Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

The series about z = ∞ for 1/(z + 2) is
1
2 + z
=
1/z
1 + 2/z
=
1
z
∞

n=0
(−2/z)
n
, for |2/z| < 1
=
∞

n=0
(−1)
n
2
n
z
−n−1
, for |z| > 2
=
−1

n=−∞

(−1)
n+1
2
n+1
z
n
, for |z| > 2
To find the expansions in the three regions, we just choose the appropriate series.
1.
f(z) =
1
1 + z
−
1
2 + z
=
∞

n=0
(−1)
n
z
n
−
∞

n=0
(−1)
n
2

n+1
z
n
, for |z| < 1
=
∞

n=0
(−1)
n

1 −
1
2
n+1

z
n
, for |z| < 1
f(z) =
∞

n=0
(−1)
n
2
n+1
− 1
2
n+1

z
n
, for |z| < 1
614
2.
f(z) =
1
1 + z
−
1
2 + z
f(z) =
−1

n=−∞
(−1)
n+1
z
n
−
∞

n=0
(−1)
n
2
n+1
z
n
, for 1 < |z| < 2

3.
f(z) =
1
1 + z
−
1
2 + z
=
−1

n=−∞
(−1)
n+1
z
n
−
−1

n=−∞
(−1)
n+1
2
n+1
z
n
, for 2 < |z|
f(z) =
−1

n=−∞

(−1)
n+1
2
n+1
− 1
2
n+1
z
n
, for 2 < |z|
Solution 12.27
Laurent Series. We assume that m is a non-negative integer and that n is an integer. The Laurent series about the
point z = 0 of
f(z) =

z +
1
z

m
is
f(z) =
∞

n=−∞
a
n
z
n
where

a
n
=
1
ı2π

C
f(z)
z
n+1
dz
615
and C is a contour going around the origin once in the positive direction. We manipulate the coefficient integral into
the desired form.
a
n
=
1
ı2π

C
(z + 1/z)
m
z
n+1
dz
=
1
ı2π


2π
0
(
e
ıθ
+
e
−ıθ
)
m
e
ı(n+1)θ
ı
e
ıθ
dθ
=
1
2π

2π
0
2
m
cos
m
θ
e
−ınθ
dθ

=
2
m−1
π

2π
0
cos
m
θ(cos(nθ) − ı sin(nθ)) dθ
Note that cos
m
θ is even and sin(nθ) is odd about θ = π.
=
2
m−1
π

2π
0
cos
m
θ cos(nθ) dθ
Binomial Series. Now we find the binomial series expansion of f(z).

z +
1
z

m

=
m

n=0

m
n

z
m−n

1
z

n
=
m

n=0

m
n

z
m−2n
=
m

n=−m
m−n even


m
(m − n)/2

z
n
616
The coefficients in the series f(z) =

∞
n=−∞
a
n
z
n
are
a
n
=


m
(m−n)/2

−m ≤ n ≤ m and m −n even
0 otherwise
By equating the coefficients found by the two methods, we evaluate the desired integral.

2π
0

(cos θ)
m
cos(nθ) dθ =

π
2
m−1

m
(m−n)/2

−m ≤ n ≤ m and m −n even
0 otherwise
Solution 12.28
First we write f(z) in the form
f(z) =
g(z)
(z − ı/2)(z − 2)
2
.
g(z) is an entire function which grows no faster that z
3
at infinity. By expanding g(z) in a Taylor series about the
origin, we see that it is a polynomial of degree no greater than 3.
f(z) =
αz
3
+ βz
2
+ γz + δ

(z − ı/2)(z − 2)
2
Since f(z) is a rational function we expand it in partial fractions to obtain a form that is convenient to integrate.
f(z) =
a
z − ı/2
+
b
z − 2
+
c
(z − 2)
2
+ d
We use the value of the integrals of f(z) to determine the constants, a, b, c and d.

|z|=1

a
z − ı/2
+
b
z − 2
+
c
(z − 2)
2
+ d

dz = ı2π

ı2πa = ı2π
a = 1
617

|z|=3

1
z − ı/2
+
b
z − 2
+
c
(z − 2)
2
+ d

dz = 0
ı2π(1 + b) = 0
b = −1
Note that by applying the second constraint, we can change the third constraint to

|z|=3
zf(z) dz = 0.

|z|=3
z

1
z − ı/2

−
1
z − 2
+
c
(z − 2)
2
+ d

dz = 0

|z|=3

(z − ı/2) + ı/2
z − ı/2
−
(z − 2) + 2
z − 2
+
c(z − 2) + 2c
(z − 2)
2

dz = 0
ı2π

ı
2
− 2 + c


= 0
c = 2 −
ı
2
Thus we see that the function is
f(z) =
1
z − ı/2
−
1
z − 2
+
2 − ı/2
(z − 2)
2
+ d,
where d is an arbitrary constant. We can also write the function in the form:
f(z) =
dz
3
+ 15 − ı8
4(z − ı/2)(z − 2)
2
.
Complete Laurent Series. We find the complete Laurent series about z = 0 for each of the terms in the partial
618
fraction expansion of f(z).
1
z − ı/2
=

ı2
1 + ı2z
= ı2
∞

n=0
(−ı2z)
n
, for | − ı2z| < 1
= −
∞

n=0
(−ı2)
n+1
z
n
, for |z| < 1/2
1
z − ı/2
=
1/z
1 − ı/(2z)
=
1
z
∞

n=0


ı
2z

n
, for |ı/(2z)| < 1
=
∞

n=0

ı
2

n
z
−n−1
, for |z| < 2
=
−1

n=−∞

ı
2

−n−1
z
n
, for |z| < 2
=

−1

n=−∞
(−ı2)
n+1
z
n
, for |z| < 2
619
−
1
z − 2
=
1/2
1 − z/2
=
1
2
∞

n=0

z
2

n
, for |z/2| < 1
=
∞


n=0
z
n
2
n+1
, for |z| < 2
−
1
z − 2
= −
1/z
1 − 2/z
= −
1
z
∞

n=0

2
z

n
, for |2/z| < 1
= −
∞

n=0
2
n

z
−n−1
, for |z| > 2
= −
−1

n=−∞
2
−n−1
z
n
, for |z| > 2
620
2 − ı/2
(z − 2)
2
= (2 − ı/2)
1
4
(1 − z/2)
−2
=
4 − ı
8
∞

n=0

−2
n



−
z
2

n
, for |z/2| < 1
=
4 − ı
8
∞

n=0
(−1)
n
(n + 1)(−1)
n
2
−n
z
n
, for |z| < 2
=
4 − ı
8
∞

n=0
n + 1

2
n
z
n
, for |z| < 2
2 − ı/2
(z − 2)
2
=
2 − ı/2
z
2

1 −
2
z

−2
=
2 − ı/2
z
2
∞

n=0

−2
n

−

2
z

n
, for |2/z| < 1
= (2 − ı/2)
∞

n=0
(−1)
n
(n + 1)(−1)
n
2
n
z
−n−2
, for |z| > 2
= (2 − ı/2)
−2

n=−∞
(−n − 1)2
−n−2
z
n
, for |z| > 2
= −(2 − ı/2)
−2


n=−∞
n + 1
2
n+2
z
n
, for |z| > 2
We take the appropriate combination of these series to find the Laurent series expansions in the regions: |z| < 1/2,
621
1/2 < |z| < 2 and 2 < |z|. For |z| < 1/2, we have
f(z) = −
∞

n=0
(−ı2)
n+1
z
n
+
∞

n=0
z
n
2
n+1
+
4 − ı
8
∞


n=0
n + 1
2
n
z
n
+ d
f(z) =
∞

n=0

−(−ı2)
n+1
+
1
2
n+1
+
4 − ı
8
n + 1
2
n

z
n
+ d
f(z) =

∞

n=0

−(−ı2)
n+1
+
1
2
n+1

1 +
4 − ı
4
(n + 1)

z
n
+ d, for |z| < 1/2
For 1/2 < |z| < 2, we have
f(z) =
−1

n=−∞
(−ı2)
n+1
z
n
+
∞


n=0
z
n
2
n+1
+
4 − ı
8
∞

n=0
n + 1
2
n
z
n
+ d
f(z) =
−1

n=−∞
(−ı2)
n+1
z
n
+
∞

n=0


1
2
n+1

1 +
4 − ı
4
(n + 1)

z
n
+ d, for 1/2 < |z| < 2
For 2 < |z|, we have
f(z) =
−1

n=−∞
(−ı2)
n+1
z
n
−
−1

n=−∞
2
−n−1
z
n

− (2 − ı/2)
−2

n=−∞
n + 1
2
n+2
z
n
+ d
f(z) =
−2

n=−∞

(−ı2)
n+1
−
1
2
n+1
(1 + (1 − ı/4)(n + 1))

z
n
+ d, for 2 < |z|
Solution 12.29
The radius of convergence of the series for f(z) is
R = lim
n→∞





k
3
/3
k
(k + 1)
3
/3
k+1




= 3 lim
n→∞




k
3
(k + 1)
3





= 3.
622
Thus f(z) is a function which is analytic inside the circle of radius 3.
1. The integrand is analytic. Thus by Cauchy’s theorem the value of the integral is zero.

|z|=1
e
ız
f(z) dz = 0
2. We use Cauchy’s integral formula to evaluate the integral.

|z|=1
f(z)
z
4
dz =
ı2π
3!
f
(3)
(0) =
ı2π
3!
3!3
3
3
3
= ı2π

|z|=1

f(z)
z
4
dz = ı2π
3. We use Cauchy’s integral formula to evaluate the integral.

|z|=1
f(z)
e
z
z
2
dz =
ı2π
1!
d
dz
(f(z)
e
z
)


z=0
= ı2π
1!1
3
3
1


|z|=1
f(z)
e
z
z
2
dz =
ı2π
3
Solution 12.30
1. (a)
1
z(1 −z)
=
1
z
+
1
1 − z
=
1
z
+
∞

n=0
z
n
, for 0 < |z| < 1
=

1
z
+
∞

n=−1
z
n
, for 0 < |z| < 1
623
(b)
1
z(1 −z)
=
1
z
+
1
1 − z
=
1
z
−
1
z
1
1 − 1/z
=
1
z

−
1
z
∞

n=0

1
z

n
, for |z| > 1
= −
1
z
∞

n=1
z
−n
, for |z| > 1
= −
−∞

n=−2
z
n
, for |z| > 1
624
(c)

1
z(1 −z)
=
1
z
+
1
1 − z
=
1
(z + 1) − 1
+
1
2 − (z + 1)
=
1
(z + 1)
1
1 − 1/(z + 1)
−
1
(z + 1)
1
1 − 2/(z + 1)
, for |z + 1| > 1 and |z + 1| > 2
=
1
(z + 1)
∞


n=0
1
(z + 1)
n
−
1
(z + 1)
∞

n=0
2
n
(z + 1)
n
, for |z + 1| > 1 and |z + 1| > 2
=
1
(z + 1)
∞

n=0
1 − 2
n
(z + 1)
n
, for |z + 1| > 2
=
∞

n=1

1 − 2
n
(z + 1)
n+1
, for |z + 1| > 2
=
−∞

n=−2

1 − 2
−n−1

(z + 1)
n
, for |z + 1| > 2
2. First we factor the denominator of f(z) = 1/(z
4
+ 4).
z
4
+ 4 = (z − 1 − ı)(z −1 + ı)(z + 1 − ı)(z + 1 + ı)
We l ook for an annulus about z = 1 containing the point z = ı where f(z) is analytic. The sin gularities at
z = 1 ±ı are a distance of 1 from z = 1; the singularities at z = −1 ± ı are at a distance of
√
5. Since f(z) is
analytic in the domain 1 < |z − 1| <
√
5 there is a convergent Laurent series in that domain.
625

Chapter 13
The Residue Theorem
Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on.
- Winston Churchill
13.1 The Residue Theorem
We will find that many integrals on closed contours may be evaluated in terms of the residues of a function. We first
define residues and then prove the Residue Theorem.
626
Result 13.1.1 Residues. Let f(z) be single-valued an analytic in a d eleted neighborhood
of z
0
. Then f(z) has the Laurent series expansion
f(z) =
∞

n=−∞
a
n
(z −z
0
)
n
,
The residue of f(z) at z = z
0
is the coefficient of the
1
z−z
0
term:

Res(f(z), z
0
) = a
−1
.
The residue at a branch point or non-isolated singularity is undefined as the Laurent series
does not exist. If f(z) has a pole of order n at z = z
0
then we can use the Residue Formula:
Res(f(z), z
0
) = lim
z→z
0

1
(n − 1)!
d
n−1
dz
n−1

(z −z
0
)
n
f(z)


.

See Exercise 13.4 for a proof of the Residue Formula.
Example 13.1.1 In Example 8.4.5 we showed that f(z) = z/ sin z has first order poles at z = nπ, n ∈ Z \{0}. Now
627
C B
Figure 13.1: Deform the contour to lie in the deleted disk.
we find the residues at these isolated singularities.
Res

z
sin z
, z = nπ

= lim
z→nπ

(z − nπ)
z
sin z

= nπ lim
z→nπ
z − nπ
sin z
= nπ lim
z→nπ
1
cos z
= nπ
1
(−1)

n
= (−1)
n
nπ
Residue Theorem. We can evaluate many integrals in terms of the residues of a function. Suppose f(z) has only
one singularity, (at z = z
0
), inside the simple, closed, positively oriented contour C. f(z) has a convergent Laurent
series in some deleted disk about z
0
. We deform C to lie in the disk. See Figure 13.1. We now evaluate

C
f(z) dz by
deforming the contour and using the Laurent series expansion of the function.
628

C
f(z) dz =

B
f(z) dz
=

B
∞

n=−∞
a
n

(z − z
0
)
n
dz
=
∞

n=−∞
n=−1
a
n

(z − z
0
)
n+1
n + 1

r
e
ı(θ+2π)
r
e
ıθ
+ a
−1
[log(z − z
0
)]

r
e
ı(θ+2π)
r
e
ıθ
= a
−1
ı2π

C
f(z) dz = ı2π Res(f(z), z
0
)
Now assume that f(z) has n singularities at {z
1
, . . . , z
n
}. We deform C to n contours C
1
, . . . , C
n
which enclose the
singularities and lie in deleted disks about the singularities in which f(z) has convergent Laurent series. See Figure 13.2.
We evaluate

C
f(z) dz by deforming the contour.

C

f(z) dz =
n

k=1

C
k
f(z) dz = ı2π
n

k=1
Res(f(z), z
k
)
Now instead let f(z) be analytic outside and on C except for isolated singularities at {ζ
n
} in the domain outside C
and perhaps an isolated singularity at infinity. Let a be any point in the interior of C. To evaluate

C
f(z) dz we make
the change of variables ζ = 1/(z − a). This maps the contour C to C

. (Note that C

is negatively oriented.) All
the points outside C are mapped to points inside C

and vice versa. We can then evaluate the integral in terms of the
singularities inside C


.
629
C
C
C
C
1
2
3
Figure 13.2: Deform the contour n contours which enclose the n singularities.

C
f(z) dz =

C

f

1
ζ
+ a

−1
ζ
2
dζ
=

−C


1
z
2
f

1
z
+ a

dz
= ı2π

n
Res

1
z
2
f

1
z
+ a

,
1
ζ
n
− a


+ ı2π Res

1
z
2
f

1
z
+ a

, 0

.
630
a
C
C’
Figure 13.3: The change of variables ζ = 1/(z − a).
Result 13.1.2 Residue Theorem. If f(z) is analytic in a compact, closed, connected
domain D except for isolated singularities at {z
n
} in the interior of D then

∂D
f(z) dz =

k


C
k
f(z) dz = ı2π

n
Res(f(z), z
n
).
Here the set of contours {C
k
} make up the positively oriented boundary ∂D of the domain
D. If the boundary of the domain is a single contour C then the formula simplifies.

C
f(z) dz = ı2π

n
Res(f(z), z
n
)
If instead f(z) is analytic outside and on C except for isolated singularities at {ζ
n
} in the
domain outside C and perhaps an isolated singularity at infinity then

C
f(z) dz = ı2π

n
Res


1
z
2
f

1
z
+ a

,
1
ζ
n
− a

+ ı2π Res

1
z
2
f

1
z
+ a

, 0

.

Here a is a any point in the interior of C.
631
Example 13.1.2 Consider
1
ı2π

C
sin z
z(z − 1)
dz
where C is the positively oriented circle of radius 2 centered at the origin. Since the integrand is si ngle-valu ed with
only isolated singularities, the Resid ue Theorem applies. The value of the integral is the sum of the residues from
singularities inside the contour.
The only places that the integrand could have singularities are z = 0 and z = 1. Sin ce
lim
z→0
sin z
z
= lim
z→0
cos z
1
= 1,
there is a removable singularity at the point z = 0. There is no residue at this point.
Now we consider the point z = 1. Since sin(z)/z is analytic and nonzero at z = 1, that point is a first order pole
of the integrand. The residue there is
Res

sin z
z(z − 1)

, z = 1

= lim
z→1
(z − 1)
sin z
z(z − 1)
= sin(1).
There is only one singular point with a residue inside the path of integration. The residue at this point is sin(1).
Thus the value of the integral is
1
ı2π

C
sin z
z(z − 1)
dz = sin(1)
Example 13.1.3 Evaluate the integral

C
cot z coth z
z
3
dz
where C is the unit circle about the origin in the positive direction.
The integrand is
cot z coth z
z
3
=

cos z cosh z
z
3
sin z sinh z
632
sin z has zeros at nπ. sinh z has zeros at ınπ. Thus the only pole inside the contour of integration is at z = 0. Since
sin z and sinh z both have simple zeros at z = 0,
sin z = z + O(z
3
), sinh z = z + O(z
3
)
the integrand has a pole of order 5 at the origin. The residue at z = 0 is
lim
z→0
1
4!
d
4
dz
4

z
5
cot z coth z
z
3

= lim
z→0

1
4!
d
4
dz
4

z
2
cot z coth z

=
1
4!
lim
z→0

24 cot(z) coth(z)csc(z)
2
− 32z coth(z)csc(z)
4
− 16z cos(2z) coth(z)csc(z)
4
+ 22z
2
cot(z) coth(z)csc(z)
4
+ 2z
2
cos(3z) coth(z)csc(z)

5
+ 24 cot(z) coth(z)csch(z)
2
+ 24csc(z)
2
csch(z)
2
− 48z cot(z)csc(z)
2
csch(z)
2
− 48z coth(z)csc(z)
2
csch(z)
2
+ 24z
2
cot(z) coth(z)csc(z)
2
csch(z)
2
+ 16z
2
csc(z)
4
csch(z)
2
+ 8z
2
cos(2z)csc(z)

4
csch(z)
2
− 32z cot(z)csch(z)
4
− 16z cosh(2z) cot(z)csch(z)
4
+ 22z
2
cot(z) coth(z)csch(z)
4
+ 16z
2
csc(z)
2
csch(z)
4
+ 8z
2
cosh(2z)csc(z)
2
csch(z)
4
+ 2z
2
cosh(3z) cot(z)csch(z)
5

=
1

4!

−
56
15

= −
7
45
633
Since taking the fourth derivative of z
2
cot z coth z really sucks, we would like a more elegant way of finding the
residue. We expand the functions in the integrand in Taylor series about the origin.
cos z cosh z
z
3
sin z sinh z
=

1 −
z
2
2
+
z
4
24
− ···


1 +
z
2
2
+
z
4
24
+ ···

z
3

z −
z
3
6
+
z
5
120
− ···

z +
z
3
6
+
z
5

120
+ ···

=
1 −
z
4
6
+ ···
z
3

z
2
+ z
6

−1
36
+
1
60

+ ···

=
1
z
5
1 −

z
4
6
+ ···
1 −
z
4
90
+ ···
=
1
z
5

1 −
z
4
6
+ ···

1 +
z
4
90
+ ···

=
1
z
5


1 −
7
45
z
4
+ ···

=
1
z
5
−
7
45
1
z
+ ···
Thus we see that the residue is −
7
45
. N ow we can evaluate the integral.

C
cot z coth z
z
3
dz = −ı
14
45

π
13.2 Cauchy Principal Value for Real Integrals
13.2.1 The Cauchy Principal Value
First we recap improper integrals. If f (x) has a singularity at x
0
∈ (a . . . b) then

b
a
f(x) dx ≡ lim
→0
+

x
0
−
a
f(x) dx + lim
δ→0
+

b
x
0
+δ
f(x) dx.
634
For integrals on (−∞. . . ∞),

∞

−∞
f(x) dx ≡ lim
a→−∞, b→∞

b
a
f(x) dx.
Example 13.2.1

1
−1
1
x
dx is divergent. We show this with the definition of improper integrals.

1
−1
1
x
dx = lim
→0
+

−
−1
1
x
dx + lim
δ→0
+


1
δ
1
x
dx
= lim
→0
+
[ln |x|]
−
−1
+ lim
δ→0
+
[ln |x|]
1
δ
= lim
→0
+
ln  − lim
δ→0
+
ln δ
The integral diverges because  and δ approach zero independently.
Since 1/x is an odd function, it appears that the area under the curve is zero. Consider what would happen if  and
δ were not independent. If they approached zero symmetrically, δ = , then the value of the integral would be zero.
lim
→0

+


−
−1
+

1


1
x
dx = lim
→0
+
(ln  − ln ) = 0
We could make the integral have any value we pleased by choosing δ = c.
1
lim
→0
+


−
−1
+

1
c


1
x
dx = lim
→0
+
(ln  − ln(c)) = −ln c
We have seen it is reasonable that

1
−1
1
x
dx
has some meaning, and if we could evaluate the integral, the most reasonable value would be zero. The Cauchy principal
value provides us with a way of evaluating such integrals. If f(x) is continuous on (a, b) except at the point x
0
∈ (a, b)
1
This may remind you of conditionally convergent series. You can rearrange the terms to make the series sum to any number.
635
then the Cauchy principal value of the integral is defined
−

b
a
f(x) dx = lim
→0
+



x
0
−
a
f(x) dx +

b
x
0
+
f(x) dx

.
The Cauchy principal value is obtained by approaching the singularity symmetrically. The principal value of the integral
may exist when the integral diverges. If the integral exists, it is equal to the principal value of the integral.
The Cauchy principal value of

1
−1
1
x
dx is defined
−

1
−1
1
x
dx ≡ lim
→0

+


−
−1
1
x
dx +

1

1
x
dx

= lim
→0
+

[log |x|]
−
−1
[log |x|]
1


= lim
→0
+
(log | − | −log ||)

= 0.
(Another notation for the principal value of an integral is PV

f(x) dx.) Since the limits of integration approach zero
symmetrically, the two halves of the integral cancel. If the limits of integration approached zero independently, (the
definition of the integral), then the two halves would both diverge.
Example 13.2.2

∞
−∞
x
x
2
+1
dx is divergent. We show this with the definition of improper integrals.

∞
−∞
x
x
2
+ 1
dx = lim
a→−∞, b→∞

b
a
x
x
2

+ 1
dx
= lim
a→−∞, b→∞

1
2
ln(x
2
+ 1)

b
a
=
1
2
lim
a→−∞, b→∞
ln

b
2
+ 1
a
2
+ 1

636
The integral diverges because a and b approach infinity independently. Now consider what would happen if a and b
were not independent. I f they approached zero symmetrically, a = −b, then the value of the integral would be zero.

1
2
lim
b→∞
ln

b
2
+ 1
b
2
+ 1

= 0
We could make the integral have any value we pleased by choosing a = −cb.
We c an assign a meaning to divergent integrals of the form

∞
−∞
f(x) dx with the Cauchy principal value. The
Cauchy principal value of the integral is defined
−

∞
−∞
f(x) dx = lim
a→∞

a
−a

f(x) dx.
The Cauchy principal value is obtained by approaching infinity symmetrically.
The Cauchy principal value of

∞
−∞
x
x
2
+1
dx is defined
−

∞
−∞
x
x
2
+ 1
dx = lim
a→∞

a
−a
x
x
2
+ 1
dx
= lim

a→∞

1
2
ln

x
2
+ 1


a
−a
= 0.
637