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
∞
X
1 − e2z
1
=
−
(2z)k
4
z
k!
!,
z4
k=1
∞
X
1 k k−4
=−
2 z
k!
k=1
so the negative powers of the expansion are
−2
1 1
11
1
− 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:
e
2z
=
∞
X
f (k) (1)
k=0
k!
k
(z − 1) =
∞
X
2k e2
k=0
k!
(z − 1)k
since Dk (e2k ) = 2k e2k . Then we find that
∞
X 2k e2
e2z
=
(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
Z
−z
e−z
d
2e
dz = 2πiRes(f (z); 0) = 2πi lim
z 2 = −2πie0 = −2πi.
2
z→0
z
dz
z
|z|=3
(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
Z
|z|=3
e−z
dz = 2πiRes(f (z); 0)
(z − 1)2
d
e−z
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
1
1 1
1 1
1 1
f (z) = z 1 + +
+
+
+ ···
z 2! z 2 3! z 3 4! z 4
11
1 1
1
+
+ ···
= z2 + z + +
2! 3! z 4! z 2
2
which shows that Res(f (z); 0) = 1/6. By the Residue Theorem, we must have
Z
z 3 e1/z dz = 2πiRes(f (z); 0) = 2πi
|z|=3
1
πi
= .
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
Res(f (z); i) = lim(z − i)
z→i
z 1/4
= i1/4 = cos(π/8) + i sin(π/8)
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 Log z
d
2 Log z
(z − i) 2
= lim
lim
2
z→i dz (z + i)2
z→i dz
(z + 1)
(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 = −eπ .
Res(f (z); π) =
cos z z=π
(4) Let ZC be the circle |z|Z= 2. Evaluate Z
the following integrals:
1
cos πz
(a)
tan z dz, (b)
dz,
dz
2
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
Z
1
−1
dz = 2πi
= −πi.
sin
2z
2
C
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
dz
=
2πi
1
+
(−e
/4
−
e
/4)
+
(−e
/4
−
e
/4)
.
2
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
R→∞ CR
f (z) dz = 0.
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
g(x)
Note that for integrals of the form
dx, the corresponding analytic funccos 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)
2
2
2
(x2 + 1)2
x4 + 1
−∞
−∞ (x + 9)(x + 4)
Z−∞
Z ∞
∞
dx
dx
(d)
, (e)
.
2
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 ∞
4π
x4
= 2πi(−2i/25) =
2 + 9)(x2 + 4)2
(x
25
−∞
provided we can show that
Z
lim
R→∞ CR
f (z) dz = 0
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
Z
z4
R4
≤ lim πR
dz
= 0.
lim R→∞
R→∞ CR (z 2 + 9)(z 2 + 4)2
(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
= 2πi
=
2
2
4
2
−∞ (x + 1)(x + 2x + 3)
provided we can show that
Z
lim
R→∞ CR
f (z) dz = 0
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
1
≤ lim πR 1 = 0.
dz
R→∞
2
R2
CR z + 2z + 3
√
(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
= 2πi (−1/8 − i/8 + 1/8) =
2
2
4
−∞ (x + 1)(x + 2x + 3)
Z
lim R→∞
provided we can show that
Z
lim
R→∞ CR
f (z) dz = 0
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)
=
,
0
<
b
<
a;
(b)
dx = πe−a ,
2 + a2 )(x2 + b2 )
2 + b2 )ab
2+1
(x
(a
x
−∞
−∞
0 < a;
Z ∞
Z ∞
√
cos ax
πe−ab
x sin 2x
−2 3
(c)
dx
=
,
0
<
a,
0
<
b;
(d)
dx
=
πe
;
2
2 2
2
b
−∞ (x + b )
−∞ x + 3
Z ∞ 3
x sin ax
dx = −π(a − 1)e−2a , 0 < a;
(e)
2 + 4)2
(x
Z −∞
∞
9
5
x3 sin x
dx = − πe−3 + e−1 π.
(f)
2 + 1)(x2 + 9)
(x
64
64
0
(7) UseZresidues to find the principal
values of the integrals
below:
Z ∞
Z ∞
∞
sin x
(x + 1) cos x
cos x
(a)
dx,
dx,
dx, 0 < b.
2
2
2
2
−∞ 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
dx = = 2πi − e
i
2
2
−∞ x + 4x + 5
= =(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
iz
e
1
dz
→ 0 as R → ∞
≤ πR
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
−1−2i
(−1 + i)e
/[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
dx = < 2πi (1 + i)e
2
2
−∞ x + 4x + 5
= 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 < .
R
R
0
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
Z θ
iz
iθ + 1)eiReiθ
(z
+
1)e
(Re
iθ
dz = iRe dθ
2 + 4z + 5
iθ )2 + 4Reiθ + 5
z
(Re
CR
0
Z π
iθ
(R + 1)|eiRe |
≤
R dθ
(R − 2)2 + 1
0
Z π
R2 + R
=
e−R sin θ dθ
(R − 2)2 − 1 0
R2 + R
π
<
→ 0 as R → ∞.
2
(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
cos t =
z + z −1
,
2
to obtain the contour integral
Z
R
|z|=1
sin t =
z − z −1
,
2i
z + z −1 z − z −1
,
2
2i
6
dt =
dz
.
iz
dz
iz
(8) Use residue to evaluate
Z 2π
1
2π
(a)
dθ =
,
5 + 4 sin θ
3
0
√
1
2 dθ = π 2,
1 + sin θ
2π
Z
0
Z
0
2π
cos2 3θ
3π
dθ =
.
5 − 4 cos 2θ
8
Comments: (a) We make the substitutions indicated in the above review:
2π
Z
0
1
dθ =
5 + 4 sin θ
Z
1
dz
−1 )/(2i)] iz
5
+
4[(z
−
z
|z|=1
Z
1
= −i
dz
2
|z|=1 5z − 2i(z − 1)
Z
1
dz
= −i
|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
=
dz
−1 )/(2i)] iz
5
+
4[(z
−
z
2
(z
+
2i)(z
+ i/2)
|z|=1
|z|=1
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
0
2π
Z
1
dz
−1 )/(2i)]2 iz
1
+
[(z
−
z
|z|=1
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
dz
1 3
−1
|z|=1 z − 4 (z − 2z + z )
Z
1
= −4i
dz
3 − 2z + z −1 )
4z
−
(z
|z|=1
Z
1
= −4i
dz
3
−1
|z|=1 −z + 6z − z
Z
z
dz.
= 4i
4
2
|z|=1 z − 6z + 1
1
dθ =
1 + sin2 θ
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
w = 3 ± 2 2. p
Since w2 = z, the roots of
p roots
√
√
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
dz = 4i (2πi [Res(g(z); z+ ) + Res(g(z); z− )])
4 − 6z 2 + 1
z
|z|=1
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
8