Math20001 dec 2015

Methods of Complex Functions
Lecture Notes Bristol MATH20001
Karoline Wiesner, School of Mathematics, University of Bristol
2015 (updated 10 November 2015)
Reading list
There 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’s
web 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 their
relevance in physics I highly recommend The Road To Reality: A
Complete Guide to the Laws of the Universe by Roger Penrose, in
particular Chapters 4, 5, 7, and 8.
Some course information:
• Drop-in sessions Tuesdays 3.30pm-4.30pm, ofﬁce 3.1a, beginning
week 8.
• Homework given out every Friday, hand-in following Friday 10am
in 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 to
be announced.
• Maths Cafe with Dan Taylor Lewis: Friday 12pm - 1pm, Week
commencing 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.

methods of complex functions lecture notes bristol math20001 2

1 Complex numbers
section 1 is a quick review of what you
know 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 actually
negative. Yet, this is a situation similar to when people were looking for a
square root of the number 2 which has no square root within the system of
rational numbers. In that case they resolved the situation by extending their
system of numbers from the rationals (Q) to a larger system, the system of
reals (R). We will do the same by extending the number system of reals by
introducing a single quantity, called ‘i , which is to square to −1, and adjoin
it to the system of reals, allowing combinations of i with real numbers to
form 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 as
ordered pairs of real numbers.
Definition 1.1 (The complex numbers). The complex numbers are the
set of ordered pairs of real numbers (x, y) ∈ R2 with addition and
multiplication 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 multiplication
as well as the distributive law follow easily from the same properties of real
numbers. The additive identity, or zero, is given by (0, 0), and hence the
additive inverse of (x, y) is (−x, −y). The multiplicative identity is (1, 0). To
find the multiplicative inverse of any nonzero (x1, y1) we set
(x1, y1)(x2, y2) = (1, 0), (3)
which has the solution
x2 =
x1
x2
1 + y2
1
, y2 =
−y1
x2
1 + y2
1
. (4)
This shows that the set of complex numbers, denoted C, together with
addition and multiplication as defined in Definition 1.1 form a field. Two complex numbers are equal if and
only if their real and imaginary parts
are equal.
We associate complex numbers of the form (x, 0) with the corresponding
real numbers x. It follows that (x1, 0) + (x2, 0) = (x1 + x2, 0) corresponds
to 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 complex
numbers 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 use
the 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 and
z2 = 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 the
product of z1 and z2. Expanding the factors using the ordinary rules of
algebra we get
(x1 + iy1)(x2 + iy2) = x1x2 + i2
y1y2 + i(y1x2 + x1y2) (8)
= (x1x2 − y1y1) + i(y1x2 + x1y2) (9)
where we have applied the rule i2 = −1 and the final answer for the product
of 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 see
this, we solve (x + iy)2 = a + bi, by multiplying out the square and compare
the real and imaginary parts. That is, we set x2 − y2 = a and 2xy = b which
is equivalent to 4x4 − 4ax2 − b2 = 0 and y = b/2x. Solving first for x2, we
find the two solutions are given by
x = ±
a +
√
a2 + b2
2
(10)
y = ±
−a +
√
a2 + b2
2
sign(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 admits
a solution in the complex numbers. Indeed, we will see that any polynomial
of 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. We
define
the real part of z: Re z = x;
the imaginary part of z: Im z = y ;
the conjugate of z: ¯z = x − iy;
the modulus of z: |z| = x2 + y2;
the argument of z: tan(arg z) = y/x, x, y = 0 .

Note that Im z and Re z are real num-
bers.
Both ¯z and z∗ are used to denote the
conjugate of z.
A special ﬂavour of complex analysis arises because one may think of the
complex numbers C both algebraically as a number system and geometri-
cally as a vector space. It is essential therefore to have a good geometrical
intuition for the complex plane.
To each complex number z = x + iy we associate the point (x, y) in the
Cartesian plane. Real numbers are thus associated with points on the x -axis,
called the real axis while the purely imaginary numbers iy correspond to
points 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 on
the Geometrical Interpretation of Imagi-
nary Quantities”. The mathematician A.
Legendre (1752-1833) mentioned it in a
letter to Francois Francais, a professor
of mathematics. It was published as
part of an article in 1813 in the Annales
de Mathémathiques giving the basics of
complex 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 useful
for 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 their
moduli 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 the
arguments:
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 = −
1
2
+ i
√
3
2
, z3 = −
1
2
− i
√
3
2
.
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 inscribed
in 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 one
vertex 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 defined
modulo 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 numbers
z1 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 number
z ∈ 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. Another
common convention for the principle argument sets Arg z ∈ [0, 2π). Note
that any convention for the principle argument will introduce a discontinuity
into the arg function.
Using these definitions we can identify regions of the complex plane with
subsets of the complex numbers. Let’s look at some examples.

Examples 1. {z : Re z > 0} is represented geometrically by the right
half-plane.
2. {z : z = ¯z} 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 an
open set: A point z0 in a set S is called
interior point if there is a neighbour-
hood of z0 completely contained in S.
If every point in a set S is an interior
point S is an open set.
There is also an alternative way of
defining a closed set: A point z0 in a
set S is called boundary point if every
neighbourhood of z0 contains at least
one point in S and at least one point not
in S. S is closed if it contains all of its
boundary points.
Definition 1.4 (Open disk). A domain D(z0, R) of radius R > 0 centred
on some point z0 and defined as D(z0, R) = {z : |z − z0| < R} is called
an open disk.
Definition 1.5 (Open set, closed set). A (possibly infinite) union or a
finite 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 R
such 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 curve
does 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 ( ) is a simple closed curve, a figure eight (∞) is a
closed curve but not simple, a straight line from some point a to some point
b = 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 any
two 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 be
simply connected if any closed curve in S can be shrunk continuously
inside 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 not
connected.
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 case
that the series diverges, which is to say that it does not settle down to a
particular finite value as we add up more and more of its terms. For an
example, 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 the
terms 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 particular
finite 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 limiting
value 4/3, so the series is now convergent.
Figure 1: In the complex plane, the
functions (1 − z2)−1 and (1 + z2)−1
have the same circle of convergence,
there being poles for the former at
z = ±1 and poles for the latter at
z = ±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 diverge
for that value of x. When we substitute x = 1/2, the answer is 4/3, and the
series 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 ask
whether 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 that
we had before, namely
1 − x2
+ x4
− x6
+ x8
− · · · = (1 + x2
)−1
, (23)
the difference being merely a change of sign in alternate terms. This series
behaves slightly different from the one we studied first. We get the following
behaviour 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 in
x = 1 into the function (1 + x2)−1 we get the number 1/2 which is not the
liit of the series which merely fluctuates between 0 and 1. To get a better
understanding 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 two
places x = 1 and x = −1; but, with (1 + x2)−1, we can see no singularity
at these places and, indeed, no real singularities at all. However, in terms of
the complex variable z, we see that these two functions are much more on a
par with one another. We have noted the singularities of (1 − z2)−1 at two
points z = ±1, of unit distance from the origin along the real axis; but now
we 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 imaginary
axis.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 a
simple type called poles, something we
will 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 it
converges 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 value
of z such that |z − z0| > |ν − z0|. In other words, one can always find a
circle of radius R in the complex plane (where R can be 0 or ∞ or anything
in between) centred at z0 with the property that if the complex number z
lies 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 that
value of z. Whether or not the series converges when z lies actually on the
circle of convergence does not have a general answer. R is called the radius
of convergence. The fixed parameter z0 is the point of expansion, the centre
of 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 is
1.
Figure 2: In the complex plane, the
functions (1 − z2)−1 and (1 + z2)−1
have the same circle of convergence for
a power expansion around the origin.
The poles for the former are z = ±1
and 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 any
z ∈ 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 can
be said about the case {z : |z − z0| = R}.
Examples Find the radius of convergence R. If R < ∞ what happens for
points in the set {z : |z − z0| = R}?

1. ∑∞
n=1 nzn
2. ∑∞
n=1(zn/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 series
also diverges for |z| = 1 since |n1n| = n → ∞ as n → ∞.
2. ∑∞
n=1(zn/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
=
1
n2
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 can
immediately be extended to the complex numbers. In other words, we can
define our first elementary complex functions in terms of their power series
expansion. The most important one are listed here and can be assumed for
the 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 convergence
is ∞. 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 are
represented as a power series. But instead we will start with a more general
point of view. A function of a complex number is, to start with, simply a
mapping from the point (x, y) ∈ R2 to a point (u, v) ∈ R2. This means that
u(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 continuous
derivatives 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 → w = u(x, y) + iv(x, y) (37)
which we may also regard as a map f : R2 → R2,
(x, y) → (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 to
those 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 real
case. 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 by
fx = ux + ivx and fy = uy + ivy respectively, where we assume the latter
exist.
Definition 3.1. A complex function f : A → C, whith domain A ∈ C, is
said to be differentiable at z ∈ A if
lim
h→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 limit
must exist irrespective of the manner in which h approaches 0 in the complex
plane.
Definition 3.2 (Holomorphic function). A complex function is called
holomorphic at z0 ∈ C if it is differentiable for all z in an open disk
D(z0, R).
You say analytic, I say holomorphic.
In some text books on complex analysis
the terms holomorphic and analytic
are used interchangeably, for a good
reason. A function is analytic if it is
locally (i.e. in a neighbourhood) given
by a Taylor series and hence (infinitely)
differentiable. However, in R, not every
function differentiable in a neigh-
bourhood (i.e. holomorphic) can be
represented by a Taylor series. Hence,
in R analytic implies holomorphic but
not vice versa. It is a major theorem
of complex analysis that the two sets
coincide 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 differentiable
at 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 = 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 in
exactly the same way as for real variables. We will say more about differen-
tiability later on. In the first place, notice that the identity
zk
1 − zk
z1 − z
= zk−1
1 + zk−2
1 z + · · · + zk−1
(40)
holds. If we now let z1 tend to z, we get
d
dz
zk
= lim
z1→z
zk
1 − zk
z1 − z
= kzk−1
. (41)
In the same way we get, for a general polynomial of order k,
d
dz
Pk(z) = lim
z1→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 be
differentiated, 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 domain
A = {z : |z − z0| < R} ⊂ C. Then P(z) may be differentiated term-by-
term inside A . That is, the limit
P (z) = lim
z1→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 arbitrarily
many 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
d
dz
ez
= ez
, (47)
d
dz
cos(z) = − sin(z) , (48)
d
dz
sin(z) = cos(z) , (49)
d
dz
cosh(z) = sinh(z) , (50)
d
dz
sinh(z) = cosh(z) . (51)
By differentiating Equation 36 term-by-term we get
d
dz
log(1 + z) =
1
1 + z
z = −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 must
exist regardless of the manner in which h approaches 0 in the complex plane.
Now, consider the following f (z) = ¯z, that is u(x, y) = x and v(x, y) = −y.
To compute the limit, let us set h = r, r ∈ R. We find
lim
r→0
x + r − iy − (x − iy)
r
= 1 (53)
while setting h = is, s ∈ R, we find that
lim
is→0
x − iy − is − (x − iy)
is
= −1 . (54)
Hence, the limit h → 0 is not unique and so does not exist – the function
f (z) = ¯z is not differentiable anywhere in C.
From this example, it becomes clear that complex differentiability is more
restrictive than real differentiability. In order to insure differentiability of a
function f restrictions apply. A generalisation of the above example leads to
the 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 fy
are continuous at z. Without this latter
condition the CR-equations are merely
necessary. Consider the example (see
Fig. 3)
f (z) = f (x, y) =
xy(x+iy)
x2+y2 , z = 0
0 , z = 0
Figure 3: f (z) = xy(x+iy)
x2+y2 , z = z. The
vertical axis depicts the real part of z
while the colours depict the argument
of z.
Theorem 3.3 (Cauchy-Riemann equations). Let f : A → C be a complex
function 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 continuous
on 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 that
there are two equivalent ways to compute the derivative of a complex-valued
function,
f = fx = −i fy (57)
Example We already know that the complex function f (z) = z2 or more
generally f (z) = zn for any integer n is differentiable. Therefore the Cauchy-
Riemann equations must hold. Confirm this.
Solution. From
f (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 hence
for all z ∈ C.
Example Confirm that the complex function f (z) = z · ¯z is not complex
differentiable 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 of
infinite power series. In the following, we define a complex analogue to the
real 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 the
desired properties:
1. |ez| = ex.
2. ez = 0. Property 2 follows from 1. since ex = 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 = 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, see
Fig. 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. It
is 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, there
always does exist z such that w = ez, so we can define log w = z. Since
ew = 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 are
easily proven:
1. If z = 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 the
nonpositive real axis, i.e. D∗ = C (−∞, 0], and
d
dz
log(z) =
1
z
for z ∈ D∗
. (65)
But there is a catch here: for a given w there is more than one z which
satisfies the equation log w = z for the reason that if w = ez holds then
w = ez+2πki also holds for any fixed integer k. Hence we define a unique
branch 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 logarithm
in the real numbers.
Any other convention for Arg z would yield a different branch of the
logarithm.

3.4 Conformal Mappings
Complex differentiability is different from real differentiability as we saw in
the example of the complex function f (z) = ¯z which is nowhere differen-
tiable.
To understand the Example it is helpful to view matters not algebraically
but 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 draw
the graph of a complex-valued function,
and hence we cannot generalise this
particular conception of the derivative
in 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
,
x
y
→
u(x, y)
v(x, y)
.
We know from multi-variable calculus that the Jacobian matrix
J =
ux uy
vx vy
(66)
describes the ‘local’ behaviour of such a map T. If a map T obtained from a
complex function f is differentiable in the sense of real-valued multi-variable
calculus then the following statements are equivalent.
1. f is complex differentiable at z0.
2. h → f (z0 + h) − f (z0) is locally (i.e. at z0) the composition of a rotation
and an expansion or contraction.
3. The Jacobian matrix of the map T satisfies
ux uy
vx 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 → ¯z is not complex differentiable because it is a reflection, an
operation which cannot be represented as a combination of rotation and
expansion or contraction. To see that the limit does not exist note that
(f (z + h) − f (z))/h = ¯h/h which equals +1 if h is real and −1 if h is purely
imaginary. Because of the geometric significance
of complex differentiability it is more
meaningful to ask whether a function
is differentiable in an open set rather
than whether it is differentiable at a
single point. This is why the notion of
holomorphic is more useful than the
notion of analytic alone.
Let us stay with the Jacobian matrix for a little while longer. The Jacobian
determinant D of the map T is
D = uxvy − uyvx (67a)
= u2
x + v2
x using the CR-equations (67b)
= |f |2
, (67c)
If we assume that |f |2 = 0, then the function f maps a neighbourhood of a
point z uniquely and reversibly on to a neighbourhood of a point ξ in a way
that 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 complex
plane.

Definition 3.6 ((Smooth) Path or curve). A path or curve in the complex
plane is the range of the continuous function γ : [a, b] → C given by
γ(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 = 0∀t ∈ (a, b).
Definition 3.7 (Conformal map). A map f : A → C is called conformal
at z ∈ A if it is one-to-one in a neighbourhood of z and for every pair
of smooth paths γ1, γ2 intersecting at z the angle between γ1 and γ2
at z is equal to the angle between the images f (γ1) and f (γ2) at z in
magnitude and sense. If f is conformal at every z ∈ A then f is a
conformal 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 ∈ A
and f (z0) = 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 conformal
since the Jacobian is only a linear approximation to the function f (z) near the
point 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 that
the integral of a function may be regarded as the ‘primitive’ function or
‘indefinite integral’ of the original function. We will obtain a corresponding
relation 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 in
the complex plane. We subdivide the curve into k sections with correspond-
ing intersection points z0, z1, . . . , zk. Let zi be any point on the curve between
zi and zi+1. We now form the sum
Sk =
k
∑
n=1
f (zn)(zn − zn−1) . (68)
Making the partitioning of the curve finer and finer such that the length of
the largest section tends to zero we obtain a sum which is independent on
the exact partitioning and the curve as long as the entire curve is inside A. A
sketch 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 zn = xn + iyn. 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 curve
γ from a to b, denoted γ f (z)dz. Thus,
γ
f (z)dz =
γ
(udx − vdy) + i
γ
(vdx + udy) . (71)
We conclude that a complex-valued function f on a real interval [a, b] is
called integrable, if Re f, Im f are integrable functions in the sense of real
analysis. We define the definite integral, also called contour integral or line
integral 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 integral
along a smooth path γ is defined as
γ
f (z)dz =
b
a
f (γ(t))
dγ(t)
dt
dt . (72)

Examples 1. Suppose z = x + iy and f (z) = x2 + iy2, and consider the curve
γ(t) = t + it, 0 ≤ t ≤ 1. Find γ (t). Then compute the integral γ f (z)dz.
Solution.
γ (t) = 1 + i and
γ
f (z)dz =
1
0
(t2
+ it2
)(1 + i)dt = (1 + i)2
1
0
t2
dt = 2i/3 .
2. Let
f (z) =
1
z
and take
γ(t) = R cos t + iR sin t, 0 ≤ t ≤ 2π, R = 0 .
Find f in terms of (x, y). Then compute γ f (z)dz.
Solution.
f (x, y) =
x
x2 + y2
− i
y
x2 + y2
,
γ
f (z)dz =
2π
0
cos t
R
− i
sin t
R
(−R sin t + iR cos t)dt
=
2π
0
idt = 2πi .
That is, the integral of 1/z around any circle of non-zero radius centred at
the 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
a
f (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
γ
f (z)dz ≤ ML . (75)
Examples 1. Let C be the unit circle and suppose |f | ≤ 1 on C. Then
M = 1, L = 2π, and
c
f (z)dz ≤ 2π .
Compare this to the integral of f (z) = 1/z along the unit circle centred at
the origin.

2. Let C be given by C(t) = 2eit, 0 ≤ t ≤ 2π. Then,
C
ez
z2 + 1
dz ≤
4πe2
3
.
4.2 Independence of Path
An essential result in complex analysis is the fact that path integrals in the
complex plane are independent of the path and only dependent on the end
points.
To begin with, let f be a continuous function7 in a domain A. A function 7
A function is continuous on a set S if
the 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 a
domain A ⊂ C and have an anti-derivative F continuous on A. Then for
path γ ⊂ A joining z0 and z1 in A, we have
γ
f (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
C
f (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 = 1 and closed curve C
not 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 on
A and C ⊂ A. Find the anti-derivative F. This way, conﬁrm that
C
f (z)dz = 0 .

Solution. F = log(z − a) is continuous on any domain not containing
points 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 holomorphic
on a closed region containing C and C and the points between them.
Then,
C
f =
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 the
function is holomorphic, by means of the values the function takes on the
boundary C.
In the following we assume that the function f (z) is holomorphic in the
simply-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 the
point z0. First we make a new curve C by deforming the closed curve C to
a unit circle centred at the point z0. Then we known from the Deformation
Property that
C
g(z)dz =
C
g(z)dz . (80)
Let us rewrite the RHS,
C
g(z)dz =
C
f (z0)
z − z0
dz +
C
f (z) − f (z0)
z − z0
dz . (81)
We know from a previous example that
C
f (z0)
z − z0
dz = f (z0)
C
1
z − z0
dz = f (z0)2πi . (82)
Hence, we have
C
g(z)dz = f (z0)2πi +
C
f (z) − f (z0)
z − z0
dz (83)
and will now let the radius r of C go to zero, r → 0. Neither the LHS nor the
ﬁrst term on the RHS do depend on r. Hence, the second term must remain
unchanged when r → 0. Since f is continuous on A there must exist an M
such that
f (z) − f (z0)
z − z0
=
|f (z) − f (z0)|
r
≤
M
r
. (84)

We know from the ML-Inequality that then
C
f (z) − f (z0)
z − z0
dz ≤
M
r
L(C ) =
M
r
2πr = 2πM . (85)
Since f is continuous M → 0 as r → 0. Thus, we ﬁnd that
lim
r→0 C
f (z) − f (z0)
z − z0
dz = 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 on
a simply connected domain A, let C be a simple, closed, positively
oriented 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 expressed
as a line integral along a simple closed curve around the point.
Example Compute the integral
γ
e2z + sin z
z − 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 is
equal to 2πie2π.
It is interesting to note that, if C is a circle C : z = z0 + reiθ with centre z0
then
f (z0) =
1
2π
2π
0
f (z0 + reiθ
)dθ .
In other words, the value of a complex function at the centre of a circle is
equal to the mean of its values on the circumference, provided that the closed
area of the circle is a region in which the function is holomorphic. This is
known 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, the
chief of which is the proof of the fact that every holomorphic function can
be expanded in a power series. That is, that every holomorphic function is
analytic. We have the following theorem.

Theorem 4.5 (Taylor’s Theorem). Let f : A → C be holomorphic on
some open disk D(z0, R) ⊂ A. Then f can be expanded in a power
series 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 the
variable 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 on
the boundary of the circle. Hence,
|ζ − z0| > |z − z0|.
f (z) =
1
2πi C
f (ζ)
ζ − z
dζ . (90)
We rewrite 1
ζ−z using the geometric series,
1
ζ − z
=
1
ζ − z + z0 − z0
(91)
=
1
ζ − z0
1
1 − z−z0
ζ−z0
(92)
=
1
ζ − z0
1 +
z − z0
ζ − z0
+
z − z0
ζ − 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 assume
without proving it here.
f (z) =
1
2πi C
f (ζ)
ζ − z0
∞
∑
n=0
z − z0
ζ − z0
n
dζ (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 many
times within its circle of convergence. We also know now that this is true for
integration. Hence, integration and differentiation of complex holomorphic
functions can be carried out without restriction.
Every power series expansion is the Taylor series of the function which it
represents. 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 let
C ⊂ A be a simple, closed, positively oriented curve and z0 ∈ int(C) be
a 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 set
containing the unit circle (in fact, it is entire) we can use Eq. 100 with z0 = 0
and k = 3:
C
sin(3z)
z4
dz =
2πi
3!
f (0) = −9πi . (101)
We have the following consequence of Cauchy’s Integral Formula for
Derivatives.
Theorem 4.7 (Liouville’s Theorem.). Any bounded entire function is
constant.
Liouville’s Theorem can be shown by upper bounding f (n)(z) using the
ML-Inequality and showing that that bound goes to zero for any n > 0 and
radius of integration R → ∞. Liouville’s Theorem provides another
proof for cos(z) and sin(z) being
unbounded 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 in
a power series has consequences for functions which are not holomorphic
at 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 constant
term 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)m
q(z) , (102)
where q(z0) = 0. A point z0 for which Equation 102 holds is said to be a zero
of order m of the function f.
Deﬁnition 5.1 (Zero of order m). Let f : A → C, be holomorphic at
z0 ∈ 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) = 0. A zero of order 1 is called a simple
zero.
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 differentiable
except at the points where f vanishes, that is, except precisely at the zeros of
f. Using Equation 102 we can write
g(z) =
1
f (z)
=
1
(z − z0)m
1
q(z)
=
1
(z − z0)m
r(z) , (103)
where r(z) is the reciprocal of q. The point z0 is a singularity of the function
g. In particular if f is non-zero in an open disk around z0, i.e. excluding the
point z0 itself, we call this singularity isolated. We give that type of disk a
name of its own, a punctured disk, also called a deleted neighbourhood.
Deﬁnition 5.2 ((Isolated) singularity / pole of order m). A complex
function g : A → C has a singularity (or pole) of order m at the point
z0 ∈ A if 1/g has a zero of order m. For m = 0 we call it a removable
singularity. If there exists an R > 0 such that g is holomorphic in a
punctured 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 the
singularity is called essential.10 10
The canonical example of an essential
singularity is the point z = 0 of the
function e1/z. Expanding it out into a
power series will reveal the essential
nature of the singularity.
Since the function q(z) in Equation 103 is well defined for all points in the
open disk D(z0, R) and q(z0) = 0 the function r(z) is holomorphic within
that disk. Hence, there exists a power series expansion
r(z) = ˜c0 + ˜c1(z − z0) + ˜c2(z − z0)2
+ . . . (104)
around the point z0 and we can write
g(z) = ˜c0(z − z0)−m
+ ˜c1(z − z0)−m+1
+ · · · + ˜cm + ˜cm+1(z − z0) + . . . (105)
= c−m(z − z0)−m
+ · · · + c−1(z − z0)−1
+ c0 + c1(z − z0) + . . . (106)
with new coefficients ˜cn = cn−m. The expansion in Equation 106 is also called
the Laurent expansion of the function g. The radius of the convergence
of the Laurent expansion in Equation 106 is the same as that of the Taylor
expansion in Equation 104. For m = 0 the singularity is remov-
able and the Laurent expansion of g
becomes 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 will
receive 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 all
isolated. 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
lim
z→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 integer
N such that
lim
z→z0
(z − z0)N
f (z) = 0 (108)
but lim
z→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 expansion
of 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 = 0
and 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 in
complex function theory, Cauchy’s Residue Theorem.
Definition 5.4 (Residue). Let f : A → C, A ⊂ C, be holomorphic in
a punctured disk D(z0, R) {z0} of an isolated singularity z0 so that
it 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 [ 1
1−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 at
z0 then all the negative powers of the
power 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) = lim
z→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 at
z0, then
Res [f; z0] = lim
z→z0
(z − z0)f (z) . (112)
If f = A(z)/B(z) has a simple pole at z0 and A(z), B(z) are differentiable
at 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 C
except at a ﬁnite number of points z1, z2, . . . , zk ∈ Int(C), then
C
f (z)dz = 2πi
k
∑
j=1
Res [f; zj] .
Examples From previous examples for singularities and residues we can
immediately 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 2i
to −2 and back in a straight line to 2. The singularities of the integrand occur
at the fourth roots of −1, i.e. at eπi/4, e3πi/4, e5πi/4, e7πi/4. But only eπi/4 and
e3πi/4 are inside the curve. Hence, we have
C
dz/(z4
+ 1) = 2πi Res (1/(z4
+ 1); eπi/4
) + 2πi Res (1/(z4
+ 1); e3πi/4
)
(114)
= 2πi
1
4(eπi/4)3
+
1
4(e3πi/4)3
= 2πi
1
4
e−3πi/4
+ e−πi/4
(115)
= 2πi
1
4
−eπi/4
+ e−πi/4
=
πi
2
(−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 as
improper real integrals. Some of these integrals occur in physical and engi-
neering applications, and often cannot be evaluated by using the methods of
real 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) = 0 and deg Q− deg P ≥ 2. To show this, let CR be the
closed contour consisting of the real line segment from −R to R and the
upper semi-circle ΓR centered at the origin and of radius R. Let’s make R
large enough to enclose all zeroes of Q lying in the upper half-plane.
In this case, we note that
∞
−∞
P(x)
Q(x)
dx = lim
R→∞
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
P
Q
; zj , (118)
where the points zj are the zeroes of Q in the upper half-plane. We now
upper-bound the integral ΓR
P/Q using the ML-inequality.
ΓR
P(z)
Q(z)
dz ≤ πR
A
R2
(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
∞
−∞
dx
x4 + 1
(121)
The zeros of Q(z) = z4 + 1 in the upper half-plane are z1 = eiπ/4 and
z2 = e3iπ/4, each of which is a simple pole. The residues are given by the
value of 1/4z3 at the poles. Thus,
∞
−∞
dx
x4 + 1
= 2πi
1
4e3πi/4
+
1
4e9πi/4
= 2πi
1
4
√
2
((−1 − i) + (1 − i)) =
π
√
2
2
.
(122)
∞
−∞
R(x) cos(x)dx or
∞
−∞
R(x) sin(x)dx
Another kind of real integrals which can be solved easily using Cauchy’s
Residue Theorem are integrals of the form
∞
−∞ R(x) cos(x)dx or
∞
−∞ R(x) sin(x)dx
where R(x) = P(x)/Q(x), P and Q are polynomials, Q(x) = 0 and deg Q >
deg P. We use a similar argument as above and refer to the same contours
CR and ΓR. Consider the integral12 12
Integrating R(z) cos(z) along the
contour CR is not helpful in this case
since cos(z) is not bounded.

CR
R(z)eiz
dz (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 have
sin θ ≥ 2θ/π.
ΓR
R(z)eiz
dz ≤
π
0
R(eiθ
) eiR(cos θ+i sin θ)
Rdθ ≤ A
π
0
e−R sin θ
dθ (124)
= 2A
π/2
0
e−R sin θ
dθ ≤ 2A
π/2
0
e−2Rθ/π
dθ (125)
= π
A
R
(1 − e−R
) → 0 as R → ∞ . (126)
Hence,
ΓR
R(z)eiz
dz → 0 as R → ∞ (127)
and thus
CR
R(z)eiz
dz →
∞
−∞
R(x)eix
dx as R → ∞. (128)
Then, by applying the Residue Theorem, we get
∞
−∞
R(x)eix
dx = 2πi ∑
zj∈UHP
Res R(z)eiz
; zj . (129)
Using Euler’s formula we ﬁnd the integrals
∞
−∞ R(x) cos(x)dx and
∞
−∞ R(x) sin(x)dx
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 = 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
b
e−b
. (132)
∞
−∞
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)eiax
dx = 2πi ∑
zj∈UHP
Res R(z)eiaz
; zj , a > 0 , (133)
∞
−∞
R(x)eiax
dx = −2πi ∑
zj∈LHP
Res 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. For
a < 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
ΓR
R(z)eiaz
dz ≤ 2A
−π/2
0
e|a|R2θ/π
dθ (138)
= 2A
π
|a|R2
eR2θ/π −π/2
0
(139)
= π
A
|a|R
(e−|a|R
− 1) → 0 as R → ∞ . (140)
The claim follows.
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 computed
by 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 means
of the partial fraction decomposition. It is often much simpler to apply the
Residue Theorem, by interpreting the given integral as an integral along a
suitable 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 and
sin θ = (eiθ − e−iθ)/(2i) we can write cos θ and sin θ in the new parametrisa-
tion as
cos θ =
1
2
z +
1
z
and sin θ =
1
2i
z −
1
z
. (141)
To reparametrise the integral, we differentiate eiθ along C:
dz
dθ
= iz and hence dθ = −i
dz
z
. (142)
Now, the integral
2π
0 R(cos θ, sin θ)dθ can be rewritten
2π
0
R(cos θ, sin θ)dθ =
C={|z|=1}
R(
z + z−1
2
,
z − z−1
2i
)
dz
iz
. (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 the
two poles is inside of the circle of
integration.2π
0
dθ
2 + cos(θ)
=
2
i C
dz
z2 + 4z + 1
(144)
= 4π Res
1
z2 + 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
∞
0
cos x2
dx =
∞
0
sin x2
dx =
1
2
π
2
(146)
To derive them we need the following two results: Inequality 147 can easily be seen
by drawing the graph of sin x for
0 ≤ x ≤ θ/2 and that the straight line
between 0 and θ/2.
2
π
θ < sin θ , 0 < θ <
π
2
and (147)
L =
∞
0
e−x2
dx =
√
π
2
. (148)
To establish Eq. 148 we note ﬁrst that
L2
=
∞
0
e−x2
dx
∞
0
e−y2
dy =
∞
0
∞
0
e−(x2
+y2
)
dxdy (149)
and then use polar coordinates to obtain
L2
=
π/2
0
∞
0
e−r2
rdrdθ = −
1
2
e−r2 ∞
0
×
π
2
=
π
4
. (150)
Now, to establish Equations 146, we consider the entire function f (z) =
eiz2
and 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 we
know that
γ
eiz2
dz =
OA
eiz2
dz +
AB
eiz2
dz +
BO
eiz2
dz = 0 . (151)
We can rewrite these integrals using the properties of the contour γ:
γ
eiz2
dz =
ρ
0
eix2
dx +
π/4
0
eiρ2
exp 2iθ
ρieiθ
dθ +
0
ρ
eir2
exp πi/2
eπi/4
dr = 0 .
(152)
Now, in view of Eq. 147, we have
π/4
0
eiρ2
exp 2iθ
ρieiθ
dθ ≤ ρ
π/4
0
eiρ2
(cos(2θ)+i sin(2θ)
dθ
≤ ρ
π/4
0
e−ρ2
sin(2θ)
dθ ≤ ρ
π/4
0
e−ρ2
(4θ/π)
dθ
=
π
4
1 − e−ρ2
ρ
→ 0 as ρ → ∞ , (153)
whereas Eq. 148 gives
lim
ρ→∞
0
ρ
eir2
exp πi/2
eπi/4
dr = e−πi/4
∞
0
e−r2
dr = −
1 + i
√
2
√
π
2
. (154)
Combining Eqs. 152 - 154, we get
lim
ρ→∞
ρ
0
eix2
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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