253
Chapter 9
Miscellaneous Applications
In this chapter we consider selected methods where complex variable techniques can
be employed to solve various types of problems. In particular, we consider the summation
of series, complex potentials, two-dimensional fluid flow and a special mapping called the
Schwarz-Christoffel transformation.
Summation of series
cos πz
The function π cot πz = π
has singularities at the points z = n for n = 0, ±1, ±2, . . .
sin πz
where the function sin πz is zero. These singularities are simple poles because L’Hospital’s rule
z−n
1
for n = 0, ±1, . . . . If we assume that f(z) is a meromorphic function
=
gives lim
z→n sin πz
π cos πn
which has poles at the points z1 , z2, z3 , . . ., zm not equal to any of the integer positions on the
x-axis, then we can employ the residue theorem and integrate the function g(z) = π cot πz f(z)
around the square contour CN bounded by the lines y = N + 1/2, y = −N − 1/2, x = N + 1/2,
x = −N − 1/2, illustrated in the figure 9-1, to obtain the following result
Z
................
...... ....
....
π cot πz f(z) dz = 2πi
(
CN
N
X
Res [ g, n ] +
n=−N
m
X
)
Res [ g, zk ] ,
where g = π cot πz f(z)
(9.1)
k=1
The residue of g = π cot πz f(z) at the integer value z = n, is given by
Res [ g, n ] = lim (z − n)π cot πz f(z) = lim
z→n
z→n
z−n
lim π cos πz f(z) = f(n)
sin πz z→n
(9.2)
for n = 0, ±1, ±2, . . ., ±N. Now if we can show that
lim
N →∞
Z
................
.... ...
......
π cot πz f(z) dz = 0,
(9.3)
CN
then as N → ∞, the equation (9.1) simplifies to give the result
∞
X
n=−∞
f(n) = −
m
X
Res [ g, zk ]
where g = g(z) = π cot πz f(z)
(9.4)
k=1
This is an application of the residue theorem for the summation of an infinite series. If in
the limit as N → ∞ the function f(z) has an infinite number of poles inside the contour CN ,
then we must let m → ∞ in the equation (9.4).
254
Figure 9-1.
Square contour CN containing integers −N, . . . , N on real axis.
In order to employ the summation formula given by equation (9.2) we must assume that
the functions cot πz and f(z) are bounded on the contour CN in such a way that the contour
integral given by equation (9.3) approaches zero as N increases without bound. Toward this
purpose we consider the following three inequalities for the magnitude of cot πz when z is on
the contour CN .
(i) For z = N + 12 + iy with −1/2 ≤ y ≤ 1/2 we have
| cot πz| = | cot π(N +
1
π
π
+ iy)| = | cot( + iπy)| = | tanh πy| ≤ tanh
2
2
2
and for z = −N − 12 + iy and −1/2 ≤ y ≤ 1/2 we have
| cot πz| = | cot π(−N −
1
π
π
+ iy)| = | cot( − iπy)| = | tanh πy| ≤ tanh
2
2
2
(ii) For y > 1/2, we have
i πz
e
+ e−i πz e−πy + eπy 1 + e−2πy
1 + e−π
π
=
=
| cot πz| = i πz
≤
= coth
e
− e−i πz eπy − e−πy 1 − e−2πy
1 − e−π
2
(iii) For y < −1/2, we have
| cot πz| = | cot π(−N −
1
π
π
+ iy)| = | cot( + iπy)| = | tanh πy| ≤ tanh
2
2
2
255
An examination of the graphs of the hyperbolic tangent
and hyperbolic cotangent function shows that at x = π/2
we have the inequality coth π/2 > tanh π/2. Therefore, one
can say that the function cot πz is bounded for values of z
on the contour CN and one can write | cot πz| ≤ coth π/2. We
also assume that the modulus of the function f(z), for z on
M
the contour CN , satisfies the inequality |f(z)| ≤ |z|
k where
M and k > 1 are constants.
The above assumptions allow us to calculate the magnitude of the contour integral on
the left-hand side of equation (9.1) in the limit as N increases without bound. One finds
Z
.......
.... .
lim .................
N →∞ C
N
Z
...............
.... ....
π cot πz f(z) dz ≤ lim .....
N →∞ C
π | cot πz| |f(z)| dz ≤ lim π
N →∞
N
M
π
(8N + 4) coth = 0
Nk
2
where k > 1 and 8N + 4 is the length of the closed contour CN .
In the above development of a summation formula for infinite series, by use of the residue
theorem and special contour of integration, we employed the function π cot πz. This function
can be replaced by other types of functions having singularities on the real axis. Some
example summation formulas are
∞
X
(−1)n f(n) = −
n=−∞
∞
X
f
n=−∞
∞
X
n
(−1) f
n=−∞
where the summation
X
2n + 1
2
2n + 1
2
X
Res [ π csc πz f(z), zk ]
(9.5)
Res [ π tan πz f(z), zk ]
(9.6)
Res [ π sec πz f(z), zk ]
(9.7)
k
=
X
k
=
X
k
denotes a summation over all the poles of f(z). These summation
k
formula are derived in a manner very similar to that of equation (9.4).
Example 9-1.
(Summation of series)
1
1 1
1
For S =
= + +
+ · · ·, use the results from the equation (9.4) to find S .
2
n +1
2 5 10
n=1
1
π cot πz
Solution: Let f(z) = 2
and g(z) = π cot πz f(z) = 2
. The function f(z) has simple
z +1
z +1
poles at z1 = i and z2 = −i and so the residues of g(z) at these poles are given by
∞
X
π cot πz
π
π cot πz
= − coth π
=
2
z +1
z + i z=i
2
π cot πz
π
π cot πz
Res [ g(z), −i ] = lim (z + i) 2
= − coth π
=
z→−i
z +1
z − i z=−i
2
Res [ g(z), i ] = lim(z + i)
z→i
256
Using equation (9.4) we obtain
∞
X
f(n) =
n=−∞
∞
X
X
1
=−
Res [ g, zk ] = π coth π
2
n +1
n=−∞
2
k=1
Note that f(−n) = f(n) and so f(n) is an even function of n. Consequently, one can write
∞
X
f(n) =
n=−∞
−1
X
f(n) + f(0) +
n=−∞
∞
X
f(n) = π coth π
(9.8)
n=1
In the first summation in equation (9.8) replace n by −n and write
∞
X
f(n) + 1 +
n=1
∞
X
f(n) = 2
n=1
∞
X
f(n) + 1 = π coth π
n=1
which simplifies to
∞
X
f(n) =
n=1
∞
X
1
1
= (π coth π − 1)
2+1
n
2
n=1
The argument principle
Recall that if f(z) has a zero of order n1 at the point z = z1 , then f(z) can be expressed
in the form
f(z) = (z − z1 )n1 G(z)
(9.9)
where G(z) is analytic in some neighborhood of the point z1 . The function f(z) has the
derivative
f 0 (z) = n1 (z − z1 )n1 −1G(z) + (z − z1 )n1 G0(z)
(9.10)
Note that the ratio f 0 (z)/f(z) can be expressed
f 0 (z)
G0(z)
n1
+
=
f(z)
z − z1
G(z)
If we construct a small circle of radius r1 about the point z1 where
inside the small circle, then we can define the contour integral
I1 =
(9.11)
G0 (z)
G(z)
is analytic everywhere
Z
Z
n1
f 0 (z)
1 .................
G0 (z)
1 ....................
..........
..........
+
dz =
dz = n1
2πi |z−z1 |=r1 f(z)
2πi |z−z1 |=r1 z − z1
G(z)
(9.12)
In equation (9.12) the first integral gives the order n1 and the second integral is zero because
of the Cauchy theorem.
Similarly, if f(z) has a pole at the point z = ζ1 , which is a pole of order m1 , then f(z) can
be represented in the form
f(z) =
cm1
c1
H(z)
+···+
+ b0 + b1(z − ζ1 ) + · · · =
m
1
(z − ζ1 )
z − ζ1
(z − ζ1 )m1
(9.13)
257
This function has the derivative
f 0 (z) =
m1 H(z)
H 0 (z)
−
(z − ζ1)m1
(z − ζ1 )m1 +1
(9.14)
If we construct a small circle of radius r2 about the point z = ζ1 where
everywhere inside the circle, then we can define the integral
H 0 (z)
H(z)
Z
Z
0
H (z)
f 0 (z)
1 ..................
1 ....................
m1
.
.
.
........
........
dz = −m1
P1 =
dz =
−
2πi |z−ζ1 |=r2 f(z)
2πi |z−ζ1 |=r2 H(z)
z − ζ1
is analytic
(9.15)
Here the first integral is zero by Cauchy’s theorem and the second integral produces the
order m1 of the pole with a sign change.
Let |z −z1 | = r1 and |z −ζ1 | = r2 be disjoint circles and construct a simple closed path which
encircles both of the smaller circles centered at z1 and ζ1 so that f(z) is analytic everywhere
inside C except for the pole at ζ1 or order m1 . One can then write
Z
1 .................... f 0 (z)
.........
I=
dz = I1 + P1 = n1 − m1
2πi C f(z)
(9.16)
which represents the difference between the order of the zero and the order of the pole.
If there are N -zeros inside the simple closed
curve C at the points z1 , z2, . . . , zN with multiplicities
n1 , n2, . . ., nN and there are M -poles inside C at the
points ζ1 , ζ2, . . . , ζM of orders m1 , m2 , . . ., mM respectively, then one can surround these zeros and poles
with small disjoint circles Czj for j = 1, 2, . . ., N and
Cζj for j = 1, 2, . . ., M , and then employ the extended
Cauchy theorem to show
Z
Z
Z
N
M
X
X
1 .................... f 0 (z)
1 ................... f 0 (z)
1 ..................... f 0 (z)
..........
.
.
.........
.
.
.
.....
dz =
dz +
dz
2πi C f(z)
2πi
f(z)
2πi
C
Cζ f(z)
j=1
z
j=1
j
=
N
X
j=1
nj −
M
X
j
(9.17)
mj = N − P
j=1
where N is the number of zeros of f(z) (counting multiplicities) inside C and P is the summation of the orders associated with the poles of f(z) within the simple closed contour C .
If ω = f(z) with dω = f 0 (z) dz , then equation (9.17) can be expressed
Z
Z
1 .................... f 0 (z)
1 ..................... dω
.........
.........
dz =
2πi C f(z)
2πi C 0 ω
(9.18)
where C 0 is the image of the curve C under the mapping ω = f(z). If z = z(t), for t0 ≤ t ≤ t1 is
the parametric representation of the curve C , then ω = ω(t) = f(z(t)), for t0 ≤ t ≤ t1 gives the