Complex Variables . . . . . . . . Review Problems (Residue Calculus – Comments) . . . . . . . .

Fall 2012

Initial Draft

(1) Show that the singular point of f (z) is a pole; determine its order m and its residue B:
(a) (1 − e2z )/z 4 , (b) e2z /(z − 1)2 .
Comments: (a) The order of the pole is m = 3 and the residue is B = −4/3. To verify
this, we use the Taylor expansion e2z = 1 + (2z) + 2!1 (2z)2 + 3!1 (2z)3 + · · · to obtain
∞
!,
1 − e2z X 1
4
= − (2z)k z4
z k!
k=1
∞
X 1 k k−4
=− 2 z
k!
k=1

so the negative powers of the expansion are

1 1 1 11
−2 − 22 − 23 .
z3 2! z 2 3! z
Hence the order of the pole at z = 0 is 3 and the residue given by the coefficient of 1/z is
−23 /3! = −4/3.
(b) The order of the pole is m = 2 and the residue is B = 2e2 . We use the Taylor expansion
of e2z about z = 1:
∞ ∞
2z
X f (k) (1) k
X 2k e2
e = (z − 1) = (z − 1)k
k! k!
k=0 k=0

since Dk (e2k ) = 2k e2k . Then we find that

∞
e2z X 2k e2
= (z − 1)k−2
(z − 1)2 k!
k=0

so the negative powers are

e2 (z − 1)−2 + 2e2 (z − 1)−1 .

We conclude that 1 is a pole of order 2 and its residue is 2e2 .

(2) Evaluate the following integrals around the circle |z| = 3:
(a) e−z /z 2 , (b) e−z /(z − 1)2 , (c) z 2 e1/z .
Comments: These integrals can all be found using the Residue Theorem.
(a) Let f (z) = e−z /z 2 which has a unique pole at z = 0 of order 2. By the Residue Theorem,
we have
e−z −z
Z  
d 2e
2
dz = 2πiRes(f (z); 0) = 2πi lim z 2 = −2πie0 = −2πi.
|z|=3 z z→0 dz z

(b) Let f (z) = e−z /(z − 1)2 . Then f (z) has a unique pole at z = 0 which has order 2. By
1
 applying the Residue Theorem, we find that

e−z
Z
dz = 2πiRes(f (z); 0)
|z|=3 (z − 1)2
e−z
 
d 2
= 2πi lim (z − 1)
z→0 dz (z − 1)2
= −2πie0 = −2πi

(c) Let f (z) = z 2 e1/z which has a unique singularity at z = 0 which is essential. Note that

 
2 1 1 1 1 1 1 1
f (z) = z 1 + + + + + ···
z 2! z 2 3! z 3 4! z 4
1 11 1 1
= z2 + z + + + + ···
2! 3! z 4! z 2

which shows that Res(f (z); 0) = 1/6. By the Residue Theorem, we must have

Z
1 πi
z 3 e1/z dz = 2πiRes(f (z); 0) = 2πi = .
|z|=3 6 3

Review: Identifying the order of a pole if f (z) = p(z)/q(z). Assume that p(a) 6= 0,
q(a) = 0 but q 0 (a) 6= 0. Then write

p(z) p(z)
= 0 1
q(z) q (a)(z − a) + 2! q 00 (a)(z − a)2 + · · ·
1 p(z)
= .
z − a q 0 (a) + 2!1 q 00 (a)(z − a) + · · ·

Note that both p(z) and q 0 (a) + 2!1 q 00 (a)(z − a) + · · · are analytic functions around z = a
and are both nonzero at z = a. Hence the order of the pole is indeed 1.

(3) Evaluate the following residues:

(a) f (z) = z 1/4 /(z − i) where z 1/4 (principal branch); Res(f (z); i).
(b) f (z) = (Logz)/(z 2 + 1)2 (principal branch); Res(f (z); i).
(c) f (z) = ez / sin z; Res(f (z); π).
Comments: (a) f has a simple pole at z = i so we find that

z 1/4
Res(f (z); i) = lim(z − i) = i1/4 = cos(π/8) + i sin(π/8)
z→i z−i
2
 since i = eπ/2 . (b) f (z) has a pole of order 2 at −1 so the residue is given as
 
d 2 Log z d Log z
lim (z − i) 2 2
= lim
z→i dz (z + 1) z→i dz (z + i)2

(z + i)2 /z − 2(z + i)Log z

= lim
z→i (z + i)4
 
2Log z 1
= lim −
z→i (z + i)3 z(z + i)2
2Log i 1
= 3
−
(2i) i(2i)2
1 1
= π+ i
8 4
(c) f (z) has a simple pole at z = π since sin z has only simple zeros since its derivative is
nonzero there. Hence the residue is given as
ez 

Res(f (z); π) = = −eπ .
cos z z=π

(4) Let ZC be the circle |z|Z= 2. Evaluate Zthe following integrals:

1 cos πz
(a) tan z dz, (b) dz, 2
dz
C C sin 2z C z(z + 1)
Comments: (a) Let f (z) = tan z. f (z) has simple poles at integer multiples of π/2; in
particular, the two poles at ±π/2 are lie inside the circle |z| = 2. Now Res(f (z); π/2) = −1
and Res(f (z); −π/2) = −1. By the Residue Theorem, we have
Z
tan(z) dz = 2πi(−2) = −4πi.
C
(b) Let f (z) = 1/ sin(2z). Then f (z) has three simple poles inside the circle |z| = 2 at 0 and
±π/2. We find that Res(f (z); 0) = 1/2, Res(f (z); π/2) = −1/2, and Res(f (z); −π/2) =
−1/2. By the Residue Theorem, we find that
−1
Z
1
dz = 2πi = −πi.
C sin 2z 2
cos πz
(c) Let f (z) = z(z 2 +1) which has three simple poles at 0 and ±i. We find that Res(f (z); 0) =

1, Res(f (z); i) = −eπ /4 − e−π /4, and Res(f (z); −i) = −eπ /4 − e−π /4. By the Residue
Theorem, we have
Z
cos πz π −π π −π


2
dz = 2πi 1 + (−e /4 − e /4) + (−e /4 − e /4) .
C z(z + 1)

Review: applying the Residue Theorem to evaluate integrals of the form

Z ∞
I= f (x) dx
−∞
when f can be extended as an analytic function in a domain that includes the real line and
the upper half plane except for finitely many poles with positive imaginary part.
Let CR denote the semicircle |z| = R with =z ≥ 0; let ΓR be the closed semicircle
consisting of CR and the line segment [−R, R] oriented in the positive sense. Then
Z R Z 
I = lim f (x) dx + f (z) dz
R→∞ −R CR
3
 provided the limit Z
lim f (z) dz = 0.
R→∞ CR
In many cases, we can show that this limit is 0 by using the M L-inequality while in more
subtle cases we need to use Jordan’s Lemma(that we describe below in Problem # 7.
Z ∞
sin x
Note that for integrals of the form g(x) dx, the corresponding analytic func-
−∞ cos x
tion used is g(z)eiz because the contribution of the semicircle in the upper half plane can
usually be shown goes to 0 in the limit.

(5) UseZresidues to compute Z

∞ ∞ Z ∞
dx dx x4
(a) , (b) , (c) ,
Z−∞ (x2 + 1)2 −∞ x4 + 1 2 2
−∞ (x + 9)(x + 4)
2
∞ Z ∞
dx dx
(d) 2
, (e) 2 2
.
−∞ x + 2x + 3 −∞ (x + 1)(x + 2x + 3)
Comments:√ (a) π/4
(b) π/(2 2).
z4
(c) Let f (z) = (z 2 +9)(z 2 +4)2 whose poles that lie in the upper half plane are 3i of order 1 and

2i of order 2. We have Res(f (z); 3i) = −27i/50 and Res(f (z); 2i) = 23i/50. The Residue
Theorem will give
Z ∞
x4 4π
2 + 9)(x2 + 4)2
= 2πi(−2i/25) =
−∞ (x 25
provided we can show that
Z
lim f (z) dz = 0
R→∞ CR

where CR is the semicircle |z| = R with =z ≥ 0. To check this, we use the M L-inequality.
On CR we have
z4 R4
 
 
≤
 (z 2 + 9)(z 2 + 4)2  (R2 − 9)(R2 − 4)2 , R > 3.


Hence we conclude that

z4 R4
Z 
 
lim  dz  ≤ lim πR = 0.
R→∞ CR (z 2 + 9)(z 2 + 4)2  R→∞ (R2 − 9)(R2 − 4)2
√
(d) Let f (z) = 1/(z 2 + 2z + 3) has one √ pole −1 + i 2 in the upper half plane which is
simple. The residue of f (z) there is −i 2/4. The Residue Theorem yields
Z ∞ √ √
dx −i 2 π 2
2 2
= 2πi =
−∞ (x + 1)(x + 2x + 3) 4 2
provided we can show that
Z
lim f (z) dz = 0
R→∞ CR

where CR is the semicircle |z| = R with =z ≥ 0. To check this, we use the M L-inequality.
On CR we have
 
 1  1 1
 z 2 + 2z + 3  ≤ (R + 1)2 + 2 ≤ R2 , R > 0.
 
4
 Hence we conclude that
Z 
 1  ≤ lim πR 1 = 0.

lim  2
dz
R→∞ CR z + 2z + 3
 R→∞ R2
√
(e) Let f (z) = (z 2 +1)(z12 +2z+3) . f (z) has two poles i and −1 + i 2 in the upper half plane
√
which are both simple. Then Res(f (z); i) = −1/8 − i/8 and Res(f (z); −1 + i 2) = 1/8.
The Residue Theorem shows that
Z ∞
dx π
2 2
= 2πi (−1/8 − i/8 + 1/8) =
−∞ (x + 1)(x + 2x + 3) 4
provided we can show that
Z
lim f (z) dz = 0
R→∞ CR

where CR is the semicircle |z| = R with =z ≥ 0. To check this, we use the M L-inequality.
On CR we have
 
 1  1 1
 (z 2 + 1)(z 2 + 2z + 3)  ≤ (R2 + 1) [(R + 1)2 + 2] ≤ R4 , R > 0.
 

Hence we conclude that

Z 
 1  ≤ lim πR 1 = 0.

lim 

2 2
R→∞ CR (z + 1)(z + 2z + 3)  R→∞ R4

(6) Use residues to compute

Z ∞ Z ∞
cos x dx π(−beb + aea )ea−b cos ax
(a) 2 + a2 )(x2 + b2 )
= 2 + b2 )ab
, 0 < b < a; (b) 2+1
dx = πe−a ,
−∞ (x (a −∞ x
0 < a;
Z ∞ Z ∞
cos ax πe−ab x sin 2x √
−2 3
(c) 2 2 2
dx = , 0 < a, 0 < b; (d) 2
dx = πe ;
−∞ (x + b ) b −∞ x + 3
Z ∞ 3
x sin ax
(e) 2 + 4)2
dx = −π(a − 1)e−2a , 0 < a;
(x
Z −∞
∞
x3 sin x 9 5
(f) 2 + 1)(x2 + 9)
dx = − πe−3 + e−1 π.
0 (x 64 64
(7) UseZresidues to find the principal values of the integrals below:
∞ Z ∞ Z ∞
sin x (x + 1) cos x cos x
(a) 2
dx, 2
dx, 2 2
dx, 0 < b.
−∞ x + 4x + 5 −∞ x + 4x + 5 −∞ (x + a) + b

Comments: (a) Write f (z) = eiz /(z 2 + 4z + 5) which has simple poles at −2 ± i. Futher,
Res(f (z); −2 + i) = e−1−2i /[2(−2 + i) + 4] = − 21 e−1−2i i. We want to justify that
Z ∞   
sin x 1 −1−2i
2
dx = = 2πi − e i
−∞ x + 4x + 5 2
= =(πe−1−2i ) = πe−1 sin(−2) ' −1.05089 9905.
To do this, let CR denote the semicircle |z| = R with =z ≥ 0. We consider
Z 
 eiz  1
dz  ≤ πR → 0 as R → ∞
 
2
 |z|=R z + 4z + 5 

(R − 2)2 − 1
5
 iθ
since |eiz | = |eiRe | = e−R sin θ ≤ 1 since 0 ≤ θ ≤ π. Further, write z 2 + 4z + 5 as (z + 2)1 + 1
so by repeated uses of the reversed triangle inequality we get
|(Reiθ + 2)2 + 1| ≥ |(Reiθ + 2)2 | − 1
≥ |Reiθ + 2|2 − 1
≥ (|Reiθ | − 2|2 − 1
≥ (R − 2)2 − 1.
iz
(b) Let f (z) = z(z+1)e
2 +4z+5 . It has a unique pole in the upper half plane at −2 + i with residue

(−1 + i)e −1−2i /[2i] = − 21 (−1 + i)e−1−2i i = 12 (1 + i)e−1−2i . We will argue that
Z ∞   
(x + 1) cos x 1 −1−2i
2
dx = < 2πi (1 + i)e
−∞ x + 4x + 5 2
= πe−1 < ((1 + i)(cos(2) − i sin(2)))
= πe−1 (sin(2) + cos(2)) ≈ 0.56994 76249
We need to use Jordan’s Lemma:
Z π
π π
e−R sin θ dθ ≤ 1 − e−R < .


0 R R
Let CR be the portion of the circle |z| = R with =z ≥ 0. So CR is parametrized as z = Reiθ ,
dz = iReiθ dθ with 0 ≤ θ ≤ π. Then we have the bounds
Z iz
 Z θ iθ + 1)eiReiθ

 (z + 1)e   (Re iθ

dz  =  iRe dθ
  
 z 2 + 4z + 5 (Re iθ )2 + 4Reiθ + 5
CR  0 
Z π iθ
(R + 1)|eiRe |
≤ R dθ
0 (R − 2)2 + 1
Z π
R2 + R
= e−R sin θ dθ
(R − 2)2 − 1 0
R2 + R π
< 2
→ 0 as R → ∞.
(R − 2) − 1 R

Review: let R(s, t) be a rational function in s and t. Then integrals of the form
Z 2π
R(cos t, sin t) dt
0

can be evaluated using the residue theorem by making the substitutions

z + z −1 z − z −1 dz
cos t = , sin t = , dt =
2 2i iz
to obtain the contour integral
z + z −1 z − z −1
Z  
dz
R , .
|z|=1 2 2i iz

6
(8) Use residue to evaluate
Z 2π
1 2π
Z 2π
1 √ Z 2π
cos2 3θ 3π
(a) dθ = , 2 dθ = π 2, dθ = .
0 5 + 4 sin θ 3 0 1 + sin θ 0 5 − 4 cos 2θ 8
Comments: (a) We make the substitutions indicated in the above review:
Z 2π Z
1 1 dz
dθ = −1 )/(2i)] iz
0 5 + 4 sin θ |z|=1 5 + 4[(z − z
Z
1
= −i 2
dz
|z|=1 5z − 2i(z − 1)
Z
1
= −i dz
|z|=1 −2i(z + 2i)(z + i/2)
Z
1 1
= dz
2 |z|=1 (z + 2i)(z + i/2)

Let f (z) = 1/[(z + 2i)(z + i/2))] which has only one pole −i/2 that lies inside the unit
circle. The corresponding residue equals 1/(−i/2 + 2i) = − 23 i. We conclude that
Z Z
1 dz 1 1
−1 )/(2i)] iz
= dz
|z|=1 5 + 4[(z − z 2 |z|=1 (z + 2i)(z + i/2)
1
= 2πi Res(f (z); −i/2)
2  
2 2
= πi − i = π.
3 3

(b) We make the indicated substitution outlined in the above review:

Z 2π Z
1 1 dz
dθ = −1 )/(2i)]2 iz
0 1 + sin2 θ |z|=1 1 + [(z − z
Z
1 dz
= −i 1 −1 2
|z|=1 1 − 4 (z − z ) z
Z
1 dz
= −i 1 2 −2
|z|=1 1 − 4 (z − 2 + z ) z
Z
1
= −i 1 3 −1
dz
|z|=1 z − 4 (z − 2z + z )
Z
1
= −4i 3 − 2z + z −1 )
dz
|z|=1 4z − (z
Z
1
= −4i 3 −1
dz
|z|=1 −z + 6z − z
Z
z
= 4i 4 2
dz.
|z|=1 z − 6z + 1

Let q(z) = z 4 − 6z 2 + 1, the denominator polynomial. We can√find its roots by the quadratic
formula applied to w2 − 6w + 1 which has p roots w = 3 ± 2 2. p Since w2 = z, the roots of
√ √
q(z) are two sets of values given as: ± 3 + 2 2 and z± = ± 3 − 2 2. The latter two
roots z± lie inside the unit circle while the other do not so they can be discarded below.
7
By the Residue Theorem, we have
Z
z
4i 4 − 6z 2 + 1
dz = 4i (2πi [Res(g(z); z+ ) + Res(g(z); z− )])
|z|=1 z
where g(z) = z/(z 4 − 6z 2 + 1). We next find these residues:

z+ 
Res(g(z); z+ ) = 3

4z − 12z  z=z+
z+
= 3
4z+ − 12z+
1 1
= 2
4 z+ − 3
1 1
= √
4 (3 − 2 2) − 3
1 1 1
= √ =− √ .
4 −2 2 8 2
Since z+ = −z− , we have their have the same squares. By examining the above calculation
of the residue at z+ , we find that the residue at z− possesses the same value. Finally we
now can evaluate the integral:
4i (2πi [Res(g(z); z+ ) + Res(g(z); z− )]) = −8π[2 · Res(g(z); z+ )]
 
1
= −16π − √
8 2
1 √
= 2π √ = π 2.
2