Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

9.4 Exercises
Exercise 9.1
Consider two functions, f(x, y) and g(x, y). They are said to be functionally dependent if there is a an h(g) such that
f(x, y) = h(g(x, y)).
f and g will be functionally dependent if and only if their Jacobian vanishes.
If f and g are functionally dependent, then the derivatives of f are
f
x
= h

(g)g
x
f
y
= h

(g)g
y
.
Thus we have
∂(f, g)
∂(x, y)
=




f
x
f

y
g
x
g
y




= f
x
g
y
− f
y
g
x
= h

(g)g
x
g
y
− h

(g)g
y
g
x
= 0.

If the Jacobian of f and g vanishes, then
f
x
g
y
− f
y
g
x
= 0.
This is a first order partial differential equation for f that has the general solution
f(x, y) = h(g(x, y)).
Prove that an analytic function u(x, y) + ıv(x, y) can be written in terms of a function of a complex variable,
f(z) = u(x, y) + ıv(x, y).
Exercise 9.2
Which of the following functions are the real part of an analytic function? For those that are, find the harmonic
conjugate, v(x, y), and find the analytic function f(z) = u(x, y) + ıv(x, y) as a function of z.
1. x
3
− 3xy
2
− 2xy + y
2.
e
x
sinh y
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3.
e
x

(sin x cos y cosh y −cos x sin y sinh y)
Exercise 9.3
For an analytic function, f(z) = u(r, θ) + ıv(r, θ) prove that under suitable restrictions:
f(z) = 2u

z
1/2
, −
ı
2
log z

+ const.
455
9.5 Hints
Hint 9.1
Show that u(x, y) + ıv(x, y) is functionally dependent on x + ıy so that you can write f(z) = f(x + ıy) = u(x, y) +
ıv(x, y).
Hint 9.2
Hint 9.3
Check out the derivation of Equation 9.2.
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9.6 Solutions
Solution 9.1
u(x, y) + ıv(x, y) is functionally dependent on z = x + ıy if and only if
∂(u + ıv, x + ıy)
∂(x, y)
= 0.
∂(u + ıv, x + ıy)
∂(x, y)

=




u
x
+ ıv
x
u
y
+ ıv
y
1 ı




= −v
x
− u
y
+ ı (u
x
− v
y
)
Since u and v satisfy the Cauchy-Riemann equations, this vanishes.
= 0
Thus we see that u(x, y) + ıv(x, y) is functionally dependent on x + ıy so we can write

f(z) = f(x + ıy) = u(x, y) + ıv(x, y).
Solution 9.2
1. Consider u(x, y) = x
3
− 3xy
2
− 2xy + y. The Laplacian of this function is
∆u ≡ u
xx
+ u
yy
= 6x −6x
= 0
Since the function is harmonic, it is the real part of an analytic function. Clearly the analytic function is of the
form,
az
3
+ bz
2
+ cz + ıd,
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with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products
yields,
a

x
3
+ ı3x
2
y − 3xy

2
− ıy
3

+ b

x
2
+ ı2xy − y
2

+ c(x + ıy) + ıd.
By inspection, we see that the analytic function is
f(z) = z
3
+ ız
2
− ız + ıd.
The harmonic conjugate of u is the imaginary part of f(z),
v(x, y) = 3x
2
y − y
3
+ x
2
− y
2
− x + d.
We can also do this problem with analytic continuation. The derivatives of u are
u

x
= 3x
2
− 3y
2
− 2y,
u
y
= −6xy − 2x + 1.
The derivative of f(z) is
f

(z) = u
x
− ıu
y
= 3x
2
− 2y
2
− 2y + ı(6xy − 2x + 1).
On the real axis we have
f

(z = x) = 3x
2
− ı2x + ı.
Using analytic continuation, we see that
f


(z) = 3z
2
− ı2z + ı.
Integration yields
f(z) = z
3
− ız
2
+ ız + const
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2. Consider u(x, y) =
e
x
sinh y. The Laplacian of this function is
∆u =
e
x
sinh y +
e
x
sinh y
= 2
e
x
sinh y.
Since the function is not harmonic, it is not the real part of an analytic function.
3. Consider u(x, y) =
e
x
(sin x cos y cosh y − cos x sin y sinh y). The Laplacian of the function is

∆u =
∂
∂x
(
e
x
(sin x cos y cosh y − cos x sin y sinh y + cos x cos y cosh y + sin x sin y sinh y))
+
∂
∂y
(
e
x
(−sin x sin y cosh y − cos x cos y sinh y + sin x cos y sinh y − cos x sin y cosh y))
= 2
e
x
(cos x cos y cosh y + sin x sin y sinh y) −2
e
x
(cos x cos y cosh y + sin x sin y sinh y)
= 0.
Thus u is the real part of an analytic function. The derivative of the analytic function is
f

(z) = u
x
+ ıv
x
= u

x
− ıu
y
From the derivatives of u we computed before, we have
f(z) = (
e
x
(sin x cos y cosh y − cos x sin y sinh y + cos x cos y cosh y + sin x sin y sinh y))
− ı (
e
x
(−sin x sin y cosh y − cos x cos y sinh y + sin x cos y sinh y − cos x sin y cosh y))
Along the real axis, f

(z) has the value,
f

(z = x) =
e
x
(sin x + cos x).
By analytic continuation, f

(z) is
f

(z) =
e
z
(sin z + cos z)

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We obtain f(z) by integrating.
f(z) =
e
z
sin z + const.
u is the real part of the analytic function
f(z) =
e
z
sin z + ıc,
where c is a real constant. We find the harmonic conjugate of u by taking the imaginary part of f .
f(z) =
e
x
(cosy + ı sin y)(sin x cosh y + ı cos x sinh y) + ıc
v(x, y) =
e
x
sin x sin y cosh y + cos x cos y sinh y + c
Solution 9.3
We consider the analytic function: f(z) = u(r, θ) + ıv(r, θ). Recall that the complex derivative in terms of polar
coordinates is
d
dz
=
e
−ıθ
∂
∂r

= −
ı
r
e
−ıθ
∂
∂θ
.
The Cauchy-Riemann equations are
u
r
=
1
r
v
θ
, v
r
= −
1
r
u
θ
.
We differentiate f(z) and use the partial derivative in r for the right side.
f

(z) =
e
−ıθ

(u
r
+ ıv
r
)
We use the Cauchy-Riemann equations to right f

(z) in terms of the derivatives of u.
f

(z) =
e
−ıθ

u
r
− ı
1
r
u
θ

(9.6)
Now consider the function of a complex variable, g(ζ):
g(ζ) =
e
−ıζ

u
r

(r, ζ) − ı
1
r
u
θ
(r, ζ)

=
e
ψ−ıξ

u
r
(r, ξ + ıψ) − ı
1
r
u
θ
(r, ξ + ıψ)

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This function is analytic where f(ζ) is analytic. It is a simple calculus exercise to show that the complex derivative in
the ξ direction,
∂
∂ξ
, and the complex derivative in the ψ direction, −ı
∂
∂ψ
, are equal. Since these partial derivatives are
equal and continuous, g(ζ) is analytic. We evaluate the function g(ζ) at ζ = −ı log r. (Substitute θ = −ı log r into

Equation 9.6.)
f


r
e
ı(−ı log r)

=
e
−ı(−ı log r)

u
r
(r, −ı log r) −ı
1
r
u
θ
(r, −ı log r)

rf


r
2

= u
r
(r, −ı log r) −ı

1
r
u
θ
(r, −ı log r)
If the expression is non-singular, then it defines the analytic function, f

(z), on a curve. The analytic continuation to
the complex plane is
zf


z
2

= u
r
(z, −ı log z) −ı
1
z
u
θ
(z, −ı log z).
We integrate to obtain an expression for f (z
2
).
1
2
f


z
2

= u(z, −ı log z) + const
We make a change of variables and solve for f(z).
f(z) = 2u

z
1/2
, −
ı
2
log z

+ const.
Assuming that the above expression is non-singular, we have found a formula for writing the analytic function in terms
of its real part, u(r, θ). With the same method, we can find how to write an analytic function in terms of its imaginary
part, v(r, θ).
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Chapter 10
Contour Integration and the Cauchy-Goursat
Theorem
Between two evils, I always pick the one I never tried before.
- Mae West
10.1 Line Integrals
In this section we will recall the definition of a line integral in the Cartesian plane. In the next section we will use
this to define the contour integral in the complex plane.
Limit Sum Definition. First we develop a limit sum definition of a line integral. Consider a curve C in the Cartesian
plane joining the points (a
0

, b
0
) and (a
1
, b
1
). We partition the curve into n segments with the points (x
0
, y
0
), . . . , (x
n
, y
n
)
where the first and last points are at the endpoints of the curve. We define the differences, ∆x
k
= x
k+1
− x
k
and
∆y
k
= y
k+1
− y
k
, and let (ξ
k

, ψ
k
) be points on the curve between (x
k
, y
k
) and (x
k+1
, y
k+1
). This is shown pictorially
in Figure 10.1.
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(x ,y )
0
0
(ξ ,ψ )
0
0
1
1
(ξ ,ψ )
1
1
(x ,y )
2
2
(ξ ,ψ )
2
2

(ξ ,ψ )
n−1
n−1
(x ,y )
n
n
(x ,y )
n−1
n−1
(x ,y )
y
x
Figure 10.1: A curve in the Cartesian plane.
Consider the sum
n−1

k=0
(P (ξ
k
, ψ
k
)∆x
k
+ Q(ξ
k
, ψ
k
)∆y
k
) ,

where P and Q are continuous functions on the curve. (P and Q may be complex-valued.) In the limit as each of the
∆x
k
and ∆y
k
approach zero the value of the sum, (if the limit exists), is denoted by

C
P (x, y) dx + Q(x, y) dy.
This is a line integral along the curve C. The value of the line integral depends on the functions P(x, y) and Q(x, y),
the endpoints of the curve and the curve C. We can also write a line integral in vector notation.

C
f(x) · dx
Here x = (x, y) and f(x) = (P(x, y), Q(x, y)).
463
Evaluating Line Integrals with Parameterization. Let the curve C be parametrized by x = x(t), y = y(t)
for t
0
≤ t ≤ t
1
. Then the differentials on the curve are dx = x

(t) dt and dy = y

(t) dt. Using the parameterization we
can evaluate a line integral in terms of a definite integral.

C
P (x, y) dx + Q(x, y) dy =


t
1
t
0

P (x(t), y(t))x

(t) + Q(x(t), y(t))y

(t)

dt
Example 10.1.1 Consider the line integral

C
x
2
dx + (x + y) dy,
where C is the semi-circle from (1, 0) to (−1, 0) in the upper half plane. We parameterize the curve with x = cos t,
y = sin t for 0 ≤ t ≤ π.

C
x
2
dx + (x + y) dy =

π
0


cos
2
t(−sin t) + (cos t + sin t) cos t

dt
=
π
2
−
2
3
10.2 Contour Integrals
Limit Sum Definition. We develop a limit sum definition for contour integrals. It will be analogous to the definition
for line integrals except that the notation is cleaner in complex variables. Consider a contour C in the complex plane
joining the points c
0
and c
1
. We partition the contour into n segments with the points z
0
, . . . , z
n
where the first and
last points are at the endpoints of the contour. We define the differences ∆z
k
= z
k+1
−z
k
and let ζ

k
be points on the
contour between z
k
and z
k+1
. Consider the sum
n−1

k=0
f(ζ
k
)∆z
k
,
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where f is a continuous function on the contour. In the limit as each of the ∆z
k
approach zero the value of the sum,
(if the limit exists), is denoted by

C
f(z) dz.
This is a contour integral along C.
We can write a contour integral in terms of a line integral. Let f(z) = φ(x, y). (φ : R
2
→ C.)

C
f(z) dz =


C
φ(x, y)(dx + ı dy)

C
f(z) dz =

C
(φ(x, y) dx + ıφ(x, y) dy) (10.1)
Further, we can write a contour integral in terms of two real-valued line integrals. Let f(z) = u(x, y) + ıv(x, y).

C
f(z) dz =

C
(u(x, y) + ıv(x, y))(dx + ı dy)

C
f(z) dz =

C
(u(x, y) dx −v(x, y) dy) + ı

C
(v(x, y) dx + u(x, y) dy) (10.2)
Evaluation. Let the contour C b e parametrized by z = z(t) for t
0
≤ t ≤ t
1
. Then the differential on the contour

is dz = z

(t) dt. Using the parameterization we can evaluate a contour integral in terms of a definite integral.

C
f(z) dz =

t
1
t
0
f(z(t))z

(t) dt
Example 10.2.1 Let C be the positively oriented unit circle about the origin in the complex plane. Evaluate:
1.

C
z dz
2.

C
1
z
dz
3.

C
1
z

|dz|
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In each case we parameterize the contour and then do the integral.
1.
z =
e
ıθ
, dz = ı
e
ıθ
dθ

C
z dz =

2π
0
e
ıθ
ı
e
ıθ
dθ
=

1
2
e
ı2θ


2π
0
=

1
2
e
ı4π
−
1
2
e
ı0

= 0
2.

C
1
z
dz =

2π
0
1
e
ıθ
ı
e
ıθ

dθ = ı

2π
0
dθ = ı2π
3.
|dz| =


ı
e
ıθ
dθ


=


ı
e
ıθ


|dθ| = |dθ|
Since dθ is positive in this case, |dθ| = dθ.

C
1
z
|dz| =


2π
0
1
e
ıθ
dθ =

ı
e
−ıθ

2π
0
= 0
10.2.1 Maximum Modulus Integral Bound
The absolute value of a real integral obeys the inequality





b
a
f(x) dx




≤


b
a
|f(x)||dx| ≤ (b − a) max
a≤x≤b
|f(x)|.
466
Now we prove the analogous result for the modu lus of a contour integral.





C
f(z) dz




=





lim
∆z→0
n−1

k=0

f(ζ
k
)∆z
k





≤ lim
∆z→0
n−1

k=0
|f(ζ
k
)||∆z
k
|
=

C
|f(z)||dz|
≤

C

max
z∈C
|f(z)|


|dz|
=

max
z∈C
|f(z)|


C
|dz|
=

max
z∈C
|f(z)|

× (length of C)
Result 10.2.1 Maximum Modulus Integral Bound.





C
f(z) dz





≤

C
|f(z)||dz| ≤

max
z∈C
|f(z)|

(length of C)
10.3 The Cauchy-Goursat Theorem
Let f(z) be analytic in a compact, closed, connected domain D. We consider the integral of f(z) on the boundary of
the domain.

∂D
f(z) dz =

∂D
ψ(x, y)(dx + ı dy) =

∂D
ψ dx + ıψ dy
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Recall Green’s Theorem.

∂D
P dx + Q dy =

D
(Q

x
− P
y
) dx dy
If we assume that f

(z) is continuous, we can apply Green’s Theorem to the integral of f(z) on ∂D.

∂D
f(z) dz =

∂D
ψ dx + ıψ dy =

D
(ıψ
x
− ψ
y
) dx dy
Since f(z) is analytic, it satisfies the Cauch y-Rie mann equation ψ
x
= −ıψ
y
. The integrand in the area integral,
ıψ
x
− ψ
y
, is zero. Thus the contour integral vanishes.


∂D
f(z) dz = 0
This is known as Cauchy’s Theorem. The assumption that f

(z) is continuous is not necessary, but it makes the
proof much simpler because we can use Green’s Theorem. If we remove this restriction the result is known as the
Cauchy-Goursat Theorem. The proof of this result is omitted.
Result 10.3.1 The Cauchy-Goursat Theorem. If f(z) is analytic in a compact, closed,
connected domain D then the integral of f(z) on the boundary of the domain vanishes.

∂D
f(z) dz =

k

C
k
f(z) dz = 0
Here the set of contours {C
k
} make up the positively oriented boundary ∂D of the domain
D.
As a special case of the Cauchy-Goursat theorem we can consider a simply-connected region. For this the boundary
is a Jordan curve. We can state the theorem in terms of this curve instead of referring to the boundary.
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Result 10.3.2 The Cauchy-Goursat Theorem for Jordan Curves. If f(z) is analytic
inside and on a simple, closed contour C, then

C

f(z) dz = 0
Example 10.3.1 Let C be the unit circle about the origin with positive orientation. In Example 10.2.1 we calculated
that

C
z dz = 0
Now we can evaluate the integral without parameterizing the curve. We simply note that the integrand is analytic
inside and on the circle, which is simple and closed. By the Cauchy-Goursat Theorem, the integral vanishes.
We cannot apply the Cauchy-Goursat theorem to evaluate

C
1
z
dz = ı2π
as the integrand is not analytic at z = 0.
Example 10.3.2 Consider the domain D = {z | |z| > 1}. The boundary of the domain is the unit circle with negative
orientation. f(z) = 1/z is analytic on D and its boundary. However

∂D
f(z) dz does not vanish and we cannot apply
the Cauchy-Goursat Theorem. This is because the domain is not compact.
10.4 Contour Deformation
Path Independence. Consider a function f(z) that is analytic on a simply connected domain a contour C in that
domain with end points a and b. The contour integral

C
f(z) dz is independent of the path connecting the end points
and can be denoted

b

a
f(z) dz. This result is a direct consequence of the Cauchy-Goursat Theorem. Let C
1
and C
2
be two different paths connecting the points. Let −C
2
denote the second contour with the opposite orientation. Let
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C be the contour which is the union of C
1
and −C
2
. By the Cauchy-Goursat theorem, the integral along this contour
vanishes.

C
f(z) dz =

C
1
f(z) dz +

−C
2
f(z) dz = 0
This implies that the integrals along C
1
and C
2

are equal.

C
1
f(z) dz =

C
2
f(z) dz
Thus contour integrals on simply connected domains are independent of path. This result does not hold for multiply
connected domains.
Result 10.4.1 Path Independence. Let f(z) be analytic on a simply connected domain.
For points a and b in the domain, the contour integral,

b
a
f(z) dz
is independent of the path conne cting the points.
Deforming Contours. Consider two simple, closed, positively oriented contours, C
1
and C
2
. Let C
2
lie completely
within C
1
. If f(z) is analytic on and between C
1
and C

2
then the integrals of f(z) along C
1
and C
2
are equal.

C
1
f(z) dz =

C
2
f(z) dz
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Again, this is a direct consequence of the Cauchy-Goursat Theorem. Let D be the domain on and between C
1
and C
2
.
By the Cauchy-Goursat Theorem the integral along the boundary of D vanishes.

C
1
f(z) dz +

−C
2
f(z) dz = 0


C
1
f(z) dz =

C
2
f(z) dz
By following this li ne of reasoning, we see that we can deform a contour C without changing the value of

C
f(z) dz
as long as we stay on the domain where f(z) is analytic.
Result 10.4.2 Contour Deformation. Let f(z) be analytic on a domain D. If a set of
closed contours {C
m
} can be continuously deformed on the domain D to a set of contours
{Γ
n
} then the integrals along {C
m
} and {Γ
n
} are equal.

{C
m
}
f(z) dz =

{Γ

n
}
f(z) dz
10.5 Morera’s Theorem.
The converse of the Cauchy-Goursat theorem is Morera’s Theorem. If the integrals of a continuous function f(z)
vanish along all possible simple, closed contours in a domain, then f(z) is analytic on that domain. To prove Morera’s
Theorem we will assume that first partial derivatives of f(z) = u(x, y) + ıv(x, y) are continuous, although the result
can be derived without this restriction. Let the simple, closed contour C be the boundary of D which is contained in
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the domain Ω.

C
f(z) dz =

C
(u + ıv)(dx + ı dy)
=

C
u dx −v dy + ı

C
v dx + u dy
=

D
(−v
x
− u
y

) dx dy + ı

D
(u
x
− v
y
) dx dy
= 0
Since the two integrands are continuous and vanish for all C in Ω, we conclude that the integrands are identically zero.
This implies that the Cauchy-Riemann equations,
u
x
= v
y
, u
y
= −v
x
,
are satisfied. f(z) is analytic in Ω.
The converse of the Cauchy-Goursat theorem is Morera’s Theorem. If the integrals of a continuous function f(z)
vanish along all possible simple, closed contours in a domain, then f(z) is analytic on that domain. To prove Morera’s
Theorem we will assume that first partial derivatives of f(z) = φ(x, y) are continuous, although the result can be
derived without this restriction. Let the simple, closed contour C be the boundary of D which is contained in the
domain Ω.

C
f(z) dz =


C
(φ dx + ıφ dy)
=

D
(ıφ
x
− φ
y
) dx dy
= 0
Since the integrand, ıφ
x
− φ
y
is continuous and vanishes for all C in Ω, we conclude that the integrand is identically
zero. This implies that the Cauchy-Riemann equation,
φ
x
= −ıφ
y
,
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is satisfied. We conclude that f(z) is analytic in Ω.
Result 10.5.1 Morera’s Theorem. If f(z) is continuous in a simply c onnected domain Ω
and

C
f(z) dz = 0
for all possible si mple , closed contours C in the domain, then f(z) is analytic in Ω.

10.6 Indefinite Integrals
Consider a function f(z) which is analytic in a domain D. An anti-derivative or indefinite integral (or simply integral)
is a function F(z) which satisfies F

(z) = f (z). This integral exis ts and is unique up to an additive constant. Note
that if the domain is not connected, then the additive constants in each connected component are independent. The
indefinite integrals are denoted:

f(z) dz = F(z) + c.
We will prove existence later by writing an indefinite integral as a contour integral. We briefly consid er uniqueness
of the indefinite integral here. Let F (z) and G(z) be integrals of f(z). Then F

(z) − G

(z) = f(z) − f(z) = 0.
Although we do not prove it, it certainly makes sense that F (z) − G(z) is a constant on each connected component
of the domain. I ndefi nite integrals are unique up to an additive constant.
Integrals of analytic functions have all the nice properties of integrals of functions of a real variable. All the formulas
from integral tables, including things like integration by parts, carry over directly.
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10.7 Fundamental Theorem of Calculus via Primitives
10.7.1 Line Integrals and Primitives
Here we review some concepts from vector calculus. An alagous to an in tegral in fu nctions of a single variable is
a primitive in functions of several variables. Consider a function f(x). F (x) is an integral of f(x) if and only if
dF = f dx. Now we move to functions of x and y. Let P (x, y) and Q(x, y) be defined on a simply connected domain.
A primitive Φ satisfies
dΦ = P dx + Q dy.
A necessary and sufficient condition for the existence of a primitive is that P
y
= Q

x
. The definite integral can be
evaluated in terms of the primitive.

(c,d)
(a,b)
P dx + Q dy = Φ(c, d) − Φ(a, b)
10.7.2 Contour Integrals
Now consider integral along the contour C of the function f(z) = φ(x, y).

C
f(z) dz =

C
(φ dx + ıφ dy)
A primitive Φ of φ dx + ıφ dy exists if and only if φ
y
= ıφ
x
. We recognize this as the Cauch-Riemann equation,
φ
x
= −ıφ
y
. Thus a primitive exists if and only if f(z) is analytic. If so, then
dΦ = φ dx + ıφ dy.
How do we find the primitive Φ that satisfies Φ
x
= φ and Φ
y

= ıφ? Note that choosing Ψ(x, y) = F (z) where F (z)
is an anti-derivative of f(z), F

(z) = f (z), does the trick. We express the complex derivative as partial derivatives in
the coordinate directions to show this.
F

(z) = f(z) = ψ(x, y), F

(z) = Φ
x
= −ıΦ
y
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From this we see that Φ
x
= φ and Φ
y
= ıφ so Φ(x, y) = F (z) is a primitive. Since we can evaluate the line integral
of (φ dx + ıφ dy),

(c,d)
(a,b)
(φ dx + ıφ dy) = Φ(c, d) − Φ(a, b),
We can evaluate a definite integral of f in terms of its indefinite integral, F .

b
a
f(z) dz = F(b) −F(a)
This is the Fundamental Theorem of Calculus for functions of a complex variable.

10.8 Fundamental Theorem of Calculus via Complex Calculus
Result 10.8.1 Constructing an Indefinite Integral. If f(z) is analytic in a simply con-
nected domain D and a is a point in the domain, then
F (z) =

z
a
f(ζ) dζ
is analytic in D and is an indefinite integral of f(z), (F

(z) = f(z)).
Now we consider anti-derivatives and definite integrals without using vector calculus. From real variables we know
that we can construct an integral of f(x) with a definite integral.
F (x) =

x
a
f(ξ) dξ
Now we will prove the analogous property for functions of a complex variable.
F (z) =

z
a
f(ζ) dζ
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Let f(z) be analytic in a simply connected domain D and let a be a point in the domain. To show that F(z) =

z
a
f(ζ) dζ

is an integral of f(z), we apply the limit definition of differentiation.
F

(z) = lim
∆z→0
F (z + ∆z) − F(z)
∆z
= lim
∆z→0
1
∆z


z+∆z
a
f(ζ) dζ −

z
a
f(ζ) dζ

= lim
∆z→0
1
∆z

z+∆z
z
f(ζ) dζ
The integral is independent of path. We choose a straight line connecting z and z + ∆z. We add and subtract

∆zf(z) =

z+∆z
z
f(z) dζ from the expression for F

(z).
F

(z) = lim
∆z→0
1
∆z

∆zf(z) +

z+∆z
z
(f(ζ) − f(z)) dζ

= f(z) + lim
∆z→0
1
∆z

z+∆z
z
(f(ζ) − f(z)) dζ
Since f(z) is analytic, it is certainly continuous. This means that
lim

ζ→z
f(ζ) = 0.
The limit term vanishes as a result of this continuity.
lim
∆z→0




1
∆z

z+∆z
z
(f(ζ) − f(z)) dζ




≤ lim
∆z→0
1
|∆z|
|∆z| max
ζ∈[z z+∆z]
|f(ζ) − f(z)|
= lim
∆z→0
max
ζ∈[z z+∆z]

|f(ζ) − f(z)|
= 0
Thus F

(z) = f(z).
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This results demonstrates the existence of the indefinite integral. We will use this to prove the Fundamental Theorem
of Calculus for functions of a complex variable.
Result 10.8.2 Fundamental Theorem of Calculus. If f(z) is analytic i n a simp ly con-
nected domain D then

b
a
f(z) dz = F (b) − F (a)
where F (z) is any indefinite integral of f(z).
From Result 10.8.1 we know that

b
a
f(z) dz = F(b) + c.
(Here we are considering b to be a variable.) The case b = a determines the constant.

a
a
f(z) dz = F(a) + c = 0
c = −F (a)
This proves the Fundamental Theorem of Calculus for functions of a complex variable.
Example 10.8.1 Consider the integral

C

1
z − a
dz
where C is any closed contour that goes around the p oint z = a once in the positive direction. We use the Fundamental
Theorem of Calculus to evaluate the integral. We start at a point on the contour z −a = r
e
ıθ
. When we traverse the
contour once in the positive direction we end at the point z − a = r
e
ı(θ+2π)
.

C
1
z − a
dz = [log(z − a)]
z−a=r
e
ı(θ+2π)
z−a=r
e
ıθ
= Log r + ı(θ + 2π) −(Log r + ıθ)
= ı2π
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