CUST 2023-2024

Maths for Physics 2

Session 13

Cauchy-Riemann equations and complex

integration
Summary

In this session we continue our study of complex analysis. In particular, we deepen

our study of differentiability and derive a useful set of equations that a complex
function must satisfy in order to be differentiable: these equations are the so-called
Cauchy-Riemann equations. We also discuss how integration can be defined for
complex functions.

i
Contents

1 The Cauchy-Riemann equations 1

1.1 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 The Cauchy-Riemann equations as a necessary condition of differen-
tiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 The Cauchy-Riemann equations as a sufficient condition of differen-
tiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Some remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Complex integration 7
2.1 Curves and contours in the complex plane . . . . . . . . . . . . . . . 7
2.2 Contour integral for smooth contours . . . . . . . . . . . . . . . . . . 10

ii
Chapter 1

The Cauchy-Riemann equations

We recall that a complex function f (z) can, upon writing z as

z = x + iy , (1.1)

be written in the form

f (z) = u(x, y) + iv(x, y) , (1.2)

where u and v are thus real functions of the two real variables x and y.

1.1 The theorem

We have the following important theorem, which expresses an equivalent property of
differentiability of a complex function at a point:

Cauchy-Riemann equations

The complex function

f (z) = u(x, y) + iv(x, y) (1.3)

is differentiable at a point z = x + iy of a region in the complex

plane if and only if the partial derivatives

∂u ∂u ∂v ∂v
, , , ,
∂x ∂y ∂x ∂y

are continuous at the point z = x + iy and in addition satisfy

the Cauchy-Riemann equations at this point z, namely

∂u ∂v ∂v ∂u
= and =− . (1.4)
∂x ∂y ∂x ∂y

1
CHAPTER 1. THE CAUCHY-RIEMANN EQUATIONS 2

In order to prove this theorem, we must hence proceed in two steps:

i) we must first show that if f is differentiable, then the Cauchy-Riemann equa-

tions are satisfied: in other words, we must first show that the Cauchy-Riemann
equations are a necessary condition of differentiability;

ii) then we must show that if u and v are continuous and satisfy the Cauchy-
Riemann equations, then f is differentiable: in other words, we must then show
that the Cauchy-Riemann equations are a sufficient condition of differentiabil-
ity.

1.2 The Cauchy-Riemann equations as a necessary

condition of differentiability

Im
Im
n
n

Dx
Dz so
Dz
i
dy 70

S Y
Re
s
Re

Figure 1: Illustration of the two different limits ∆z → 0 taken either i) along the
positive real axis, or ii) along the positive imaginary axis.

Here we assume that the function f (z) is differentiable at the point z = x + iy in

a region of the complex plane. By definition of the derivative f ′ (z), we must hence
have that the limit

f (z + ∆z) − f (z)
f ′ (z) ≡ lim (1.5)
∆z→0 ∆z
CHAPTER 1. THE CAUCHY-RIEMANN EQUATIONS 3

exists. This means in particular that the limit (1.5) must return the same number
f ′ (z) independently of the direction in which ∆z approaches zero. Therefore, let’s
consider the two particular cases, illustrated in figure 1, where:

i) ∆z approaches zero along the positive real axis, i.e. ∆z = ∆x > 0;

ii) ∆z approaches zero along the positive imaginary axis, i.e. ∆z = i∆y, with
∆y > 0.

Therefore, writing (1.5) in these two cases and writing z = x + iy, we have

f (x + iy + ∆x) − f (x + iy) f (x + iy + i∆y) − f (x + iy)

lim = lim ,
∆x→0 ∆x ∆y→0 i∆y

where the left-hand side corresponds to case i) above and the right-hand side corre-
sponds to case ii), so that we get, with f = u + iv,

u(x + ∆x, y) + iv(x + ∆x, y) − u(x, y) − iv(x, y)

lim
∆x→0 ∆x

u(x, y + ∆y) + iv(x, y + ∆y) − u(x, y) − iv(x, y)

= lim ,
∆y→0 i∆y

that is, writing separately the equality of the real parts and the imaginary parts on
both sides of the equation,

u(x + ∆x, y) − u(x, y) v(x, y + ∆y) − v(x, y)

lim = lim (1.6)
∆x→0 ∆x ∆y→0 ∆y

and

v(x + ∆x, y) − v(x, y) u(x, y + ∆y) − u(x, y)

lim = − lim . (1.7)
∆x→0 ∆x ∆y→0 ∆y

Now, note that the limits involved in (1.6) and (1.7) are merely the partial
derivatives of u and v! Indeed, the definition of the partial derivative of a function
g(x, y) is

∂g g(x + ∆x, y) − g(x, y) ∂g g(x, y + ∆y) − g(x, y)

≡ lim and ≡ lim .
∂x ∆x→0 ∆x ∂y ∆x→0 ∆y

Therefore, (1.6) and (1.7) are simply

∂u ∂v
= (1.8)
∂x ∂y
CHAPTER 1. THE CAUCHY-RIEMANN EQUATIONS 4

and

∂v ∂u
=− . (1.9)
∂x ∂y

These two equations (1.8)-(1.9) precisely correspond to the Cauchy-Riemann equa-

tions (1.4): the latter are thus indeed a necessary condition of differentiability.

1.3 The Cauchy-Riemann equations as a sufficient

condition of differentiability
Now we assume that the real and imaginary parts u and v of f are continuous,
and so are their partial derivatives, and that in addition the Cauchy-Riemann equa-
tions (1.4) are satisfied: so here we want to show that these assumptions imply that
f is differentiable.
First, let’s write down what the continuity of u and v means. To do this, let’s
construct the quantities

 ∆u ≡ u(x + ∆x, y + ∆y) − u(x, y)
. (1.10)
∆v ≡ v(x + ∆x, y + ∆y) − v(x, y)


That is, ∆u and ∆v quantify the amount by which u and v change if we go from the
point (x, y) to the point (x + ∆x, y + ∆y). Using a standard result of real analysis
for real functions of real variables, we know that the continuity of u and v at the
point (x, y) ensures that
 ∂u ∂u
p
 ∆u = ∂x
∆x + ∂y
∆y + ϵ1 ∆x2 + ∆y 2
p , (1.11)
∂v ∂v 2 2
∆v = ∆x + ∆y + ϵ2 ∆x + ∆y

∂x ∂y

where ϵ1 and ϵ2 are such that

lim ϵ1 = lim ϵ2 = 0 . (1.12)

∆x→0 ∆x→0
∆y→0 ∆y→0

If we now define the quantity

∆f ≡ f (z + ∆z) − f (z)

= u(x + ∆x, y + ∆y) − u(x, y) + i [v(x + ∆x, y + ∆y) − v(x, y)] , (1.13)
CHAPTER 1. THE CAUCHY-RIEMANN EQUATIONS 5

we hence have in view of (1.10)

∆f ≡ ∆u + i∆v . (1.14)

Therefore, dividing (1.14) by ∆z yields, in view of (1.11),

p
∆f ∆u ∆v ∂u ∆x ∂u ∆y ∆x2 + ∆y 2
= +i = + + ϵ1
∆z ∆z ∆z ∂x ∆z ∂y ∆z ∆z
" p #
∂v ∆x ∂v ∆y 2
∆x + ∆y 2
+i + + ϵ2 . (1.15)
∂x ∆z ∂y ∆z ∆z

Now, note that

p
|∆z| = |∆x + i∆y| = ∆x2 + ∆y 2 ,

so that
p
∆x2 + ∆y 2 |∆z|
= ,
∆z ∆z

and thus, since any complex number w can be written as w = |w|eiArg w ,

|∆z|
= e−iArg ∆z . (1.16)
∆z

Substituting (1.16) into (1.15) hence yields

 
∆f ∂u ∆x ∂u ∆y ∂v ∆x ∂v ∆y
= + +i + + (ϵ1 + iϵ2 )e−iArg ∆z . (1.17)
∆z ∂x ∆z ∂y ∆z ∂x ∆z ∂y ∆z

We now use our assumption that the functions u and v must satisfy the Cauchy-
Riemann equations (1.4), which we repeat here for clarity, i.e we must have

∂u ∂v ∂v ∂u
= and =− ,
∂x ∂y ∂x ∂y

so that we get for (1.17)

∆f ∂u ∂v
= +i + (ϵ1 + iϵ2 )e−iArg ∆z . (1.18)
∆z ∂x ∂x

Finally, we take the limit ∆z → 0 in (1.18) and we get, in view of (1.12), and
recalling the definition (1.13) of ∆f ,

∆f f (z + ∆z) − f (z) ∂u ∂v
lim ≡ lim = +i . (1.19)
∆z→0 ∆z ∆z→0 ∆z ∂x ∂x
CHAPTER 1. THE CAUCHY-RIEMANN EQUATIONS 6

Since the (real) partial derivatives ∂u/∂x and ∂v/∂x in the right-hand side in (1.19)
exist by assumption, the limit in the left-hand side in (1.19) must hence also exist:
since the latter is nothing but the definition (1.5) of the derivative of f at the
point z, this shows that f is indeed differentiable at the point z, that is that the
Cauchy-Riemann equations are thus indeed a sufficient condition of differentiability.

This completes the proof of the theorem.

1.4 Some remarks

The Cauchy-Riemann equations (1.4) are very useful in practice: indeed, because
of their equivalence with differentiability, they give us a convenient criterion for
checking differentiability. Therefore, if we are given a function f (z) and we want to
show whether f is differentiable at some point z, we will in practice rarely show this
by showing that the limit (1.5) exists, which can be cumbersome. Rather, we will
use the Cauchy-Riemann equations (1.4), and thus proceed as follows:

i) given f (z), we first write it in the form

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

ii) we then compute the partial derivatives ∂u/∂x, ∂u/∂y, ∂v/∂x and ∂v/∂y;

iii) finally, we check whether the Cauchy-Riemann equations (1.4) are satisfied at
the point z or not.

Notation: instead of writing ∂u/∂x, it is sometimes convenient to use the shorter

(and commonly used) notations

∂u
≡ ∂x u ≡ ux ,
∂x

and similarly for ∂u/∂y, etc. . .

The Cauchy-Riemann equations are also typically a method of choice to show
that a function f is analytic at a point z (which hence requires to show that f
is differentiable, and thus that the Cauchy-Riemann equations are satisfied, in a
neighborhood of z).
Chapter 2

Complex integration

So far, we discussed in some details how we can differentiate complex functions.

Therefore, our next natural question is: how can we integrate complex functions?
This will be done by integrating a complex function along a contour (i.e. some kind
of curve) in the complex plane.
First, let’s consider the simplest case where we have a complex-valued function
f of a real variable t, i.e. f (t) ∈ C but t ∈ R. We assume that in addition t takes
values between a and b, i.e. t ∈ [a, b] with a < b. Writing f = u + iv, we hence have
in this case

f (t) = u(t) + iv(t) ,

where u and v here are thus real functions of the single real variable t. Since u, v
and t are real, we can use the results from real analysis to construct their integrals,
from which we then immediately get the definition of the integral of f (t): we thus
say that f is integrable on the interval [a, b] if both functions u and v are integrable
on [a, b], and we write
 b  b  b
dt f (t) = dt u(t) + i dt v(t) .
a a a

We then extend this definition of complex integration in order to integrate a

complex function on a contour in the complex plane.

2.1 Curves and contours in the complex plane

As is schematized in figure 2, a curve (or arc) γ in the complex plane is simply a

set of points in the complex plane that can be described by a certain parametrization,
with, say, a parameter t: that is, to each value of the parameter t corresponds a

7
CHAPTER 2. COMPLEX INTEGRATION 8

Im n
3161
My

X
a
Z
0 Re

Figure 2: Schematic representation of a curve.

complex number z(t) on this curve. Let’s say that t ∈ [a, b]: the curve γ is then the
set of all points z(t) obtained for any t ∈ [a, b]. We can write

z(t) = x(t) + iy(t) , t ∈ [a, b] ,

so that to any t corresponds a couple of real numbers (x(t), y(t)). A curve also
embeds a notion of direction: indeed, the points z(t) are ordered in the direction of
increasing t: therefore, the point z(a) can be called the initial point of the curve γ,
and the point z(b) can be called the final point of the curve γ. We say that γ is a
continuous curve if both x(t) and y(t) are continuous. Furthermore, we say that γ
is a differentiable curve if both x(t) and y(t) are differentiable.

Simple
Not closed

b Not simple
closed

Figure 3: Illustration of the concepts of simple and closed curves.

CHAPTER 2. COMPLEX INTEGRATION 9

Let’s now introduce a bit of terminology. A curve γ is said to be simple if it has

no point of intersection1 except possibly at the endpoints t = a and t = b. A curve
for which z(a) = z(b) is said to be closed. A simple closed curve is typically referred
to as a Jordan curve. These notions are illustrated in figure 3.
As mentioned above, a curve embeds a notion of direction. When a curve γ is
closed, we say that the positive direction is such that the interior points are on the
left of γ. We will typically orient our curves in the positive direction. This notion
of direction is illustrated in figure 4.

Positrection Negatyrection

X
r X U
points
interior interior
in
points

o o

Figure 4: Illustration of the notions of positive and negative direction of a closed

curve γ.

We say that a curve γ is a smooth curve if z ′ (t) (i.e. the derivative of the function
z(t) with respect to the parameter t) is continuous for t ∈ [a, b]. We then say that
the curve γ is a contour if it consists in connected smooth curves (i.e., smooth
curves that are “glued together” at some points on the contour, as is schematized in
figure 5).

A simple (i.e. no intersections) closed (i.e. same endpoints) contour is often

referred to as a Jordan contour. Finally, a domain D is said to be simply connected
if every simple closed contour encloses only points of D.
1
That is, there exist no distinct values t1 and t2 of the parameter t for which z(t1 ) = z(t2 ).
CHAPTER 2. COMPLEX INTEGRATION 10

ME

Figure 5: Schematic representation of a contour.

2.2 Contour integral for smooth contours

Let’s now consider a complex function f (z) and a smooth contour γ, the latter being
described by the points

z(t) = x(t) + iy(t) , t ∈ [a, b] . (2.1)

We hence have the following definition of a complex integral:

Contour integral

We define the contour integral of f on (or along) the contour γ by

  b
dz f (z) ≡ dt z ′ (t)f [z(t)] . (2.2)
γ a

Intuitively, we can interpret the complex infinitesimal element dz to be simply ob-

tained from

dz
= z ′ (t) =⇒ dz = dt z ′ (t) .
dt

To go from the left-hand side to the right-hand side in (2.2), we chose the particular
parametrization (2.1) of the contour γ. We can however choose any parametrization
we want: in practice, we merely choose the most convenient parametrization that
allows for the most convenient expression for the integral in the right-hand side
of (2.2).
The standard properties of real integration still apply to contour integrals in the
complex plane. For instance, we have linearity, i.e. for two complex functions f and
CHAPTER 2. COMPLEX INTEGRATION 11

g and two numbers α, β ∈ C we have

  
dz [αf (z) + βg(z)] = α dz f (z) + β dz g(z) . (2.3)
γ γ γ

Furthermore, similarly to the real properties

 a  b  c  b  c
dt f (t) = − dt f (t) and dt f (t) = dt f (t) + dt f (t) ,
b a a a b

we have
    
dz f (z) = − dz f (z) and dz f (z) = dz f (z) + dz f (z) .
−γ γ γ1 ∪γ2 γ1 γ2
(2.4)

In addition, we also have the complex version of the fundamental theorem of calculus:

Fundamental theorem of calculus

Let f (z) and F (z) be two functions such that F (z) is analytic
and f (z) is continuous in a domain D, and such that in
addition we have F ′ (z) = f (z) (i.e. F is an antiderivative of f ).
Let γ be a contour that is entirely contained in D, with
endpoints z1 and z2 . Then we have

dz f (z) = F (z2 ) − F (z1 ) . (2.5)
γ

Proof: to simplify, let’s assume that γ is described by a parametrization z(t) such

that z(t) is continuous. Let’s first rewrite the definition (2.2) of the contour integral,
i.e.
  b
dz f (z) = dt z ′ (t)f [z(t)] . (2.6)
γ a

Now, since F ′ (z) = f (z) we have from the chain rule

d
F [z(t)] = z ′ (t)F ′ [z(t)] = z ′ (t)f [z(t)] ,
dt

so that we get for (2.6)

  b
d
dz f (z) = dt F [z(t)] = [F [z(t)]]ba = F [z(b)] − F [z(a)] = F (z2 ) − F (z1 ) ,
γ a dt

which is indeed precisely (2.5). ■

CHAPTER 2. COMPLEX INTEGRATION 12

Let’s now assume that the contour γ is closed : for closed contours, we use a
specific notation for the contour integral (2.2) and we write

dz f (z) ≡ integral over a closed contour γ . (2.7)

γ

Since γ is closed, we hence have by definition z2 = z1 . Therefore, a direct conse-

quence of the fundamental theorem of calculus (2.5) is that

dz f (z) = 0 , (2.8)
γ

for any closed contour in the domain D where f satisfies the conditions of the
theorem.
Furthermore, an other important consequence of the fundamental theorem of
calculus (2.5) is that, for any two contours γ1 and γ2 that have the same endpoints
z1 and z2 (see figure 6), we have
 
dz f (z) = dz f (z) .
γ1 γ2

In other words, the integral is independent of the actual path that is followed in
the complex plane to join the two points z1 and z2 . This hence leaves open the
possibility to deform a contour into an other contour that is easier to parametrize
and/or more suitable in order to compute the integral: for instance, straight lines
are obviously easier to parametrize than curved contours.

Eu

Figure 6: Schematic representation of two contours γ1 and γ2 that have the same
endpoints z1 and z2 .