Solutions for Math 311 Assignment #11

(1) Let C denote the circle |z| = 1 oriented counterclockwise.(a) Show that∫

C

zn exp

(1

z

)dz =

2πi

(n+ 1)!

for n = 0, 1, 2, ....(b) Show that∫

C

exp

(z +

1

z

)dz = 2πi

∞∑n=0

1

n!(n+ 1)!.

Proof. (a) Since

ez =∞∑n=0

zn

n!

for |z| <∞,

zn exp

(1

z

)= zn

∞∑m=0

z−m

m!=

∞∑m=0

zn−m

m!

for 0 < |z| <∞. Therefore,∫C

zn exp

(1

z

)dz =

∞∑m=0

1

m!

∫C

zn−m

m!dz.

And since∫Czn−mdz = 0 if n−m 6= −1 and 2πi if n−m = −1,∫C

zn exp

(1

z

)dz =

2πi

(n+ 1)!.

(b) Since

exp

(z +

1

z

)= exp(z) exp

(1

z

)=∞∑n=0

zn

n!exp

(1

z

)we obtain∫

C

exp

(z +

1

z

)dz =

∞∑n=0

1

n!

∫C

zn exp

(1

z

)dz =

∞∑n=0

2πi

n!(n+ 1)!

�

1

2

(2) Let the degrees of the polynomials

P (z) = a0 + a1z + a2z2 + ...+ anz

n (an 6= 0)

and

Q(z) = b0 + b1z + b2z2 + ...+ bmz

m (bm 6= 0)

be such that m ≥ n+ 2. Show that if all the zeros of Q(z) areinterior to a simple closed contour C, then∫

C

P (z)

Q(z)dz = 0.

Proof. Since P (z)/Q(z) is analytic outside C,∫C

P (z)

Q(z)dz = −2πiResz=∞

P (z)

Q(z)= 2πiResz=0

(1

z2P (1/z)

Q(1/z)

)Note that

1

z2P (1/z)

Q(1/z)=

1

z2a0 + a1z

−1 + a2z−2 + ...+ anz

−n

b0 + b1z−1 + b2z−2 + ...+ bmz−m

= zm−n−2a0z

n + a1zn−1 + a2z

n−2 + ...+ anb0zm + b1zm−1 + b2zm−2 + ...+ bm

is analytic at 0 since m ≥ n+ 2 and an, bm 6= 0. Therefore,

Resz=0

(1

z2P (1/z)

Q(1/z)

)= 0

and hence ∫C

P (z)

Q(z)dz = 0

�

(3) Write the principal part of each of the following functions atits isolated singular point and determine whether the point is apole, a removable singular point, or an essential singular point:

(a) z exp

(1

z

);

(b)sin z

z;

(c)1

(2− z)3.

3

Solution. (a) The function z exp(1/z) has a singularity at 0and

z exp

(1

z

)= z

∞∑n=0

z−n

n!=∞∑n=0

z1−n

n!= z + 1 +

∞∑n=2

1

(n!)zn−1

= z + 1 +∞∑n=1

1

(n+ 1)!zn

So it has an essential singularity at 0 with the principal part

∞∑n=1

1

(n+ 1)!zn

(b) The function sin z/z has a singularity at 0 and

sin z

z=

1

z

∞∑n=0

(−1)n

(2n+ 1)!z2n+1 =

∞∑n=0

(−1)n

(2n+ 1)!z2n

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 functionsis a pole. Determine the order of the pole and the residue ofthe function at the pole.

(a)1− cosh z

z3;

(b)1− e2z

z4;

(c)e2z

(z − 1)2.

4

Solution. (a) At z = 0, the Laurent series of the function is

1− cosh z

z3=

1

z3

(1− 1

2(ez + e−z)

)=

1

z3

(1− 1

2

(∞∑n=0

zn

n!+∞∑n=0

(−1)nzn

n!

))

=1

z3

(1−

∞∑n=0

z2n

(2n)!

)

= −∞∑n=1

z2n−3

(2n)!= − 1

2z−∞∑n=2

z2n−3

(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

1− exp(2z)

z4=

1

z4

(1−

∑n=0

(2z)n

n!

)= − 1

z4

∑n=1

2nzn

n!

= −∑n=1

2nzn−4

n!= − 2

z3− 2

z2− 4

3z−∑n=4

2nzn−4

n!

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

exp(2z)

(z − 1)2=e2e2(z−1)

(z − 1)2=

e2

(z − 1)2

∞∑n=0

2n(z − 1)n

n!

= e2∞∑n=0

2n(z − 1)n−2

n!

=e2

(z − 1)2+

2e2

z − 1+ e2

∞∑n=2

2n(z − 1)n−2

n!

So it has a pole of order m = 2 with residue B = 2e2 at 1.

(5) Find the value of the integral∫C

3z3 + 2

(z − 1)2(z2 + 9)dz

taken counterclockwise around the circle (a) |z − 2| = 2 (b)|z| = 4.

5

Solution. (a) In the disk |z−2| < 2, (3z3+2)/((z−1)2(z2+9))has one singularity at 1. Therefore,∫|z−2|=2

3z3 + 2

(z − 1)2(z2 + 9)dz = 2πiResz=1

3z3 + 2

(z − 1)2(z2 + 9)

= 2πi

(3z3 + 2

z2 + 9

)′z=1

=8πi

5.

(b) In the disk |z| < 4, (3z3 + 2)/((z − 1)2(z2 + 9)) has threesingularities at 1, −3i and 3i. Therefore,∫|z|=4

3z3 + 2

(z − 1)2(z2 + 9)dz = 2πi

(Resz=1

3z3 + 2

(z − 1)2(z2 + 9)

+ Resz=−3i3z3 + 2

(z − 1)2(z2 + 9)

+ Resz=3i3z3 + 2

(z − 1)2(z2 + 9)

)We have computed in part (a) that

Resz=13z3 + 2

(z − 1)2(z2 + 9)=

4

5

For its residues at ±3i, we have

Resz=−3i3z3 + 2

(z − 1)2(z2 + 9)= Resz=−3i

3z3 + 2

(z − 1)2(z + 3i)(z − 3i)

=3z3 + 2

(z − 1)2(z − 3i)

∣∣∣∣z=−3i

=11

10+

47

60i

and

Resz=3i3z3 + 2

(z − 1)2(z2 + 9)= Resz=3i

3z3 + 2

(z − 1)2(z + 3i)(z − 3i)

=3z3 + 2

(z − 1)2(z + 3i)

∣∣∣∣z=3i

=11

10− 47

60i

Therefore, ∫|z|=4

3z3 + 2

(z − 1)2(z2 + 9)dz = 6πi.

6

Alternatively, since all singularties of (3z3+2)/((z−1)2(z2+9))lie in |z| ≤ 4,∫

|z|=4

3z3 + 2

(z − 1)2(z2 + 9)dz = −2πiResz=∞

3z3 + 2

(z − 1)2(z2 + 9)

= −2πiResz=0(−z−2)3z−3 + 2

(z−1 − 1)2(z−2 + 9)

= 2πiResz=03 + 2z3

z(1− z)2(1 + 9z2)

= 6πi.

(6) Let CN denote the boundary of the square whose edges lie alongthe lines

x = ±(N +

1

2

)π and y = ±

(N +

1

2

)π

oriented counterclockwise, where N is a positive integer.(a) Show that∫

CN

dz

z2 sin z= 2πi

(1

6+ 2

N∑n=1

(−1)n

n2π2

).

(b) Show that

limN→∞

∫CN

dz

z2 sin z= 0.

(c) Show that∞∑n=1

1

n2=π2

6.

Proof. The function 1/(z2 sin z) is analytic in C\{nπ : n ∈Z}. Hence it has singularities at 0,±π,±2π, ...,±Nπ insidethe curve CN . Therefore,∫

CN

dz

z2 sin z= 2πi

N∑n=−N

Resz=nπ1

z2 sin z

7

At z = 0,

1

z2 sin z=

1

z2

(∞∑n=0

(−1)n

(2n+ 1)!z2n+1

)−1

=1

z3

(∞∑n=0

(−1)n

(2n+ 1)!z2n

)−1

=1

z3

(1−

∞∑n=1

(−1)n+1

(2n+ 1)!z2n

)−1

=1

z3

∞∑k=0

(∞∑n=1

(−1)n+1

(2n+ 1)!z2n

)k

=1

z3

(1 +

(z2

3!− ...

)+ ...

)=

1

z3+

1

6z+ ...

Therefore,

Resz=01

z2 sin z=

1

6At z = nπ with n 6= 0, 1/(z2 sin z) has a simple pole and hence

Resz=nπ1

z2 sin z= Resz=nπ

z−2

sin z=

z−2

(sin z)′

∣∣∣∣z=nπ

=1

n2π2 cos(nπ)=

(−1)n

n2π2

Therefore,∫CN

dz

z2 sin z= 2πi

(1

6+

−1∑n=−N

(−1)n

n2π2+

N∑n=1

(−1)n

n2π2

)

= 2πi

(1

6+ 2

N∑n=1

(−1)n

n2π2

)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 = | sinx cos(yi) + cos x sin(yi)|2

= | sinx cosh y + i cosx sinh y|2

= (sinx)2(cosh y)2 + (cosx)2(sinh y)2

= (sinx)2(1 + (sinh y)2) + (cos x)2(sinh y)2

= (sinx)2 + (sinh y)2

8

When x = ±(N + 1/2)π,

| sin z| ≥ | sinx| = 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

z2 sin z

∣∣∣∣ ≤ 1

(N + 1/2)2π2

Therefore,∣∣∣∣∫CN

dz

z2 sin z

∣∣∣∣ ≤ 1

(N + 1/2)2π2

∫CN

|dz| = 8

(N + 1/2)π

and hence

limN→∞

∫CN

dz

z2 sin z= 0

It follows that

2πi

(1

6+ 2

∞∑n=1

(−1)n

n2π2

)= 0

That is,∞∑n=1

(−1)n+1

n2=π2

12.

Finally, since∞∑n=1

(−1)n+1

n2=∞∑k=1

1

(2k − 1)2−∞∑k=1

1

(2k)2

=∞∑k=1

1

(2k − 1)2+∞∑k=1

1

(2k)2− 2

∞∑k=1

1

(2k)2

=∞∑n=1

1

n2− 1

2

∞∑n=1

1

n2=

1

2

∞∑n=1

1

n2,

we obtain∞∑n=1

1

n2=π2

6.

�