3.1 mustexist regardless of the manner in which h approaches 0 in the complex plane.Now, consider the following f (z) = z̄, that is u(x, y) = x and v(x, y) = −y.To compute the limit, let us set h = r, r ∈ R. We find

limr→0

x + r− iy− (x− iy)r

= 1 (53)

while setting h = is, s ∈ R, we find that

limis→0

x− iy− is− (x− iy)is

= −1 . (54)

Hence, the limit h → 0 is not unique and so does not exist – the functionf (z) = z̄ is not differentiable anywhere in C.

From this example, it becomes clear that complex differentiability is morerestrictive than real differentiability. In order to insure differentiability of afunction f restrictions apply. A generalisation of the above example leads tothe following fundamental fact in the theory of complex functions. The Cauchy-Riemann equations are a

sufficient condition for f to be differ-entiable at a point z only if fx and fyare continuous at z. Without this lattercondition the CR-equations are merelynecessary. Consider the example (seeFig. 3)

f (z) = f (x, y) =

{xy(x+iy)

x2+y2 , z 6= 0

0 , z = 0

Figure 3: f (z) = xy(x+iy)x2+y2 , z 6= z. The

vertical axis depicts the real part of zwhile the colours depict the argumentof z.

Theorem 3.3 (Cauchy-Riemann equations). Let f : A → C be a complexfunction on domain A ⊂ C, with f (x + iy) = u(x, y) + iv(x, y). Then f ′

exists if and only if the partial derivatives ux, uy, vx, vy are continuouson A and satisfy

∂u∂x

=∂v∂y

,∂u∂y

= − ∂v∂x

(55)

or, equivalently, fy = i fx . (56)

The Cauchy-Riemann equations are often abbreviated CR-equations.


An immediate consequence of the Cauchy-Riemann equations is thatthere are two equivalent ways to compute the derivative of a complex-valuedfunction,

f ′ = fx = −i fy (57)

Example We already know that the complex function f (z) = z2 or moregenerally f (z) = zn for any integer n is differentiable. Therefore the Cauchy-Riemann equations must hold. Confirm this.

Solution. Fromf (z) = (x + iy)2 = x2 − y2 + 2ixy

i.e.u(x, y) = x2 − y2 , v(x, y) = 2xy,

it follows that

ux = 2x , uy = −2y

vx = 2y , vy = 2x .

Therefore the Cauchy-Riemann equations are satisfied for all x, y and hencefor all z ∈ C.

Example Confirm that the complex function f (z) = z · z̄ is not complexdifferentiable on any open disk. Why is ‘open disk’ important here?

Solution.u(x, y) = x2 + y2 , v(x, y) = 0

and the CR equations are not satisfied except for x = y = 0.


3.3 The functions ez, sin(z), cos(z), and log(z)

We have seen the definitions of some basic complex functions in terms ofinfinite power series. In the following, we define a complex analogue to thereal function ex.

f (z) = ez = ex cos y + iex sin y . (58)

Indeed, it is easy to verify that this function f is an entire function with thedesired properties:

1. |ez| = ex.

2. ez 6= 0. Property 2 follows from 1. since ex 6= 0.Also, according to Eq. ?? above, eze−z =e0 = 1.

3. eiy = cos y + i sin y.

4. ez = a has infinitely many solutions for any a 6= 0. Property 4 is due to a = reiθ =rei(θ+2kπ), k = 0, 1, 2, . . . .5. (ez)′ = ez.

Using the definiton of ez we can define entire extensions of the real func-tions sin x and cos x by setting Note that, unlike sin x , sin z is not

bounded in modulus by 1. For example,| sin 10i| = 1/2(e10 − e−10) > 10, 000, seeFig. 4.sin z =

eiz − e−iz

2i, (59)

cos z =eiz + e−iz

2. (60)

We can confirm the following derivatives already found through term-by-term differentiation of the power series,

(sin z)′ = cos z (61)

(cos z)′ = − sin z (62)

Figure 4: The absolute value of sin z. Itis unbounded.

We now let the definition of ez provide us with an unambiguous loga-rithm, defined as the inverse of the exponential function,

z = log w if w = ez . (63)

We want the log function to behave as usual, i.e.

log(z1z2) = log(z1) + log(z2) . (64)

It is not immediately obvious that such an inverse to ez will necessarily exist.However, it turns out that, for any complex number w, apart from 0, therealways does exist z such that w = ez, so we can define log w = z. Sinceew 6= 0 for any w, log 0 is not defined. We define the logarithm as follows,

Definition 3.4 (Logarithm). For any z ∈ C \ {0}, we define the loga-rithm to be any of the infinitely many values

log z := log |z|+ i arg z = Log|z|+ iArg z + 2πki, k = 0,±1,±2, . . .

Examples

log 3 = Log3 + 2πki,


log(−1) = (2k + 1)πi,

log(1 + i) = Log√

2 + i(π/4 + 2πk), where k = 0,±1,±2, . . . .

Given this definition of the logarithm the following properties of log z areeasily proven:

1. If z 6= 0 then z = elog z,

2. log ez = z + 2πki , k ∈ Z.

3. log(z1z2) = log z1 + log z2, log(z1/z2) = log z1 − log z2.

Furthermore one can show that f (z) = log z is holomorphic in the do-main D∗ consisting of all points in the complex plane except those on thenonpositive real axis, i.e. D∗ = C \ (−∞, 0], and

ddz

log(z) =1z

for z ∈ D∗. (65)

But there is a catch here: for a given w there is more than one z whichsatisfies the equation log w = z for the reason that if w = ez holds thenw = ez+2πki also holds for any fixed integer k. Hence we define a uniquebranch of the logarithm

Definition 3.5 (Principle branch of the logarithm). For any z ∈ C \ {0},we define the principle branch of the logarithm

Logz := log |z|+ iArg z ,

where Arg z is the principle argument of z and log |z| is the logarithmin the real numbers.

Any other convention for Arg z would yield a different branch of thelogarithm.


3.4 Conformal Mappings

Complex differentiability is different from real differentiability as we saw inthe example of the complex function f (z) = z̄ which is nowhere differen-tiable.

To understand the Example it is helpful to view matters not algebraicallybut geometrically. In real analysis we have a potent

mean of visualising the derivative f ′

of a function f : R → R, namely,as the slope of the graph y = f (x).Unfortunately, due to our lack of four-dimensional imagination, we can’t drawthe graph of a complex-valued function,and hence we cannot generalise thisparticular conception of the derivativein any obvious way.

We know that for a given function f : C→ C we can write

f (x + iy) = u(x, y) + iv(x, y),

with x, y, u, v real, obtaining the map

T : R2 → R2,

(xy

)7→(

u(x, y)v(x, y)

).

We know from multi-variable calculus that the Jacobian matrix

J =

(ux uyvx vy

)(66)

describes the ‘local’ behaviour of such a map T. If a map T obtained from acomplex function f is differentiable in the sense of real-valued multi-variablecalculus then the following statements are equivalent.

1. f is complex differentiable at z0.

2. h 7→ f (z0 + h)− f (z0) is locally (i.e. at z0) the composition of a rotationand an expansion or contraction.

3. The Jacobian matrix of the map T satisfies(ux uyvx vy

)= λ

(cos θ − sin θ

sin θ cos θ

)with λ, θ ∈ R and λ ≥ 0.

4. The function f satisfies the Cauchy-Riemann conditions,ux = vy, uy = −vx.

Thus z → z̄ is not complex differentiable because it is a reflection, anoperation which cannot be represented as a combination of rotation andexpansion or contraction. To see that the limit does not exist note that( f (z + h)− f (z))/h = h̄/h which equals +1 if h is real and −1 if h is purelyimaginary. Because of the geometric significance

of complex differentiability it is moremeaningful to ask whether a functionis differentiable in an open set ratherthan whether it is differentiable at asingle point. This is why the notion ofholomorphic is more useful than thenotion of analytic alone.

Let us stay with the Jacobian matrix for a little while longer. The Jacobiandeterminant D of the map T is

D = uxvy − uyvx (67a)

= u2x + v2

x using the CR-equations (67b)

= | f ′|2 , (67c)

If we assume that | f ′|2 6= 0, then the function f maps a neighbourhood of apoint z uniquely and reversibly on to a neighbourhood of a point ξ in a waythat angles are preserved. This is captured in the term conformal mapping.To define it fully we need to say what we mean by a path in the complexplane.


Definition 3.6 ((Smooth) Path or curve). A path or curve in the complexplane is the range of the continuous function γ : [a, b] → C given byγ(t) = x(t) + iy(t), t ∈ [a, b].

A path γ is smooth if (i) γ′ exists and is continuous on [a, b], and (ii)γ′ 6= 0∀t ∈ (a, b).

Definition 3.7 (Conformal map). A map f : A → C is called conformalat z ∈ A if it is one-to-one in a neighbourhood of z and for every pairof smooth paths γ1, γ2 intersecting at z the angle between γ1 and γ2at z is equal to the angle between the images f (γ1) and f (γ2) at z inmagnitude and sense. If f is conformal at every z ∈ A then f is aconformal map in A.

In other words a mapping is conformal if angles are left unchanged by it.Indeed, the CR-equations are necessary and sufficient for a map f to be con-formal. And hence conformal is the norm in the world of complex functions.

Theorem 3.4. If a complex function f : A→ C is holomorphic at z0 ∈ Aand f ′(z0) 6= 0 then the mapping f (z) is conformal at z0.

That is, if a function f (z) is differentiable on a neighbourhood of z the corre-sponding map acting on z is locally conformal. It is only locally conformalsince the Jacobian is only a linear approximation to the function f (z) near thepoint z.

Examples 1. The map f (z) = ez is conformal everywhere in C.

2. The map f (z) = z2 is conformal everywhere in C except at z = 0.

3. The Möbius transformation f (z) = (az + b)/(cz + d) is conformal every-where in C except at z = −d/c.


4 Integration of holomorphic complex functions

The central fact of differential and integral calculus of real variables is thatthe integral of a function may be regarded as the ‘primitive’ function or‘indefinite integral’ of the original function. We will obtain a correspondingrelation for functions of a complex variable.

4.1 Definition of the line integral

We begin by extending the definition of a definite path integral from the real-valued to the complex functions. Take a smooth curve beteen two points inthe complex plane. We subdivide the curve into k sections with correspond-ing intersection points z0, z1, . . . , zk. Let z′i be any point on the curve betweenzi and zi+1. We now form the sum

Sk =k

∑n=1

f (z′n)(zn − zn−1) . (68)

Making the partitioning of the curve finer and finer such that the length ofthe largest section tends to zero we obtain a sum which is independent onthe exact partitioning and the curve as long as the entire curve is inside A. Asketch of the proof goes as follows, using results from real integrals.

We put f (z) = u(x, y) + iv(x, y), zn = xn + iyn, and z′n = x′n + iy′n. Also, let∆zn = zn − zn−1 = ∆xn + i∆yn. Then we have

Sk =k

∑n=1

(u(x′, y′)∆xn − v(x′, y′)∆yn

)+ i

k

∑n=1

(v(x′, y′)∆xn + u(x′, y′)∆yn

).

(69)

As k increases the RHS tends to the real integrals

Sk →∫

γ(udx− vdy) + i

∫γ(vdx + udy)ask→ ∞ . (70)

We call this limit the definite integral of the function f (z) along the curveγ from a to b, denoted

∫γ f (z)dz. Thus,∫

γf (z)dz =

∫γ(udx− vdy) + i

∫γ(vdx + udy) . (71)

We conclude that a complex-valued function f on a real interval [a, b] iscalled integrable, if Re f, Im f are integrable functions in the sense of realanalysis. We define the definite integral, also called contour integral or lineintegral as follows.

Definition 4.1 (Line integral). Let f : A → C be a complex function,z ∈ A ⊂ C, and t ∈ [a, b] ⊂ R be a real variable. Then the line integralalong a smooth path γ is defined as∫

γf (z)dz =

∫ b

af (γ(t))

dγ(t)dt

dt . (72)


Examples 1. Suppose z = x + iy and f (z) = x2 + iy2, and consider the curveγ(t) = t + it, 0 ≤ t ≤ 1. Find γ′(t). Then compute the integral

∫γ f (z)dz.

Solution.γ′(t) = 1 + i and∫

γf (z)dz =

∫ 1

0(t2 + it2)(1 + i)dt = (1 + i)2

∫ 1

0t2dt = 2i/3 .

2. Letf (z) =

1z

and takeγ(t) = R cos t + iR sin t, 0 ≤ t ≤ 2π, R 6= 0 .

Find f in terms of (x, y). Then compute∫

γ f (z)dz.

Solution.

f (x, y) =x

x2 + y2 − iy

x2 + y2 ,∫γ

f (z)dz =∫ 2π

0

(cos t

R− i

sin tR

)(−R sin t + iR cos t)dt

=∫ 2π

0idt = 2πi .

That is, the integral of 1/z around any circle of non-zero radius centred atthe origin (traversed counter-clockwise) is 2πi.

Recall that in real analysis we know that for real valued function f (x) :[a, b]→ R, it holds that ∣∣∣∣∫ b

af (x)dx

∣∣∣∣ ≤ ∫ b

a| f (x)| dx . (73)

An equivalent inequality exists for complex functions.

∣∣∣∣∫γ

f (z)dz∣∣∣∣ ≤ ∫

γ| f (z)| |dz| . (74)

Applying Eq. 74 to Eq. 72 yields the ML-Inequality, also known as Esti-mation lemma.

Lemma 4.1 (ML-Inequality). Let f : A→ C be continuous on domain A ⊂ C,γ be a smooth curve of length L in A, and | f (z)| ≤ M for all z ∈ γ ⊂ A. Then∣∣∣∣∫

γf (z)dz

∣∣∣∣ ≤ ML . (75)

Examples 1. Let C be the unit circle and suppose | f | ≤ 1 on C. ThenM = 1, L = 2π, and ∣∣∣∣∫c

f (z)dz∣∣∣∣ ≤ 2π .

Compare this to the integral of f (z) = 1/z along the unit circle centred atthe origin.


2. Let C be given by C(t) = 2eit, 0 ≤ t ≤ 2π. Then,∣∣∣∣∫C

ez

z2 + 1dz∣∣∣∣ ≤ 4πe2

3.

4.2 Independence of Path

An essential result in complex analysis is the fact that path integrals in thecomplex plane are independent of the path and only dependent on the endpoints.

To begin with, let f be a continuous function7 in a domain A. A function 7 A function is continuous on a set S ifthe limit limz→z0 f (z) = f (z0)∀z0 ∈ S.F such that F′(z) = f (z), ∀z ∈ A is called an anti-derivative of f .

Theorem 4.2 (Independence of Path). Let f : A → C be continuous on adomain A ⊂ C and have an anti-derivative F continuous on A. Then forpath γ ⊂ A joining z0 and z1 in A, we have∫

γf (z)dz = F(z1)− F(z0) . (76)

In particular, if γ is a closed curvein A, then∫γ

f (z)dz = 0 . (77)

Example Let C be the unit circle centred at the origin and f (z) = 1/z2.Find a domain A such that f is continuous on A and C ⊂ A. Find the anti-derivative F. This way, confirm that∫

Cf (z)dz = 0 .

Solution. F′ = (−1/z)′ = 1/z2 is continuous on the unit circle. Choose A s.t.it contains the unit circle but not the origin, e.g. A = {z : 1/2 < |z| < 3/2}.

In fact∫

C f (z)dz = 0 for any f (z) = 1/zk, integer k 6= 1 and closed curve Cnot passing through the origin.

Example Let γ be the part of the unit circle joining 1 to i in the counterclock-wise direction and f (z) = ez. Compute

∫γ f (z)dz.

Solution. ∫γ

f (z)dz = ei − e .

Example Suppose C is the circle z0 + reiθ traversed counter-clockwise,0 ≤ θ ≤ 2π, and |a− z0| > r. Find a domain A such that f is continuous onA and C ⊂ A. Find the anti-derivative F. This way, confirm that∫

Cf (z)dz = 0 .


Solution. F′ = log(z − a) is continuous on any domain not containingpoints on the negative real axis, including origin. E.g. choose e.g. A = {z :|z− z0| < |a− z0|} which contains C.

A consequence of Theorem 4.2 is the so-called deformation property.

Theorem 4.3 (Deformation Theorem). Let C and C′ be two equally ori-ented, simple, closed curves with C′ interior to C. Let f be holomorphicon a closed region containing C and C′ and the points between them.Then, ∫

Cf =

∫C′

f . (78)

4.3 Cauchy’s Integral Formula

Cauchy’s Theorem leads to a fundamental formula, again due to Cauchy,which expresses the value of a holomorphic function f (z) at any point z =

z0 in the interior of a simply connected region A, throughout which thefunction is holomorphic, by means of the values the function takes on theboundary C.

In the following we assume that the function f (z) is holomorphic in thesimply-connected region A and on its boundary C. Then the function

g(z) =f (z)

z− z0(79)

is also holomorphic everywhere in A and on its boundary C except at thepoint z0. First we make a new curve C′ by deforming the closed curve C toa unit circle centred at the point z0. Then we known from the DeformationProperty that ∫

Cg(z)dz =

∫C′

g(z)dz . (80)

Let us rewrite the RHS,∫C′

g(z)dz =∫

C′

f (z0)

z− z0dz +

∫C′

f (z)− f (z0)

z− z0dz . (81)

We know from a previous example that∫C′

f (z0)

z− z0dz = f (z0)

∫C′

1z− z0

dz = f (z0)2πi . (82)

Hence, we have ∫C

g(z)dz = f (z0)2πi +∫

C′

f (z)− f (z0)

z− z0dz (83)

and will now let the radius r of C′ go to zero, r → 0. Neither the LHS nor thefirst term on the RHS do depend on r. Hence, the second term must remainunchanged when r → 0. Since f is continuous on A there must exist an Msuch that ∣∣∣∣ f (z)− f (z0)

z− z0

∣∣∣∣ = | f (z)− f (z0)|r

≤ Mr

. (84)


We know from the ML-Inequality that then∣∣∣∣∫C′

f (z)− f (z0)

z− z0dz∣∣∣∣ ≤ M

rL(C′) =

Mr

2πr = 2πM . (85)

Since f is continuous M→ 0 as r → 0. Thus, we find that

limr→0

∫C′

f (z)− f (z0)

z− z0dz = 0 (86)

and therefore ∫C

f (z)z− z0

dz = 2πi f (z0) . (87)

We have found the following theorem.

Theorem 4.4 ( Cauchy’s Integral Formula). Let f be holomorphic ona simply connected domain A, let C be a simple, closed, positivelyoriented curve C ⊂ A, and z0 ∈ IntC. Then

f (z0) =1

2πi

∫C

f (z)z− z0

dz . (88)

We see that the value of a complex function at a point z0 can be expressedas a line integral along a simple closed curve around the point.

Example Compute the integral∫γ

e2z + sin zz− π

dz ,

where γ is the circle |z− 2| = 2 traversed counter-clockwise.

Solution. Checking the conditions of the Theorem 4.4 we find the integral isequal to 2πie2π .

It is interesting to note that, if C is a circle C : z = z0 + reiθ with centre z0then

f (z0) =1

2π

∫ 2π

0f (z0 + reiθ)dθ .

In other words, the value of a complex function at the centre of a circle isequal to the mean of its values on the circumference, provided that the closedarea of the circle is a region in which the function is holomorphic. This isknown as Gauss’ mean-value property.

4.4 Taylor’s Theorem and Cauchy’s Integral Formula for Derivatives

Cauchy’s formula has a number of important theoretical applications, thechief of which is the proof of the fact that every holomorphic function canbe expanded in a power series. That is, that every holomorphic function isanalytic. We have the following theorem.


Theorem 4.5 (Taylor’s Theorem). Let f : A → C be holomorphic onsome open disk D(z0, R) ⊂ A. Then f can be expanded in a powerseries in (z− z0) which converges in the interior of the disk,

f (z) =∞

∑n=0

cn(z− z0)n, for |z− z0| < R . (89)

To prove this let us start with the integrand in Eq. 88. If we think of z0 as thevariable here and write Cauchy’s Integral Formula as follows.8 8 Here z0 is the centre of the circle,

z is a point in the circle, and ζ is onthe boundary of the circle. Hence,|ζ − z0| > |z− z0|.

f (z) =1

2πi

∫C

f (ζ)ζ − z

dζ . (90)

We rewrite 1ζ−z using the geometric series,

1ζ − z

=1

ζ − z + z0 − z0(91)

=1

ζ − z0

11− z−z0

ζ−z0

(92)

=1

ζ − z0

(1 +

z− z0ζ − z0

+

(z− z0ζ − z0

)2+ . . .

). (93)

Putting this expression back into Cauchy’s Integral Formula and exchang-ing sum and integral9, we obtain 9 This is possible when the sum con-

verges uniformly which we assumewithout proving it here.f (z) =

12πi

∫C

f (ζ)ζ − z0

∞

∑n=0

(z− z0ζ − z0

)ndζ (94)

=1

2πi

∞

∑n=0

∫C

f (ζ)(z− z0)n

(ζ − z0)n+1 dζ (95)

=1

2πi

∞

∑n=0

(z− z0)n∫

C

f (ζ)(ζ − z0)n+1 dζ . (96)

In other words we have written f (z) as a power series in (z− z0),

f (z) =∞

∑n=0

cn(z− z0)n (97)

with coefficients

cn =1

2πi

∫C

f (ζ)(ζ − z0)n+1 dζ , (98)

which proves Taylor’s Theorem.This might look like not much of a gain. But remember that every power

series of an holomorphic function can be differentiated arbitrarily manytimes within its circle of convergence. We also know now that this is true forintegration. Hence, integration and differentiation of complex holomorphicfunctions can be carried out without restriction.

Every power series expansion is the Taylor series of the function which itrepresents. Hence, we can write

f (z) =∞

∑n=0

f (n)(z0)

n!(z− z0)

n (99)


with coefficients cn =f (n)(z0)

n! .Combining Eqs. 97–98 immediately leads us to the following generalisa-

tion of Cauchy’s integral formula.

Theorem 4.6 ( Cauchy’s Integral Formula for Derivatives). Let f : A →C be holomorphic on some simply connected domain A ⊂ C and letC ⊂ A be a simple, closed, positively oriented curve and z0 ∈ int(C) bea point in the interior of C. Then

f (k)(z0) =k!

2πi

∫C

f (z)(z− z0)k+1 dz , k = 1, 2, . . . (100)

Example Compute∫

C sin(3z)/z4dz, where C is the unit circle |z| = 1 tra-versed counter-clockwise.

Solution. Since sin 3z is holomorphic on an open, simply connected setcontaining the unit circle (in fact, it is entire) we can use Eq. 100 with z0 = 0and k = 3: ∫

C

sin(3z)z4 dz =

2πi3!

f ′′′(0) = −9πi . (101)

We have the following consequence of Cauchy’s Integral Formula forDerivatives.

Theorem 4.7 (Liouville’s Theorem.). Any bounded entire function isconstant.

Liouville’s Theorem can be shown by upper bounding f (n)(z) using theML-Inequality and showing that that bound goes to zero for any n > 0 andradius of integration R→ ∞. Liouville’s Theorem provides another

proof for cos(z) and sin(z) beingunbounded in the complex plane.


5 Zeros, poles, and residues of holomorphic functions

The fact that any function holomorphic in an open disk can be expanded ina power series has consequences for functions which are not holomorphicat points inside such a disk. Let the function f (z) vanish at a point z = z0,that is f (z0) = 0, but say it is differentiable at the point z0. Then the constantterm in its Taylor series will vanish and so will, possibly, higher order terms.We can then write f (z) in terms of a new function q(z) as

f (z) = (z− z0)mq(z) , (102)

where q(z0) 6= 0. A point z0 for which Equation 102 holds is said to be a zeroof order m of the function f .

Definition 5.1 (Zero of order m). Let f : A → C, be holomorphic atz0 ∈ A ⊂ C. f has a zero of order m at z0 if f (z0) = f ′(z0) = · · · =f (m−1)(z0) = 0 and f (m)(z0) 6= 0. A zero of order 1 is called a simplezero.

Examples 1. The function f (z) = (z− 3)2 has a zero of oder 2 at z0 = 3.

2. Find the zeroes and their order of the function f (z) = sin(z).

Solution. has simple zeroes at kπ.

3. Find the zeroes and their order of the function f (z) = z2 sin(z).

Solution. has a zero of order 3 at z0 = 0. And simple zeroes at kπ.

The reciprocal, call it g, of a differentiable function f is also differentiableexcept at the points where f vanishes, that is, except precisely at the zeros off . Using Equation 102 we can write

g(z) =1

f (z)=

1(z− z0)m

1q(z)

=1

(z− z0)m r(z) , (103)

where r(z) is the reciprocal of q. The point z0 is a singularity of the functiong. In particular if f is non-zero in an open disk around z0, i.e. excluding thepoint z0 itself, we call this singularity isolated. We give that type of disk aname of its own, a punctured disk, also called a deleted neighbourhood.

Definition 5.2 ((Isolated) singularity / pole of order m). A complexfunction g : A → C has a singularity (or pole) of order m at the pointz0 ∈ A if 1/g has a zero of order m. For m = 0 we call it a removablesingularity. If there exists an R > 0 such that g is holomorphic in apunctured disk D(z0, R) \ {z0} the singularity is called isolated.


If there is no finite m such that g can be written as in Equation 103 thesingularity is called essential.10 10 The canonical example of an essential

singularity is the point z = 0 of thefunction e1/z. Expanding it out into apower series will reveal the essentialnature of the singularity.

Since the function q(z) in Equation 103 is well defined for all points in theopen disk D(z0, R) and q(z0) 6= 0 the function r(z) is holomorphic withinthat disk. Hence, there exists a power series expansion

r(z) = c̃0 + c̃1(z− z0) + c̃2(z− z0)2 + . . . (104)

around the point z0 and we can write

g(z) = c̃0(z− z0)−m + c̃1(z− z0)

−m+1 + · · ·+ c̃m + c̃m+1(z− z0) + . . . (105)

= c−m(z− z0)−m + · · ·+ c−1(z− z0)

−1 + c0 + c1(z− z0) + . . . (106)

with new coefficients c̃n = cn−m. The expansion in Equation 106 is also calledthe Laurent expansion of the function g. The radius of the convergenceof the Laurent expansion in Equation 106 is the same as that of the Taylorexpansion in Equation 104. For m = 0 the singularity is remov-

able and the Laurent expansion of gbecomes a Taylor expansion.

This might seem like a lot of acrobatics. But the coefficient c−1 in Equa-tion 106 is so important for the integration of complex functions that it willreceive a special name: the residue of the function g at the point z0.

Examples 1. f (z) = 1/(z− 3) has an isolated singularity of order 1 at z = 3.

2. f (z) = (z + 1)/[z4(z2 + 1)] has singularities z0 = 0, i,−i which are allisolated. The order of z0 = 0 is 4, the order of z0 = i,−i is 1.

We can find the order of a singularity as follows:

i If f has an isolated singularity at z0 and if

limz→z0

(z− z0) f (z) = 0 , (107)

then the singularity is removable.

ii If f has an isolated singularity at z0 and if there exists a positive integerN such that

limz→z0

(z− z0)N f (z) 6= 0 (108)

but limz→z0

(z− z0)N+1 f (z) = 0 , (109)

then z0 is a pole of order N.

Examples For the following examples, consider the power series expansionof sin z.

1. The function f (z) = sin z/z has a removable singularity at z0 = 0.

2. Find the pole and determine its order of the function f (z) = sin z/z2.

Solution. has a pole at z0 = 0 of order 1, since limz→0(z− 0)1 f (z) = 1 6= 0and limz→0(z− 0)2 f (z) = 0.


Definition 5.3 (Meromorphic function). A function which is holomor-phic in A ⊂ C except for poles is called meromorphic.

The class of meromorphic functions includes the holomorphic functions.

5.1 Cauchy’s Residue Theorem

Let us formalise the concept of a residue and learn one of the main results incomplex function theory, Cauchy’s Residue Theorem.

Definition 5.4 (Residue). Let f : A → C, A ⊂ C, be holomorphic ina punctured disk D(z0, R) \ {z0} of an isolated singularity z0 so thatit can be expanded in a Laurent series f (z) = ∑∞

j=−∞ cj(z − z0)j. The

coefficient c−1 is called the residue of f at z0, denoted Res[ f ; z0].

Examples We can see from previous examples that

1. Res [e1/z; 0] = 1; consider the power series expansion of ez.

2. Res [ 11−z ; 0] = 0; see geometric series.

3. Res [ (z+1)2

z ; 0] = 1; factoring out the square.If f (z0) has a removable singularity atz0 then all the negative powers of thepower series expansion are zero, and so,in particular, the residue at z0 is zero.

The residue is given by

Res [ f ; z0] ≡ c−1 =1

2πi

∫C

f (z)dz. (110)

Read from right to left we have the powerful result that to evaluate an inte-gral it suffices to evaluate a residue!

If f has a pole of order N at z0 then

Res ( f ; z0) = limz→z0

1(N − 1)!

dN−1

dzN−1 [(z− z0)N f (z)] (111)

From this we get to the following special cases. If f has a simple pole atz0, then

Res [ f ; z0] = limz→z0

(z− z0) f (z) . (112)

If f = A(z)/B(z) has a simple pole at z0 and A(z), B(z) are differentiableat z0, then

Res [ f ; z0] =A(z0)

B′(z0). (113)

Examples 1. Res [csc z; 0] = 1/ cos 0 = 1 ,

2. Res [1/(z4 − 1); i] = 1/(4i3) = i/4 .

This leads us to one of the main results in complex analysis.


Theorem 5.1 (Cauchy’s Residue Theorem). Let C be a positively ori-ented, simple closed curve and f be holomorphpic on and inside Cexcept at a finite number of points z1, z2, . . . , zk ∈ Int(C), then

∫C

f (z)dz = 2πik

∑j=1

Res [ f ; zj] .

Examples From previous examples for singularities and residues we canimmediately find the values of the following integrals:

∫{|z|=1} e1/zdz = 2πi;∫

{|z|=1/2}1

1−z dz = 0;∫{|z|=1}

(z+1)2

z = 2πi .

Example Evaluate∫

dz/(z4 + 1), where C goes in a half circle from 2 via 2ito −2 and back in a straight line to 2. The singularities of the integrand occurat the fourth roots of −1, i.e. at eπi/4, e3πi/4, e5πi/4, e7πi/4. But only eπi/4 ande3πi/4 are inside the curve. Hence, we have∫

Cdz/(z4 + 1) = 2πi Res (1/(z4 + 1); eπi/4) + 2πi Res (1/(z4 + 1); e3πi/4)

(114)

= 2πi(

14(eπi/4)3 +

14(e3πi/4)3

)= 2πi

14

(e−3πi/4 + e−πi/4

)(115)

= 2πi14

(−eπi/4 + e−πi/4

)=

πi2(−2i sin(π/4)) =

π√2

(116)


6 Integrals along the real line

The theory of residues is used to compute certain types of definite as well asimproper real integrals. Some of these integrals occur in physical and engi-neering applications, and often cannot be evaluated by using the methods ofreal calculus.

6.1 Integrals of the form∫ ∞−∞

P(x)Q(x)dx

Integrals of the form∫ ∞−∞ P(x)/Q(x)dx where P and Q are polynomials will

converge if Q(x) 6= 0 and deg Q− deg P ≥ 2. To show this, let CR be theclosed contour consisting of the real line segment from −R to R and theupper semi-circle ΓR centered at the origin and of radius R. Let’s make Rlarge enough to enclose all zeroes of Q lying in the upper half-plane.

In this case, we note that∫ ∞

−∞

P(x)Q(x)

dx = limR→∞

∫ R

−R

P(x)Q(x)

dx . (117)

We know, by the Residue Theorem, that∫CR

P(z)Q(z)

dz = 2πi ∑zj∈UHP

Res[

PQ

; zj

], (118)

where the points zj are the zeroes of Q in the upper half-plane. We nowupper-bound the integral

∫ΓR

P/Q using the ML-inequality.∣∣∣∣∫ΓR

P(z)Q(z)

dz∣∣∣∣ ≤ πR

AR2 (119)

for some A ∈ R.11 Hence, taking the limit R → ∞ the integral goes to zero 11 Here, we used the fact that deg Q−deg P ≥ 2.and we can conclude that∫ ∞

−∞

P(x)Q(x)

dx = 2πi ∑zj∈UHP

Res[

P(z)Q(z)

; zj

]. (120)

Example ∫ ∞

−∞

dxx4 + 1

(121)

The zeros of Q(z) = z4 + 1 in the upper half-plane are z1 = eiπ/4 andz2 = e3iπ/4, each of which is a simple pole. The residues are given by thevalue of 1/4z3 at the poles. Thus,

∫ ∞

−∞

dxx4 + 1

= 2πi(

14e3πi/4 +

14e9πi/4

)= 2πi

14√

2((−1− i) + (1− i)) =

π√

22

.

(122)

6.2 Integrals of the form∫ ∞−∞R(x) cos(x)dx or

∫ ∞−∞R(x) sin(x)dx

Another kind of real integrals which can be solved easily using Cauchy’sResidue Theorem are integrals of the form

∫ ∞−∞R(x) cos(x)dx or


where R(x) = P(x)/Q(x), P and Q are polynomials, Q(x) 6= 0 and deg Q >

deg P. We use a similar argument as above and refer to the same contoursCR and ΓR. Consider the integral12 12 Integrating R(z) cos(z) along the

contour CR is not helpful in this casesince cos(z) is not bounded.


∫CR

R(z)eizdz (123)

Note that on ΓR, z = Reiθ and |R(z)| ≤ A/R for some A ∈ R. Then Note, that for 0 ≤ θ ≤ π/2 we havesin θ ≥ 2θ/π.∣∣∣∣∫ΓR

R(z)eizdz∣∣∣∣ ≤ ∫ π

0

∣∣∣R(eiθ)∣∣∣ ∣∣∣eiR(cos θ+i sin θ)

∣∣∣ Rdθ ≤ A∫ π

0e−R sin θdθ (124)

= 2A∫ π/2

0e−R sin θdθ ≤ 2A

∫ π/2

0e−2Rθ/πdθ (125)

= πAR(1− e−R)→ 0 as R→ ∞ . (126)

Hence, ∫ΓR

R(z)eizdz→ 0 as R→ ∞ (127)

and thus ∫CR

R(z)eizdz→∫ ∞

−∞R(x)eixdx as R→ ∞. (128)

Then, by applying the Residue Theorem, we get∫ ∞

−∞R(x)eixdx = 2πi ∑

zj∈UHPRes

[R(z)eiz; zj

]. (129)

Using Euler’s formula we find the integrals∫ ∞−∞R(x) cos(x)dx and


as the real and imaginary part, respectively, of∫ ∞−∞R(x)eixdx:

∫ ∞

−∞R(x) cos(x)dx = Re

2πi ∑zj∈UHP

Res[R(z)eiz; zj

] , (130)

and

∫ ∞

−∞R(x) sin(x)dx = Im

2πi ∑zj∈UHP

Res[R(z)eiz; zj

] . (131)

Example Let R(x) cos(x) = cos(x)x2+b2 , b 6= 0. Note that eiz/(z2 + b2) has a

simple pole at ib. The residue at this pole is 1/(2ib)e−b.

∫ ∞

−∞

cos(x)x2 + b2 dx = Re

[2πi Res

(eiz

z2 + b2 ; ib

)]=

π

be−b . (132)

6.3 Integrals of the form∫ ∞−∞R(x) cos(ax)dx or

∫ ∞−∞R(x) sin(ax)dx

There are two cases to consider, a > 0 and a < 0. We will show that∫ ∞

−∞R(x)eiaxdx = 2πi ∑

zj∈UHPRes

[R(z)eiaz; zj

], a > 0 , (133)

∫ ∞

−∞R(x)eiaxdx = −2πi ∑

zj∈LHPRes

[R(z)eiaz; zj

], a < 0 . (134)

(135)

Using Euler’s formula you find the integrals∫ ∞−∞R(x) cos(ax)dx and∫ ∞

−∞R(x) sin(ax)dx as the real and imaginary part, respectively.


For a > 0, the proof follows along the lines in the previous Section. Fora < 0 we proceed as follows. Using∣∣∣eiaz

∣∣∣ = e−aRsinθ = e|a|Rsinθ (136)

and taking the contour Γ′ = eiθ , θ ∈ [0,−π] we know that sin θ ≤ 2θ/π and

e|a|Rsinθ ≤ e|a|R2θ/π . (137)

With this we can bound the integral∣∣∣∣∫Γ′RR(z)eiazdz

∣∣∣∣ ≤ 2A∫ −π/2

0e|a|R2θ/πdθ (138)

= 2Aπ

|a|R2eR2θ/π

∣∣−π/20 (139)

= πA|a|R (e−|a|R − 1)→ 0 as R→ ∞ . (140)

The claim follows.

6.4 Integrals of the form∫ 2π

0 R(cos θ, sin θ)dθ

We consider integrals of the form∫ 2π

0 R(cos θ, sin θ)dθ where R(cos θ, sin θ)

is a rational function with real coefficients of cos θ and sin θ and whose de-nominator is non-zero on the interval [0, 2π]. Such integrals can be computedby suitable substitutions (e.g. u = tan(t/2)), which reduce the given inte-gral to the integral of a rational function, which can be computed by meansof the partial fraction decomposition. It is often much simpler to apply theResidue Theorem, by interpreting the given integral as an integral along asuitable closed curve. Let C be the positively oriented unit circle |z| = 1,parametrised by z = eiθ . Thus 1/z = e−iθ . Since cos θ = (eiθ + e−iθ)/2 andsin θ = (eiθ − e−iθ)/(2i) we can write cos θ and sin θ in the new parametrisa-tion as

cos θ =12

(z +

1z

)and sin θ =

12i

(z− 1

z

). (141)

To reparametrise the integral, we differentiate eiθ along C:

dzdθ

= iz and hence dθ = −idzz

. (142)

Now, the integral∫ 2π

0 R(cos θ, sin θ)dθ can be rewritten

∫ 2π

0R(cos θ, sin θ)dθ =

∫C={|z|=1}

R( z + z−1

2,

z− z−1

2i)

dziz

. (143)

The RHS can then be evaluated with the residue theorem.

Example 13 Let C = {z : |z = 1|}. 13 In this example, only one of thetwo poles is inside of the circle ofintegration.

∫ 2π

0

dθ

2 + cos(θ)=

2i

∫C

dzz2 + 4z + 1

(144)

= 4π Res(

1z2 + 4z + 1

;√

3− 2)=

2π√3

(145)


6.5 The Fresnel integrals

The Fresnel integrals occur in the theory of diffraction. They are given by∫ ∞

0cos x2dx =

∫ ∞

0sin x2dx =

12

√π

2(146)

To derive them we need the following two results: Inequality 147 can easily be seenby drawing the graph of sin x for0 ≤ x ≤ θ/2 and that the straight linebetween 0 and θ/2.

2π

θ < sin θ , 0 < θ <π

2and (147)

L =∫ ∞

0e−x2

dx =

√π

2. (148)

To establish Eq. 148 we note first that

L2 =

(∫ ∞

0e−x2

dx)(∫ ∞

0e−y2

dy)=∫ ∞

0

∫ ∞

0e−(x2+y2)dxdy (149)

and then use polar coordinates to obtain

L2 =∫ π/2

0

∫ ∞

0e−r2

rdrdθ = −12

e−r2 ∣∣∞0 ×

π

2=

π

4. (150)

Now, to establish Equations 146, we consider the entire function f (z) =

eiz2and the contour γ = OA + AB + BO, where O is the origin, A = ρ,

B = ρeiπ/4, and AB follows the circle of radius ρ. Since f (z) is entire weknow that ∫

γeiz2

dz =∫

OAeiz2

dz +∫

ABeiz2

dz +∫

BOeiz2

dz = 0 . (151)

We can rewrite these integrals using the properties of the contour γ:∫γ

eiz2dz =

∫ ρ

0eix2

dx +∫ π/4

0eiρ2 exp 2iθρieiθdθ +

∫ 0

ρeir2 exp πi/2eπi/4dr = 0 .

(152)

Now, in view of Eq. 147, we have∣∣∣∣∫ π/4

0eiρ2 exp 2iθρieiθdθ

∣∣∣∣ ≤ ρ∫ π/4

0

∣∣∣eiρ2(cos(2θ)+i sin(2θ)∣∣∣ dθ

≤ ρ∫ π/4

0e−ρ2 sin(2θ)dθ ≤ ρ

∫ π/4

0e−ρ2(4θ/π)dθ

=π

41− e−ρ2

ρ→ 0 as ρ→ ∞ , (153)

whereas Eq. 148 gives

limρ→∞

∫ 0

ρeir2 exp πi/2eπi/4dr = e−πi/4

∫ ∞

0e−r2

dr = −1 + i√2

√π

2. (154)

Combining Eqs. 152 - 154, we get

limρ→∞

∫ ρ

0eix2

dx =∫ ∞

0(cos x2 + i sin x2)dx =

1 + i√2

√π

2, (155)

which on comparing the real and imaginary part gives Eqs. 146.

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