Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

11.1 Cauchy’s Integral Formula
Result 11.1.1 Cauchy’s Integral Formula. If f(ζ) is analytic in a compact, closed, con-
nected domain D and z is a point in the interior of D then
f(z) =
1
ı2π

∂D
f(ζ)
ζ −z
dζ =
1
ı2π

k

C
k
f(ζ)
ζ −z
dζ. (11.1)
Here the set of contours {C
k
} make up the positively oriented boundary ∂D of the domain
D. More generally, we have
f
(n)
(z) =
n!
ı2π


∂D
f(ζ)
(ζ −z)
n+1
dζ =
n!
ı2π

k

C
k
f(ζ)
(ζ −z)
n+1
dζ. (11.2)
Cauchy’s Formula shows that the value of f(z) and all its derivatives in a domain are determined by the value of
f(z) on the boundary of the domain. Consider the first formula of the result, Equation 11.1. We deform the contour
to a circle of radius δ about the point ζ = z.

C
f(ζ)
ζ − z
dζ =

C
δ
f(ζ)
ζ − z

dζ
=

C
δ
f(z)
ζ − z
dζ +

C
δ
f(ζ) −f(z)
ζ − z
dζ
We use the result of Example 10.8.1 to evaluate the first integral.

C
f(ζ)
ζ − z
dζ = ı2πf(z) +

C
δ
f(ζ) −f(z)
ζ − z
dζ
494
The remaining integral along C
δ
vanishes as δ → 0 because f(ζ) is continuous. We demonstrate this with the maximum

modulus integral bound. The length of the path of integration is 2πδ.
lim
δ→0





C
δ
f(ζ) −f(z)
ζ − z
dζ




≤ lim
δ→0

(2πδ)
1
δ
max
|ζ−z|=δ
|f(ζ) −f(z)|

≤ lim
δ→0


2π max
|ζ−z|=δ
|f(ζ) −f(z)|

= 0
This gives us the desired result.
f(z) =
1
ı2π

C
f(ζ)
ζ − z
dζ
We derive the second formula, Equation 11.2, from the first by differentiating with respect to z. Note that the
integral converges uniformly for z in any closed subset of the interior of C. Thus we can differentiate with respect to
z and interchange the order of differentiation and integration.
f
(n)
(z) =
1
ı2π
d
n
dz
n

C
f(ζ)
ζ − z

dζ
=
1
ı2π

C
d
n
dz
n
f(ζ)
ζ − z
dζ
=
n!
ı2π

C
f(ζ)
(ζ − z)
n+1
dζ
Example 11.1.1 Consider the following integrals where C is the positive contour on the unit circle. For the third
integral, the point z = −1 is removed from the contour.
1.

C
sin

cos


z
5

dz
2.

C
1
(z −3)(3z − 1)
dz
495
3.

C
√
z dz
1. Since sin (cos (z
5
)) is an analytic function inside the unit circle,

C
sin

cos

z
5

dz = 0

2.
1
(z−3)(3z−1)
has singulari ties at z = 3 and z = 1/3. Since z = 3 is outside the contour, only the singularity at
z = 1/3 will contribute to the value of the integral. We will evaluate this integral using the Cauchy integral
formula.

C
1
(z −3)(3z − 1)
dz = ı2π

1
(1/3 −3)3

= −
ıπ
4
3. Since the curve is not closed, we cannot apply the Cauchy integral formula. Note that
√
z is single-valued and
analytic in the complex plane with a branch cut on the negative real axis. Thus we use the Fundamental Theorem
of Calculus.

C
√
z dz =

2
3

√
z
3

e
ıπ
e
−ıπ
=
2
3

e
ı3π/2
−
e
−ı3π/2

=
2
3
(−ı −ı)
= −ı
4
3
Cauchy’s Inequality. Suppose the f(ζ) is analytic in the closed disk |ζ − z| ≤ r. By Cauchy’s integral formula,
f
(n)
(z) =
n!

ı2π

C
f(ζ)
(ζ − z)
n+1
dζ,
496
where C is the circle of radius r centered about the point z. We use this to obtain an upper bound on the modulus of
f
(n)
(z).


f
(n)
(z)


=
n!
2π





C
f(ζ)
(ζ − z)

n+1
dζ




≤
n!
2π
2πr max
|ζ−z|=r




f(ζ)
(ζ − z)
n+1




=
n!
r
n
max
|ζ−z|=r
|f(ζ)|
Result 11.1.2 Cauchy’s Inequality. If f(ζ) is analytic in |ζ − z| ≤ r then




f
(n)
(z)



≤
n!M
r
n
where |f(ζ)| ≤ M for all |ζ −z| = r.
Liouville’s Theorem. Consider a function f(z) that is analytic and bounded, (|f(z)| ≤ M), in the complex plane.
From Cauchy’s inequality,
|f

(z)| ≤
M
r
for any positive r. By taking r → ∞, we see that f

(z) is identically zero for all z. Thus f(z) is a constant.
Result 11.1.3 Liouville’s Theorem. If f(z) is analytic and |f(z)| is bounded in the complex
plane then f(z) is a constant.
The Fundamental Theorem of Algebra. We will prove that every polynomial of degree n ≥ 1 has exactly n
roots, counting multiplicities. First we demonstrate that each such polynomial has at least one root. Suppose that an
497
n

th
degree polynomial p(z) has no roots. Let the lower bound on the modulus of p(z) be 0 < m ≤ |p(z)|. The function
f(z) = 1/p(z) is analytic, (f

(z) = p

(z)/p
2
(z)), and bounded, (|f(z)| ≤ 1/m), in the extended complex plane. Using
Liouville’s theorem we conclude that f(z) and hence p(z) are constants, which yields a contradiction. Therefore every
such polynomial p(z) must have at least one root.
Now we show that we can factor the root out of the polynomial. Let
p(z) =
n

k=0
p
k
z
k
.
We note that
(z
n
− c
n
) = (z − c)
n−1

k=0

c
n−1−k
z
k
.
Suppose that the n
th
degree polynomial p(z) has a root at z = c.
p(z) = p(z) − p(c)
=
n

k=0
p
k
z
k
−
n

k=0
p
k
c
k
=
n

k=0
p

k

z
k
− c
k

=
n

k=0
p
k
(z −c)
k−1

j=0
c
k−1−j
z
j
= (z − c)q(z)
Here q(z) is a polynomial of degree n − 1. By induction, we see that p(z) has exactly n roots.
Result 11.1.4 Fundamental Theorem of Algebra. Every polynomial of degree n ≥ 1 has
exactly n roots, counting multiplicities.
498
Gauss’ Mean Value Theorem. Let f(ζ) be analytic in |ζ − z| ≤ r. By Cauchy’s integral formula,
f(z) =
1
ı2π


C
f(ζ)
ζ − z
dζ,
where C is the circle |ζ −z| = r. We parameterize the contour with ζ = z + r
e
ıθ
.
f(z) =
1
ı2π

2π
0
f(z + r
e
ıθ
)
r
e
ıθ
ır
e
ıθ
dθ
Writing this in the form,
f(z) =
1
2πr


2π
0
f(z + r
e
ıθ
)r dθ,
we see that f(z) is the average value of f(ζ) on the circle of radius r about the point z.
Result 11.1.5 Gauss’ Average Value Theorem. If f(ζ) is analytic in |ζ −z| ≤ r then
f(z) =
1
2π

2π
0
f(z + r
e
ıθ
) dθ.
That is, f(z) is equal to its average value on a circle of radius r about the point z.
Extremum Modulus Theorem. Let f (z) be analytic in closed, connected domain, D. The extreme values of the
modulus of the function must occur on the boundary. If |f(z)| has an interior extrema, then the function is a constant.
We will show this with proof by contradiction. Assume that |f(z)| has an interior maxima at the point z = c. This
means that there exists an neighborhood of the point z = c for which |f(z)| ≤ |f(c)|. Choose an  so that the set
|z −c| ≤  lies inside this neighborhood. First we use Gauss’ mean value theorem.
f(c) =
1
2π

2π

0
f

c + 
e
ıθ

dθ
499
We get an upper bound on |f(c)| with the maximum mod ulu s integral bound.
|f(c)| ≤
1
2π

2π
0


f

c + 
e
ıθ



dθ
Since z = c is a maxima of |f(z)| we can get a lowe r bound on |f(c)|.
|f(c)| ≥
1

2π

2π
0


f

c + 
e
ıθ



dθ
If |f(z)| < |f(c)| for any point on |z −c| = , then the continuity of f(z) implies that |f(z)| < |f(c)| in a neighborhood
of that point which would make the value of the integral of |f(z)| strictly less than |f(c)|. Thus we conclude that
|f(z)| = |f(c)| for all |z − c| = . Since we can repeat the above procedure for any circle of radius smaller than ,
|f(z)| = |f(c)| for all |z −c| ≤ , i.e. all the points in the disk of radius  about z = c are also maxima. By recursively
repeating this procedure points in this disk, we see that |f(z)| = |f(c)| for all z ∈ D. This implies that f(z) is a
constant in the domain. By reversing the inequalities in the above method we see that the minimum modulus of f(z)
must also occur on the boundary.
Result 11.1.6 Extremum Modulus Theorem. Let f(z) be analytic in a closed, connected
domain, D. The extreme values of the modulus of the function must o c cur on the boundary.
If |f(z)| has an interior extrema, then the function is a constant.
500
11.2 The Argument Theorem
Result 11.2.1 The Argument Theorem. Let f(z) be analytic inside and on C except for
isolated poles inside the contour. Let f(z) be nonzero on C.
1

ı2π

C
f

(z)
f(z)
dz = N −P
Here N is the number of zeros and P the number of poles, counting multiplicities, of f(z)
inside C.
First we will simplify the problem and consider a function f(z) that has one zero or one pole. Let f(z) be analytic
and nonzero inside and on A except for a zero of order n at z = a. Then we can write f(z) = (z −a)
n
g(z) where g(z)
is analytic and nonzero inside and on A. The integral of
f

(z)
f(z)
along A is
1
ı2π

A
f

(z)
f(z)
dz =
1

ı2π

A
d
dz
(log(f(z))) dz
=
1
ı2π

A
d
dz
(log((z −a)
n
) + log(g(z))) dz
=
1
ı2π

A
d
dz
(log((z −a)
n
)) dz
=
1
ı2π


A
n
z −a
dz
= n
501
Now let f (z) be analytic and nonzero inside and on B except for a pole of order p at z = b. Then we can write
f(z) =
g(z)
(z−b)
p
where g(z) is analytic and nonzero inside and on B. The integral of
f

(z)
f(z)
along B is
1
ı2π

B
f

(z)
f(z)
dz =
1
ı2π

B

d
dz
(log(f(z))) dz
=
1
ı2π

B
d
dz

log((z −b)
−p
) + log(g(z))

dz
=
1
ı2π

B
d
dz

log((z −b)
−p
)+

dz
=

1
ı2π

B
−p
z −b
dz
= −p
Now consider a function f(z) that is analytic inside an on the contour C except for isolated poles at the points
b
1
, . . . , b
p
. Let f(z) be nonzero except at the isolated points a
1
, . . . , a
n
. Let the contours A
k
, k = 1, . . . , n, be simple,
positive contours which contain the zero at a
k
but no other poles or zeros of f(z). Likewise, let the contours B
k
,
k = 1, . . . , p be simple, positive contours which contain the pole at b
k
but no other poles of zeros of f(z). (See
Figure 11.1.) By deforming the contour we obtain


C
f

(z)
f(z)
dz =
n

j=1

A
j
f

(z)
f(z)
dz +
p

k=1

B
j
f

(z)
f(z)
dz.
From this we obtain Result 11.2.1.
11.3 Rouche’s Theorem

Result 11.3.1 Rouche’s Theorem. Let f(z) and g(z) be analytic inside and on a simple,
closed contour C. If |f(z)| > |g(z)| on C then f(z) and f(z) + g(z) have the same number
of zeros inside C and no zeros on C.
502
C
A
1
B
1
B
3
B
2
A
2
Figure 11.1: Deforming the contour C.
First note that since |f(z)| > |g(z)| on C, f(z) is nonzero on C. The inequality implies that |f(z) + g(z)| > 0
on C so f(z) + g(z) has no zeros on C. We well count the number of zeros of f(z) and g(z) using the Argument
Theorem, (Result 11.2.1). The number of zeros N of f(z) inside the contour is
N =
1
ı2π

C
f

(z)
f(z)
dz.
Now consider the number of zeros M of f(z) + g(z). We introduce the function h(z) = g(z)/f(z).

M =
1
ı2π

C
f

(z) + g

(z)
f(z) + g(z)
dz
=
1
ı2π

C
f

(z) + f

(z)h(z) + f (z)h

(z)
f(z) + f(z)h(z)
dz
=
1
ı2π


C
f

(z)
f(z)
dz +
1
ı2π

C
h

(z)
1 + h(z)
dz
= N +
1
ı2π
[log(1 + h(z))]
C
= N
503
(Note that since |h(z)| < 1 on C, (1 + h(z)) > 0 on C and the value of log(1 + h(z)) does not not change in
traversing the contour.) This demonstrates that f(z) and f(z) + g(z) have the same number of zeros inside C and
proves the result.
504
11.4 Exercises
Exercise 11.1
What is
(arg(sin z))



C
where C is the unit circle?
Exercise 11.2
Let C be the circle of radius 2 centered about the origin and oriented in the positive direction. Evaluate the following
integrals:
1.

C
sin z
z
2
+5
dz
2.

C
z
z
2
+1
dz
3.

C
z
2
+1
z

dz
Exercise 11.3
Let f(z) be analytic and bounded (i.e. |f(z)| < M) for |z| > R, but not necessarily analytic for |z| ≤ R. Let the
points α and β lie inside the circle |z| = R. Evaluate

C
f(z)
(z −α)(z −β)
dz
where C is any cl osed contour outside |z| = R, containing the circle |z| = R. [Hint: consider the circle at infinity] Now
suppose that in addition f(z) is analytic everywhere. Deduce that f(α) = f(β).
Exercise 11.4
Using Rouche’s theorem show that all the roots of the equation p(z) = z
6
−5z
2
+10 = 0 lie in the annulus 1 < |z| < 2.
Exercise 11.5
Evaluate as a function of t
ω =
1
ı2π

C
e
zt
z
2
(z
2

+ a
2
)
dz,
505
where C is any positively oriented contour surrounding the circle |z| = a.
Exercise 11.6
Consider C
1
, (the positively oriented circle |z| = 4), and C
2
, (the positively oriented boundary of the square whose
sides lie along the lines x = ±1, y = ±1). Explain why

C
1
f(z) dz =

C
2
f(z) dz
for the functions
1. f(z) =
1
3z
2
+ 1
2. f(z) =
z
1 −

e
z
Exercise 11.7
Show that if f(z) is of the form
f(z) =
α
k
z
k
+
α
k−1
z
k−1
+ ···+
α
1
z
+ g(z), k ≥ 1
where g is analytic inside and on C, (the positive circle |z| = 1), then

C
f(z) dz = ı2πα
1
.
Exercise 11.8
Show that if f(z) is analytic within and on a simple closed contour C and z
0
is not on C then


C
f

(z)
z −z
0
dz =

C
f(z)
(z −z
0
)
2
dz.
Note that z
0
may be either inside or outside of C.
506
Exercise 11.9
If C is the positive circle z =
e
ıθ
show that for any real constant a,

C
e
az
z
dz = ı2π

and hence

π
0
e
a cos θ
cos(a sin θ) dθ = π.
Exercise 11.10
Use Cauchy-Goursat, the generalized Cauchy integral formula, and suitable extensions to multiply-connected domains
to evaluate the following integrals. Be sure to justify your approach in each case.
1.

C
z
z
3
− 9
dz
where C is the positively oriented rectangle whose sides lie along x = ±5, y = ±3.
2.

C
sin z
z
2
(z −4)
dz,
where C is the positively oriented circle |z| = 2.
3.


C
(z
3
+ z + ı) sin z
z
4
+ ız
3
dz,
where C is the positively oriented circle |z| = π.
4.

C
e
zt
z
2
(z + 1)
dz
where C is any positive simple closed contour surrounding |z| = 1.
507
Exercise 11.11
Use Liouville’s theorem to prove the following:
1. If f(z) is entire with (f(z)) ≤ M for all z then f(z) is constant.
2. If f(z) is entire with |f
(5)
(z)| ≤ M for all z then f(z) is a polynomial of degree at most five.
Exercise 11.12
Find all functions f(z) analytic in the domain D : |z| < R that satisfy f(0) =
e

ı
and |f(z)| ≤ 1 for all z in D.
Exercise 11.13
Let f(z) =

∞
k=0
k
4

z
4

k
and evaluate the following contour integrals, providing justification in each case:
1.

C
cos(ız)f(z) dz C is the positive circle |z −1| = 1.
2.

C
f(z)
z
3
dz C is the positive circle |z| = π.
508
11.5 Hints
Hint 11.1
Use the argument theorem.

Hint 11.2
Hint 11.3
To evaluate the integral, consider the circle at infinity.
Hint 11.4
Hint 11.5
Hint 11.6
Hint 11.7
Hint 11.8
Hint 11.9
Hint 11.10
509
Hint 11.11
Hint 11.12
Hint 11.13
510
11.6 Solutions
Solution 11.1
Let f(z) be analytic insid e and on the contour C. Let f(z) be nonzero on the contour. The argument theorem states
that
1
ı2π

C
f

(z)
f(z)
dz = N − P,
where N is the number of zeros and P is the number of poles, (counting multiplicities), of f(z) inside C. The theorem
is aptly named, as

1
ı2π

C
f

(z)
f(z)
dz =
1
ı2π
[log(f(z))]
C
=
1
ı2π
[log |f(z)|+ ı arg(f(z))]
C
=
1
2π
[arg(f(z))]
C
.
Thus we could write the argument theorem as
1
ı2π

C
f


(z)
f(z)
dz =
1
2π
[arg(f(z))]
C
= N −P.
Since sin z has a single zero and no poles inside the unit circle, we have
1
2π
arg(sin(z))


C
= 1 − 0
arg(sin(z))


C
= 2π
Solution 11.2
1. Since the integrand
sin z
z
2
+5
is analytic inside and on the contour, (the only singularities are at z = ±ı
√

5 and at
infinity), the integral is zero by Cauchy’s Theorem.
511
2. First we expand the integrand in partial fractions.
z
z
2
+ 1
=
a
z −ı
+
b
z + ı
a =
z
z + ı




z=ı
=
1
2
, b =
z
z −ı





z=−ı
=
1
2
Now we can do the in tegral with Cauchy’s formula.

C
z
z
2
+ 1
dz =

C
1/2
z −ı
dz +

C
1/2
z + ı
dz
=
1
2
ı2π +
1
2

ı2π
= ı2π
3.

C
z
2
+ 1
z
dz =

C

z +
1
z

dz
=

C
z dz +

C
1
z
dz
= 0 + ı2π
= ı2π
Solution 11.3

Let C be the circle of radius r, (r > R), centered at the origin. We get an upper bound on the integral with the
Maximum Modulus Integral Bound, (Result 10.2.1).





C
f(z)
(z −α)(z −β)
dz




≤ 2πr max
|z|=r




f(z)
(z −α)(z −β)




≤ 2πr
M
(r − |α|)(r −|β|)

512
By taking the limit as r → ∞ we see that the modulus of the integral is bounded above by zero. Thus the integral
vanishes.
Now we assume that f(z) is analytic and evaluate the integral with Cauchy’s Integral Formula. (We assume that
α = β.)

C
f(z)
(z −α)(z −β)
dz = 0

C
f(z)
(z −α)(α −β)
dz +

C
f(z)
(β −α)(z −β)
dz = 0
ı2π
f(α)
α −β
+ ı2π
f(β)
β −α
= 0
f(α) = f(β)
Solution 11.4
Consider the circle |z| = 2. On this circle:

|z
6
| = 64
| −5z
2
+ 10| ≤ | − 5z
2
| + |10| = 30
Since |z
6
| < |−5z
2
+ 10| on |z| = 2, p(z) has the same number of roots as z
6
in |z| < 2. p(z) has 6 roots in |z| < 2.
Consider the circle |z| = 1. On this circle:
|10| = 10
|z
6
− 5z
2
| ≤ |z
6
| + | −5z
2
| = 6
Since |z
6
− 5z
2

| < |10| on |z| = 1, p(z) has the same numb er of roots as 10 in |z| < 1. p(z) has no roots in |z| < 1.
On the unit circle,
|p(z)| ≥ |10|− |z
6
| −|5z
2
| = 4.
Thus p(z) has no roots on the unit circle.
We conclude that p(z) has exactly 6 roots in 1 < |z| < 2.
513
Solution 11.5
We evaluate the integral with Cauchy’s Integral Formula.
ω =
1
ı2π

C
e
zt
z
2
(z
2
+ a
2
)
dz
ω =
1
ı2π


C

e
zt
a
2
z
2
+
ı
e
zt
2a
3
(z −ıa)
−
ı
e
zt
2a
3
(z + ıa)

dz
ω =

d
dz
e

zt
a
2

z=0
+
ı
e
ıat
2a
3
−
ı
e
−ıat
2a
3
ω =
t
a
2
−
sin(at)
a
3
ω =
at −sin(at)
a
3
Solution 11.6

1. We factor the denominator of the integrand.
1
3z
2
+ 1
=
1
3(z − ı
√
3/3)(z + ı
√
3/3)
There are two first order poles which could contribute to the value of an integral on a closed path. Both poles
lie inside both contours. See Fi gure 11.2. We see that C
1
can be continuously deformed to C
2
on the domain
where the integrand is analytic. Thus the integrals have the same value.
2. We consider the integrand
z
1 −
e
z
.
Since
e
z
= 1 has the solutions z = ı2πn for n ∈ Z, the integrand has singularities at these points. There is a
removable singularity at z = 0 and first order poles at z = ı2πn for n ∈ Z \{0}. Each contour contains only the

singularity at z = 0. See Figure 11.3. We see that C
1
can be continuously deformed to C
2
on the domain where
the integrand is analytic. Thus the integrals have the same value.
514
-4
-2 2
4
-4
-2
2
4
Figure 11.2: The contours and the singularities of
1
3z
2
+1
.
-6
-4
-2 2
4
6
-6
-4
-2
2
4

6
Figure 11.3: The contours and the singularities of
z
1−
e
z
.
515
Solution 11.7
First we write the integral of f(z) as a sum of integrals.

C
f(z) dz =

C

α
k
z
k
+
α
k−1
z
k−1
+ ···+
α
1
z
+ g(z)


dz
=

C
α
k
z
k
dz +

C
α
k−1
z
k−1
dz + ···+

C
α
1
z
dz +

C
g(z) dz
The integral of g(z) vanishes by the Cauchy-Goursat theorem. We evaluate the integral of α
1
/z with Cauchy’s integral
formula.


C
α
1
z
dz = ı2πα
1
We evaluate the remaining α
n
/z
n
terms with anti-derivatives. Each of these integrals vanish.

C
f(z) dz =

C
α
k
z
k
dz +

C
α
k−1
z
k−1
dz + ···+


C
α
1
z
dz +

C
g(z) dz
=

−
α
k
(k − 1)z
k−1

C
+ ···+

−
α
2
z

C
+ ı2πα
1
= ı2πα
1
Solution 11.8

We evaluate the integrals with the Cauchy integral formula. (z
0
is required to not be on C so the integrals exist.)

C
f

(z)
z −z
0
dz =

ı2πf

(z
0
) if z
0
is inside C
0 if z
0
is outside C

C
f(z)
(z −z
0
)
2
dz =


ı2π
1!
f

(z
0
) if z
0
is inside C
0 if z
0
is outside C
Thus we see that the integrals are equal.
516
Solution 11.9
First we evaluate the integral using the Cauchy Integral Formu la.

C
e
az
z
dz = [
e
az
]
z=0
= ı2π
Next we parameterize the path of integration. We use the periodicity of the cosine and sine to simplify the integral.


C
e
az
z
dz = ı2π

2π
0
e
a
e
ıθ
e
ıθ
ı
e
ıθ
dθ = ı2π

2π
0
e
a(cos θ+ı sin θ)
dθ = 2π

2π
0
e
a cos θ
(cos(sin θ) + ı sin(sin θ)) dθ = 2π


2π
0
e
a cos θ
cos(sin θ) dθ = 2π

π
0
e
a cos θ
cos(sin θ) dθ = π
Solution 11.10
1. We factor the integrand to s ee that there are singularities at the cube roots of 9.
z
z
3
− 9
=
z

z −
3
√
9

z −
3
√
9

e
ı2π/3

z −
3
√
9
e
−ı2π/3

Let C
1
, C
2
and C
3
be contours around z =
3
√
9, z =
3
√
9
e
ı2π/3
and z =
3
√
9
e

−ı2π/3
. See Figure 11.4. Let D be
the domain b etween C, C
1
and C
2
, i.e. the boundary of D is the union of C, −C
1
and −C
2
. Since the integrand
is analytic in D, the integral along the boundary of D vanishes.

∂D
z
z
3
− 9
dz =

C
z
z
3
− 9
dz +

−C
1
z

z
3
− 9
dz +

−C
2
z
z
3
− 9
dz +

−C
3
z
z
3
− 9
dz = 0
517