WA 9: Solutions

Problem 1. Suppose Γ is an analytic function in C except for a set of nonpositiveintegers, where it has simple poles. Suppose, moreover, that Γ(1) = 1 and zΓ(z) =Γ(z + 1) for all regular points. Prove that

Resz=−nΓ(z) =(−1)n

n!, n = 0, 1, . . . .

Solution. We have

Γ(z) =Γ(z + 1)

z=

Γ(z + 2)z(z + 1)

= · · · =Γ(z + n + 1)

z(z + 1) · · · (z + n − 1)(z + n).

Since Γ(z) is analytic at z = 1, we conclude that Γ(z) has a simple pole at z = −nfor every n = 0, 1, 2, . . ., and one has for n = 0 that

Resz=0Γ(z) = Γ(1) = 1 =(−1)0

0!,

and for n = 1, 2, . . . that

Resz=−nΓ(z) =Γ(z + n + 1)

z(z + 1) · · · (z + n − 1)

∣∣∣z=−n

=Γ(1)

−n(−n + 1) · · · (−1)=

(−1)n

n!.

Problem 2. If zν is a pole of 1z4+a4 , show that

Resz=zν= − zν

4a4.

Solution. Since g(z) = z4 + a4 has 4 simple zeros at the 4-th roots of −a4, thefunction f(z) = 1

g(z) = 1z4+a4 has 4 simple poles at those points. If zν is one of

these poles then

Resz=zν

1z4 + a4

=1

(z4 + a4)′|z=zν

=1

4z3ν

=zν

4z4ν

= − zν

4a4.

Problem 3. Find∫C

dz1+z3 where C is the positively oriented ellipse 2x2 +y2 = 3

2 .

Solution. The function f(z) = 11+z3 has simple poles at z1 = −1, z2 = 1

2 +√

32 i,

and z3 = 12 −

√3

2 i, of which only z2 and z3 are inside the ellipse 2x2 + y2 = 32 .

Therefore, by the Cauchy residue theorem∫C

dz

1 + z3= 2πi

(Resz=z2

11 + z3

+ Resz=z3

11 + z3

)= 2πi

(1

3z22

+1

3z23

)= −2πi

3.

Problem 4. In each case, write the Laurent series of the function at its isolatedsingular point and determine whether that point is a pole, a removable singularpoint, or an essential singular point, and find the residue at that point:

(a) ze1/z; (b) z2

1+z ; (c) 1−cosh zz3 ; (d) (2 − z)−3.

1

2

Solution. (a) The function f(z) = ze1/z has the only singular point at z = 0and its Laurent series is given by

ze1/z = z

(1 +

11!z

+1

2!z2+ · · · + 1

n!zn+ · · ·

)= z + 1 +

12z

+1

6z2+ · · · .

Thus z = 0 is an essential singular point and Resz=0

(ze1/z

)= 1

2 .

(b) The function f(z) = z2

1+z has the only singular point at z = −1 and itsLaurent series is given by

z2

1 + z=

z2 − 1 + 11 + z

= z − 1 +1

1 + z= (z + 1) − 2 +

1z + 1

.

Thus z = −1 is a simple pole and Resz=−1

(z2

1+z

)= 1.

(c) The function f(z) = 1−cosh zz3 has the only singular point at z = 0 and its

Laurent series is given by

1 − cosh z

z3=

1z3

(1 − 1 − z2

2!− z4

4!− · · · − z2n

(2n)!− · · ·

)= − 1

2z− z

24− · · · .

Thus z = 0 is a simple pole and Resz=0

(1−cosh z

z3

)= − 1

2 .

(d) The function f(z) = (2 − z)−3 has the only singular point at z = 2 and itsLaurent series is given by

(2 − z)−3 = −(z − 2)−3.

Thus z = 2 is a pole of order 3 and Resz=2

((2 − z)−3

)= 0.

Problem 5. Use the theorem involving a single residue to evaluate the integralof f(z) around the positively oriented circle |z| = 3 when

(a) f(z) = (3z+2)2

z(z−1)(2z+5) ; (b) f(z) = z3(1−3z)(1+z)(1+2z4) ; (c) f(z) = z3e1/z

1+z3 .

Solution. (a) The function f(z) = (3z+2)2

z(z−1)(2z+5) has simple poles at z = 0, z = 1,and z = −5/2, all of which lie inside the circle C3 = {z ∈ Z : |z| = 3. Therefore,∫

C3

(3z + 2)2

z(z − 1)(2z + 5)dz = 2πiResz=0

1z2

(3/z + 2)2

1/z(1/z − 1)(2/z + 5)

= 2πiResz=0(3 + 2z)2

z(1 − z)(2 + 5z)= 2πi · 9

2= 9πi.

(b) The function f(z) = z3(1−3z)(1+z)(1+2z4) has simple poles at z = −1, and at 4 roots

of order 4 of −1/2, all of which lie inside C3. Therefore,∫C3

z3(1 − 3z)(1 + z)(1 + 2z4)

dz = 2πiResz=01z2

1/z3(1 − 3/z)(1 + 1/z)(1 + 2/z4)

= 2πiResz=0z4 − 3

z(z + 1)(z4 + 2)= 2πi ·

(−3

2

)= −3πi.

3

(c) The function f(z) = z3e1/z

1+z3 has simple poles at the 3 roots of order 3 of −1and an essential singular point at z = 0, and all of these singular points lie insideC3. Therefore,∫

C3

z3e1/z

1 + z3dz = 2πiResz=0

1z2

1/z3ez

1 + 1/z3

= 2πiResz=0ez

z2(z3 + 1)= 2πi · d

dz

(ez

z3 + 1

) ∣∣∣z=0

= 2πi.