Solutions for Math 311 Assignment #11
(1) Let C denote the circle |z| = 1 oriented counterclockwise.
(a) Show that
Z
1
2πi
n
z exp
dz =
z
(n + 1)!
C
for n = 0, 1, 2, ....
(b) Show that
Z
∞
X
1
1
dz = 2πi
.
exp z +
z
n!(n
+
1)!
C
n=0
Proof. (a) Since
ez =
∞
X
zn
n=0
n!
for |z| < ∞,
∞
∞
X
X
1
z n−m
z −m
n
z exp
=z
=
z
m!
m!
m=0
m=0
n
for 0 < |z| < ∞. Therefore,
Z n−m
Z
∞
X
1
1
z
n
z exp
dz =
dz.
z
m! C m!
C
m=0
R
And since C z n−m dz = 0 if n − m 6= −1 and 2πi if n − m = −1,
Z
1
2πi
n
z exp
dz =
.
z
(n + 1)!
C
(b) Since
X
∞
1
1
zn
1
exp z +
= exp(z) exp
=
exp
z
z
n!
z
n=0
we obtain
Z
∞
∞
X
X
1
1
1
2πi
n
exp z +
dz =
z exp
dz =
z
n! C
z
n!(n + 1)!
C
n=0
n=0
Z
1
2
(2) Let the degrees of the polynomials
P (z) = a0 + a1 z + a2 z 2 + ... + an z n (an 6= 0)
and
Q(z) = b0 + b1 z + b2 z 2 + ... + bm z m (bm 6= 0)
be such that m ≥ n + 2. Show that if all the zeros of Q(z) are
interior to a simple closed contour C, then
Z
P (z)
dz = 0.
C Q(z)
Proof. Since P (z)/Q(z) is analytic outside C,
P (z)
P (z)
1 P (1/z)
dz = −2πi Resz=∞
= 2πi Resz=0
Q(z)
z 2 Q(1/z)
C Q(z)
Z
Note that
1 P (1/z)
1 a0 + a1 z −1 + a2 z −2 + ... + an z −n
=
z 2 Q(1/z)
z 2 b0 + b1 z −1 + b2 z −2 + ... + bm z −m
a0 z n + a1 z n−1 + a2 z n−2 + ... + an
= z m−n−2 m
b0 z + b1 z m−1 + b2 z m−2 + ... + bm
is analytic at 0 since m ≥ n + 2 and an , bm 6= 0. Therefore,
1 P (1/z)
Resz=0
=0
z 2 Q(1/z)
and hence
Z
C
P (z)
dz = 0
Q(z)
(3) Write the principal part of each of the following functions at
its isolated singular point and determine whether the point is a
pole, a removable
singular point, or an essential singular point:
1
(a) z exp
;
z
sin z
(b)
;
z
1
(c)
.
(2 − z)3
3
Solution. (a) The function z exp(1/z) has a singularity at 0
and
∞
∞
∞
X
X
1
z −n X z 1−n
1
z exp
=z
=
=z+1+
z
n!
n!
(n!)z n−1
n=0
n=0
n=2
=z+1+
∞
X
n=1
1
(n + 1)!z n
So it has an essential singularity at 0 with the principal part
∞
X
n=1
1
(n + 1)!z n
(b) The function sin z/z has a singularity at 0 and
∞
∞
sin z
1 X (−1)n 2n+1 X (−1)n 2n
=
z
=
z
z
z n=0 (2n + 1)!
(2n
+
1)!
n=0
So it has a removable singularity at 0 with the principal part 0.
(c) The function 1/(2 − z)3 has a pole at 2 with the principal
part
1
.
(2 − z)3
(4) Show that the singular point of each of the following functions
is a pole. Determine the order of the pole and the residue of
the function at the pole.
1 − cosh z
(a)
;
z3
2z
1−e
(b)
;
z4
e2z
(c)
.
(z − 1)2
4
Solution. (a) At z = 0, the Laurent series of the function is
1 − cosh z
1 z
1
−z
= 3 1 − (e + e )
z3
z
2
!!
∞
∞
1
1 X z n X (−1)n z n
= 3 1−
+
z
2 n=0 n! n=0
n!
!
∞
X
z 2n
1
= 3 1−
z
(2n)!
n=0
=−
∞
X
z 2n−3
n=1
∞
X z 2n−3
1
=− −
(2n)!
2z n=2 (2n)!
So it has a pole of order m = 1 with residue B = −1/2 at 0.
(b) At z = 0, the Laurent series of the function is
!
n
X
1
(2z)
1 X 2n z n
1 − exp(2z)
= 4 1−
=− 4
z4
z
n!
z n=1 n!
n=0
X 2n z n−4
X 2n z n−4
2
2
4
=−
=− 3 − 2 −
−
n!
z
z
3z n=4 n!
n=1
So it has a pole of order m = 3 with residue B = −4/3 at 0.
(c) At z = 1, the Laurent series of the function is
∞
X
exp(2z)
e2 e2(z−1)
e2
2n (z − 1)n
=
=
(z − 1)2
(z − 1)2
(z − 1)2 n=0
n!
2
=e
∞
X
2n (z − 1)n−2
n=0
n!
∞
X
e2
2e2
2n (z − 1)n−2
2
=
+
+
e
(z − 1)2 z − 1
n!
n=2
So it has a pole of order m = 2 with residue B = 2e2 at 1.
(5) Find the value of the integral
Z
3z 3 + 2
dz
2 2
C (z − 1) (z + 9)
taken counterclockwise around the circle (a) |z − 2| = 2 (b)
|z| = 4.
5
Solution. (a) In the disk |z −2| < 2, (3z 3 +2)/((z −1)2 (z 2 +9))
has one singularity at 1. Therefore,
Z
3z 3 + 2
3z 3 + 2
dz
=
2πi
Res
z=1
2 2
(z − 1)2 (z 2 + 9)
|z−2|=2 (z − 1) (z + 9)
3
0
3z + 2
8πi
= 2πi
=
.
2
z + 9 z=1
5
(b) In the disk |z| < 4, (3z 3 + 2)/((z − 1)2 (z 2 + 9)) has three
singularities at 1, −3i and 3i. Therefore,
Z
3z 3 + 2
3z 3 + 2
dz
=
2πi
Res
z=1
2 2
(z − 1)2 (z 2 + 9)
|z|=4 (z − 1) (z + 9)
3z 3 + 2
(z − 1)2 (z 2 + 9)
3z 3 + 2
(z − 1)2 (z 2 + 9)
+ Resz=−3i
+ Resz=3i
We have computed in part (a) that
Resz=1
3z 3 + 2
4
=
2
2
(z − 1) (z + 9)
5
For its residues at ±3i, we have
Resz=−3i
3z 3 + 2
3z 3 + 2
=
Res
z=−3i
(z − 1)2 (z 2 + 9)
(z − 1)2 (z + 3i)(z − 3i)
3z 3 + 2
11 47
=
+ i
=
2
(z − 1) (z − 3i) z=−3i 10 60
and
Resz=3i
3z 3 + 2
3z 3 + 2
=
Res
z=3i
(z − 1)2 (z 2 + 9)
(z − 1)2 (z + 3i)(z − 3i)
3z 3 + 2
11 47
=
=
− i
2
(z − 1) (z + 3i) z=3i 10 60
Therefore,
Z
|z|=4
3z 3 + 2
dz = 6πi.
(z − 1)2 (z 2 + 9)
6
Alternatively, since all singularties of (3z 3 +2)/((z −1)2 (z 2 +9))
lie in |z| ≤ 4,
Z
|z|=4
3z 3 + 2
3z 3 + 2
dz = −2πi Resz=∞
(z − 1)2 (z 2 + 9)
(z − 1)2 (z 2 + 9)
3z −3 + 2
(z −1 − 1)2 (z −2 + 9)
3 + 2z 3
z(1 − z)2 (1 + 9z 2 )
= −2πi Resz=0 (−z −2 )
= 2πi Resz=0
= 6πi.
(6) Let CN denote the boundary of the square whose edges lie along
the lines
1
1
x=± N+
π and y = ± N +
π
2
2
oriented counterclockwise, where N is a positive integer.
(a) Show that
!
Z
N
X
(−1)n
1
dz
= 2πi
+2
.
2
2π2
6
n
CN z sin z
n=1
(b) Show that
Z
lim
N →∞
CN
z2
dz
= 0.
sin z
(c) Show that
∞
X
π2
1
=
.
n2
6
n=1
Proof. The function 1/(z 2 sin z) is analytic in C\{nπ : n ∈
Z}. Hence it has singularities at 0, ±π, ±2π, ..., ±N π inside
the curve CN . Therefore,
Z
CN
N
X
dz
1
= 2πi
Resz=nπ 2
2
z sin z
z sin z
n=−N
7
At z = 0,
1
1
= 2
2
z sin z
z
=
1
z3
=
1
z3
=
1
z3
!−1
∞
X
(−1)n 2n+1
z
(2n + 1)!
n=0
!−1
∞
n
X
(−1)
z 2n
(2n + 1)!
n=0
!−1
∞
X
(−1)n+1 2n
1−
z
(2n
+
1)!
n=1
!k
2
∞
∞
X X
(−1)n+1 2n
z
1
z
− ... + ...
= 3 1+
(2n + 1)!
z
3!
n=1
k=0
1
1
+ ...
+
3
z
6z
Therefore,
=
1
1
=
z 2 sin z
6
At z = nπ with n 6= 0, 1/(z 2 sin z) has a simple pole and hence
1
z −2 z −2
Resz=nπ 2
= Resz=nπ
=
z sin z
sin z
(sin z)0 Resz=0
z=nπ
(−1)n
1
= 2 2
= 2 2
n π cos(nπ)
nπ
Therefore,
Z
dz
= 2πi
2
CN z sin z
= 2πi
−1
N
X
1
(−1)n X (−1)n
+
+
6 n=−N n2 π 2
n2 π 2
n=1
!
N
X
(−1)n
1
+2
6
n2 π 2
n=1
!
When z = x + yi ∈ CN , either x = ±(N + 1/2)π or y =
±(N + 1/2)π.
Note that
| sin z|2 = | sin(x + yi)|2 = | sin x cos(yi) + cos x sin(yi)|2
= | sin x cosh y + i cos x sinh y|2
= (sin x)2 (cosh y)2 + (cos x)2 (sinh y)2
= (sin x)2 (1 + (sinh y)2 ) + (cos x)2 (sinh y)2
= (sin x)2 + (sinh y)2
8
When x = ±(N + 1/2)π,
| sin z| ≥ | sin x| = 1
When y = ±(N + 1/2)π,
| sin z| ≥ | sinh y| = sinh((N + 1/2)π)
Obviously, sinh((N + 1/2)π) > 1 for N ≥ 0. Therefore,
| sin(z)| ≥ 1
for z ∈ CN . And since |z| ≥ (N + 1/2)π for z ∈ CN ,
1 1
z 2 sin z ≤ (N + 1/2)2 π 2
Therefore,
Z
Z
dz
1
8
≤
|dz| =
2
2
2
(N + 1/2) π CN
(N + 1/2)π
CN z sin z
and hence
Z
lim
N →∞
CN
dz
=0
z 2 sin z
It follows that
∞
X
(−1)n
1
+2
6
n2 π 2
n=1
2πi
That is,
∞
X
(−1)n+1
n=1
Finally, since
∞
X
(−1)n+1
n=1
n2
=
=
=
∞
X
k=1
∞
X
k=1
∞
X
n=1
we obtain
n2
=
!
=0
π2
.
12
∞
X 1
1
−
(2k − 1)2 k=1 (2k)2
∞
∞
X
X
1
1
1
+
−2
2
2
(2k − 1)
(2k)
(2k)2
k=1
k=1
∞
∞
1
1X 1
1X 1
−
=
,
n2 2 n=1 n2
2 n=1 n2
∞
X
1
π2
=
.
n2
6
n=1