Homework #11 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1
HW #11(a)
Find the singularities and determine if they are isolated for
1
1. f (z) = z cos
z 1
1
has one singularity, z0 = 0. and g(z) = z is entire. Therefore, z cos
is
Solution: cos
z
z
analytic for all z 6= 0, and z0 = 0 is an isolated singularity.
1
+ 4)
Solution: f (z) is a quotient of two functions, and the denominator is 0 only if z 3 (z + 4) = 0, so
the singularities of f (z) are
2. f (z) =
z 3 (z
z 3 = 0 =⇒ z = 0
z + 4 = 0 =⇒ z = −4
Since f (z) is analytic for all z 6= 0, −4, z0 = 0 and z0 = −4 are isolated singularities.
1
π
1 − cos 2z
Solution: The denominator is 0 only if
3. f (z) =
1 − cos
π
= 0,
2z
which is true only if
π
cos
= 1.
2z
From previous work, we know that this is true only if
π
= 2nπ, n ∈ Z
2z
1
, n 6= 0, and z = 0.
=⇒ z =
4n
1 1
1
So, the singluar points are z = 0, z = ± , ± , . . . . The singularities z =
, n = ±1, ±2, . . . are
4 8
4n
isolated singular points, since there is a deleted neighborhood of each point throughout which f (z)
is analytic.
1
z0 = 0 is not an isolated singluarity, because every deleted neighborhood of z = 0 contains at
1
least one point of the form z =
(and so, contains a singularity). Why? Given > 0, we
4n 1
1
1
can find a value N such that
and thus, z =
< just let N >
is inside the deleted
4N
4
4N
-neighborhood of z = 0.
2
HW #11(b)
2.1
p. 239: 1, 2
1. Find the residue at z = 0 of the function
1
(a)
z + z2
Solution:
1
1
=
z2 + z
z(z + 1)
1
=
z(1 − (−z))
1
1 − z + z2 − z3 + · · ·
=
z
1
= − 1 + z − z2 + · · · .
z
1
Therefore, Res 2
= 1.
z=0 z + z
1
(b) z cos
z
Solution:
1
1 1
1 1
z cos
=z 1−
+
− ···
z
2! z 2 4! z 4
11
1 1
=z−
+
− ··· .
2 z 4! z 3
1
1
Therefore, Res z cos
=− .
z=0
z
2
z − sin z
(c)
z
Solution:
z − sin z
1
= (z − sin z)
z
z
1
1 3
1 5
=
z − z − z + z − ···
z
3!
5!
1 1 3
1 5
=
z − z + ···
z 3!
5!
1
1
= z2 − z4 + · · · .
3!
5!
z − sin z
Therefore, Res
= 0.
z=0
z
2
(d)
cot z
z4
Solution:
cos z
sin z
1 − 2!1 z 2 +
=
z − 3!1 z 3 +
cot z =
1 4
4! z
1 5
5! z
− ···
.
− ···
1 3
1
− 1 z − 45
z + ···
z 3
z − 3!1 z 3 + 5!1 z 5 + · · · 1 + 0z − 2!1 z 2 + 4!1 z 4 + 0z 5 +
1+
1 7
6! z − · · ·
0 − 3!1 z 2 + 0 + 5!1 z 4 + 0z 5 + 7!1 z 7 + · · ·
1 4
− 13 z 2 + 0z 3 + 30
z + 0z 5 − 6!1 + 7!1 z 6 + · · ·
1 4
1 6
− 13 z 2 + 0z 3 + 18
z + 0z 5 − 360
z + ···
6
1 4
1
1
1
5
− 45 z + 0z − 6! + 7! − 360
z + ···
..
.
So,
1 1 1
1 3
cot z
= 4
− z − z + ···
z4
z
z 3
45
1 1
1 1
1
−
+ ··· .
= 5−
z
3 z 3 45 z
cot z
1
=− .
4
z=0 z
45
Therefore, Res
(e)
sinh z
z 4 (1 − z 2 )
Solution:
1
1
sinh z = z + z 3 + z 5 + · · ·
5!
3!
sinh z
1
1 3
1 5
= 4 z + z + z + ···
1 + z2 + z4 + z6 + · · ·
4
2
z (1 − z )
z
3!
5!
1
7
47
= 4 z + z3 + z5 + · · ·
z
6
40
1
7 1 47
= 3+
+ z + ··· .
z
6 z 40
Therefore, Res
z=0
sinh z
7
= .
z 4 (1 − z 2 )
6
2. Use Cauchy’s residue theorem to evaluate the integral of each of these functions around the circle
|z| = 3 in the positive sense.
(a)
e−z
z2
Solution: The only singularity of the function is z0 = 0. So, we need to determine the residue
at z0 = 0.
e−z
1
1 2
1 3
= 2 1 − z + z − z + ···
z2
z
2!
3!
1
1
1
1
= 2 − + − z + ··· .
z
z 2! 3!
3
e−z
= −1, and
z=0 z 2
Z
e−z
e−z
dz
=
2πiRes
= 2πi(−1) = −2πi .
2
z=0 z 2
|z|=3 z
Therefore, Res
(b)
e−z
(z − 1)2
Solution: The only singularity of the function is z0 = 1. So, we need to determine the residue
at z0 = 1.
e−z
e−(z−1+1)
=
(z − 1)2
(z − 1)2
e−1 e−(z−1)
(z − 1)2
e−1
1
1
2
3
=
1 − (z − 1) + (z − 1) − (z − 1) + · · ·
(z − 1)2
2!
3!
−1
−1
−1
−1
e
e
e
e
−
+
−
(z − 1) + · · · .
=
2
(z − 1)
z−1
2!
3!
=
e−z
= −e−1 , and
z=1 (z − 1)2
Therefore, Res
Z
|z|=3
e−z
e−z
1
2πi
dz
=
2πiRes
=
2πi
−
= −
.
z=1 (z − 1)2
(z − 1)2
e
e
1
(c) z 2 e z
1
Solution: The only singularity of z 2 e z is z0 = 0. So, we need to determine the residue at
z0 = 0.
1
1
1 1
1 1
1 1
z2e z = z2 1 + +
+
+
+
·
·
·
z 2! z 2 3! z 3 4! z 4
1
11
1 1
+
+ ··· .
= z2 + z + +
2! 3! z 4! z 2
1
Therefore, Res z 2 e z =
z=0
(d)
1
1
= , and
3!
6
Z
πi
1
2 z1
2 z1
=
z e dz = 2πiRes z e = 2πi
.
z=0
6
3
|z|=3
z+1
z 2 − 2z
Solution:
f (z) =
z+1
z+1
=
,
2
z − 2z
z(z − 2)
so f (z) has two singularities, z0 = 0 and z0 = 2, both of which are interior to the contour.
This means that we need to find the residue at both points.
4
z0 = 0:
z+1
1
1
=
+
z(z − 2)
z − 2 z(z − 2)
1
1
=−
z−2
2 1 − z2
z z2 z3
1
1+ +
=−
+
+ ···
2
2
4
8
2
3
z
1 z z
−
− ··· .
=− − −
2 4
8
16
1
1 z z2
1
z3
− − −
=
−
− ···
z(z − 2)
z
2 4
8
16
11 1 z
z2
=−
− − −
− ··· .
2 z 4 8 16
So,
1 1 3 3z 3z 2
z+1
=−
− −
−
− ··· .
z(z − 2)
2z 4
8
16
z+1
1
=⇒ Res
=− .
z=0 z(z − 2)
2
z0 = 2:
(z − 2) + 3
z+1
=
z(z − 2)
z(z − 2)
1
3
= +
.
z z(z − 2)
1
1
=
z
(z − 2) + 2
1
=
2 − (−(z − 2))
1
=
2
1
1 − − z−2
2
!
1
=
2
z−2
1−
+
2
z−2
2
2
−
z−2
2
!
3
+ ···
.
3
3
1
=
·
z(z − 2)
z−2 z
!
3
1
z−2
z−2 2
z−2 3
=
·
1−
+
−
+ ···
z−2 2
2
2
2
1
3
3 3
3
2
=
− + (z − 2) − (z − 2) + · · · .
2 z−2 2 4
8
So,
z+1
3
1
3
= ·
− +
z(z − 2)
2 z−2 4
z+1
=⇒ Res
z=2 z(z − 2)
5
3
3
(z − 2) − (z − 2)2 + · · · .
8
16
3
= .
2
Therefore,
Z
|z|=3
z+1
z+1
z+1
dz = 2πi Res
+ Res
z=0 z(z − 2)
z=2 z(z − 2)
z(z − 2)
1 3
= 2πi − +
2 2
= 2πi .
2.2
Problem 2
Evaluate the following integrals
Z
(a)
|z|= 12
dz
z(1 − z)3
Solution: The singularities of
1
are z0 = 0 and z0 = 1. Only z0 = 0 is inside the circle
z(1 − z)3
1
|z| = . Therefore, in order to evaluate the integral, we need to find the residue at z0 = 0.
2
1
1
1
= ·
3
z(1 − z)
z (1 − z)3
1
= (1 + z + z 2 + · · · )3
(1 − z)3
= (1 + z + z 2 + · · · )(1 + z + z 2 + · · · )(1 + z + z 2 + · · · )
= (1 + 2z + 3z 2 + · · · )(1 + z + z 2 + · · · )
= 1 + 3z + 6z 2 + · · · .
So,
1
1
= + 3 + 6z + · · ·
3
z(1 − z)
z
1
=⇒ Res
= 1.
z=0 z(1 − z)3
Therefore,
Z
|z|= 21
Z
(b)
|z|= 12
dz
= 2πi(1) = 2πi .
z(1 − z)3
ez
dz
z(1 − z)3
Solution: From part (a), we see that z0 = 0 is the only singularity inside |z| =
6
1
. So, we need to
2
find the residue of the function at z0 = 0.
ez
1
=
· ez
3
z(1 − z)
z(1 − z)3
1
1 2
=
+ 3 + 6z + · · ·
1 + z + z + ···
z
2!
19
1
= + 4 + z + ···
z
2
ez
=⇒ Res
= 1.
z=0 z(1 − z)3
Therefore,
Z
|z|= 21
3
ez
dz = 2πi(1) = 2πi .
z(1 − z)3
HW #11(c)
3.1
p. 243: 1, 2
1. In each case, write the principal part of the function at the isolated singular point and identify the
type of singularity.
1
(a) ze z
Solution: The only singular point is z0 = 0. And,
1
1
1 1
1 1
z
+
+ ···
ze = z 1 + +
z 2! z 2 3! z 3
11
1 1
=z+1+
+
+ ··· .
2! z 3! z 3
1
Therefore, the principal part of ze z is
∞
X
1 1
1
1
11
+
+ ··· =
,
3
2! z 3! z
(n + 1)! z n
n=1
and z0 = 0 is an essential singularity.
z2
(b)
1+z
Solution: The only singularity is z0 = −1. And,
z2
(z + 1 − 1)2
=
1+z
z+1
(z + 1)2 − 2(z + 1) + 1
=
z+1
1
=z+1−2+
.
z+1
z2
is
1+z
1
,
z+1
Therefore, the principal part of
and z0 = −1 is a simple pole.
7
sin z
z
Solution: The only singularity is z0 = 0. And,
sin z
1 3
1
1 5
z − z + z − ···
=
z
z
3!
5!
1
1
= 1 − z2 + z4 − · · · .
3!
5!
sin z
Therefore, the pricnipal part of
is 0, and z0 = 0 is a removable singularity.
z
cos z
(d)
z
Solution: The only singularity is z0 = 0. And,
cos z
1 2
1
1 4
1 − z + z − ···
=
z
z
2!
4!
1
1 3
1
= − z + z − ··· .
z 2!
4!
cos z
1
Therefore, the pricnipal part of
is , and z0 = 0 is a simple pole.
z
z
1
(e)
(2 − z)3
Solution: z0 = 2 is the only singularity. The principal part is
(c)
−
1
,
(z − 2)3
and so, z0 = 2 is a pole of order 3.
2. Show that the singular point is a pole. Determine the order m of that pole and the corresponding
residue.
1 − cosh z
z3
Solution: z0 = 0 is the only singularity. Since
1
1
1 2
1 4
(1 − cosh z) = 3 1 − 1 + z + z + · · ·
z3
z
2!
4!
1
1 2
1 4
= 3 − z − z − ···
z
2!
4!
11
1
=−
− z − ··· ,
2! z 4!
11
1
and the principal part is −
, z0 = 0 is a simple pole (so, m = 1), and the resiude is B = − .
2! z
2
2z
1−e
(b)
z4
Solution: z0 = 0 is the only singularity. Since,
1
1
1
1
1
2z
2
3
4
1−e
= 4 1 − 1 + 2z + (2z) + (2z) + (2z) + · · ·
z4
z
2!
3!
4!
1
4
2
= 4 −2z − 2z 2 − z 3 − z 4 − · · ·
z
3
3
2
2
41 2
=− 3 − 2 −
− − ··· ,
z
z
3z 3
(a)
8
41
2
2
and the principal part is −
− 2 − 3 , z0 = 0 is a pole of order 3 (so, m = 3), and the
3z
z
z
4
residue is B = − .
3
2z
e
(c)
(z − 1)2
Solution: z0 = 1 is the only singularity. Since
e2z = e2(z−1+1) = e2 e2(z−1) ,
1
e2z
e2
1
2
3
1 + 2(z − 1) + (2(z − 1)) + (2(z − 1)) + · · ·
=
(z − 1)2
(z − 1)2
2!
3!
2
2
2e
4
e
+
+ 2e2 + e2 (z − 1) + · · · ,
=
(z − 1)2 z − 1
3
2
2e
e2
the principal part is
+
.
z − 1 (z − 1)2
Therefore, z0 = 1 is a pole of order 2 (so, m = 2) and the residue is B = 2e2 .
3.2
p. 248: 1, 3, 4
1. In each case, show that the singularity is a pole. Determine the order m and the residue B of each
pole.
(a)
z2 + 2
z+1
Solution: z0 = −1 is the only singularity, and
z2 + 2
z2 + 2
=
z+1
z − (−1)
φ(z)
=
.
(z − (−1))1
Since φ(−1) = (−1)2 + 2 = 3 6= 0, z0 = −1 is a pole of order 1, and B = φ(−1) = 3.
3
z
(b)
2z + 1
1
Solution: z0 = − is the only singularity, and
2
3
z
z3
=
2z + 1
(2z + 1)3
z3
=
3
23 z + 12
=
=⇒ φ(z) =
9
z3
8
(z + 12 )3
z3
.
8
.
Then, z0 = −
1
is a pole of order m = 3. We have that
2
φ(z) =
z3
3z 2 00
3z
=⇒ φ0 (z) =
, φ (z) =
.
8
8
4
Therefore,
3(− 12 )
φ00 − 21
3
= 4 =− .
B=
2!
2
16
(c)
ez
z2 + π2
Solution: Since
ez
ez
=
,
z2 + π2
(z + πi)(z − πi)
there are two singularities, z0 = −πi and z0 = πi.
z0 = −πi:
ez
ez
= z−πi .
2
2
z +π
z + πi
So, z0 = −πi is a pole of order m = 1, and
φ(z) =
ez
.
z − πi
Therefore,
B = φ(−πi) =
e−πi
−πi − πi
−1
−2πi
1
= − i.
2π
=
z0 = πi:
ez
ez
= z+πi .
2
2
z +π
z − πi
So, z0 = πi is a pole of order m = 1, and
φ(z) =
ez
.
z + πi
Therefore,
B = φ(πi) =
−1
2πi
1
=
i.
2π
=
10
eπi
πi + πi
3. Find the value of
Z
C
3z 3 + 2
dz
(z − 1)(z 2 + 9)
taken counterclockwise aorund the circle
(a) |z − 2| = 2
Solution: The singularities of the integrand are z0 = 1, ±3i. Only z0 = 1 lies inside |z−2| = 2.
So, writing
3z 3 +2
3z 3 + 2
2
f (z) =
= z +9 ,
2
(z − 1)(z + 9)
z−1
we see that
3z 3 + 2
φ(z) = 2
,
z +9
and z0 = 1 is a simple pole. Therefore,
Res f (z) = φ(1) =
z=1
and
3(1)3 + 2
1
= ,
2
1 +9
2
1
= πi .
f (z) dz = 2πi
2
C
Z
(b) |z| = 4
Solution: In this case, all three singularities are interior to C.
z0 = 1: Already done above.
z0 = −3i:
f (z) =
=
3z 3 + 2
(z − 1)(z − 3i)(z + 3i)
3z 3 +2
(z−1)(z−3i)
z + 3i
.
So, z0 = −3i is a simple pole, and
φ(z) =
3z 3 + 2
.
(z − 1)(z − 3i)
Therefore,
Res f (z) = φ(−3i)
z=−3i
=
=
=
=
=
=
11
3(−3i)3 + 2
(−3i − 1)(−3i − 3i)
2 + 81i
6i(1 + 3i)
2 + 81i
1 − 3i
·
6i(1 + 3i) 1 − 3i
245 + 75i
6i(10)
245 + 75i
60i
5 49
− i.
4 12
z0 = 3i:
f (z) =
=
3z 3 + 2
(z − 1)(z + 3i)(z − 3i)
3z 3 +2
(z−1)(z+3i)
z − 3i
.
So, z0 = 3i is a simple pole, and
φ(z) =
3z 3 + 2
.
(z − 1)(z + 3i)
Therefore,
Res f (z) = φ(3i)
z=3i
=
=
=
=
=
=
3(3i)3 + 2
(3i − 1)(3i + 3i)
2 − 81i
6i(−1 + 3i)
2 − 81i
−1 − 3i
·
6i(−1 + 3i) −1 − 3i
−245 + 75i
6i(10)
−245 + 75i
60i
5 49
+ i.
4 12
So,
Z
f (z) dz = 2πi
C
Z
4. Find the value of the integral
C
1
+
2
5 49
5 49
− i +
+ i
= 6πi .
4 12
4 12
dz
taken counterclockwise around the circle.
+ 4)
z 3 (z
(a) |z| = 2
Solution: The singularities of the integrand are z0 = 0 and z0 = −4. Only z0 = 0 is interior
to C. Therefore,
1
1
f (z) = 3
= z+4
,
z (z + 4)
z3
and z0 = 0 is a pole of order 3. So,
φ(z) =
1
,
z+4
and
Res f (z) =
z=0
φ00 (0)
.
2!
φ0 (z) = −(z + 4)−2
φ00 (z) = 2(z + 4)−3
2
=⇒ φ00 (0) = 3 .
4
12
Therefore,
Res f (z) =
z=0
1
,
64
and so
Z
f (z) = 2πi
=⇒
C
1
64
=
π
i.
32
(b) |z + 2| = 3
Solution: In this case, both singularities are inside C. We already determined Res f (z) =
z=0
1
,
64
so we need only find Res f (z). Since
z=−4
f (z) =
1
1
z3
=
,
z 3 (z − 4)
z−4
z0 = −4 is a simple pole, and
φ(z) =
1
,
z3
so
Res f (z) = φ(−4) = −
z=−4
1
.
64
Then,
Z
f (z) dz = 2πi
C
4
1
1
= 0.
+ −
64
64
HW #11(d) – p. 255: 1, 2, 4
1
1. Show that the point z = 0 is a simple pole of f (z) = csc z =
and that the residue there is 1
sin z
by appealing to
(a) Theorem 2 in Sec. 76.
Solution: Since
f (z) =
p(z)
1
=
,
sin z
q(z)
we see that p and q are entire; p(0) = 1 6= 0; q(0) = 0; and q 0 (0) = cos 0 = 1 6= 0. Therefore,
Res f (z) =
z=0
p(0)
1
= = 1. X
0
q (0)
1
(b) the Laurent series for csc z found in Ex. 2 in Sec. 67.
Solution:
" #
1
1
1 2
1
csc z = + z +
−
z3 + · · ·
z 3!
3!
5!
=⇒ Res f (z) = 1. X.
z=0
2. Show that
13
z − sinh z
i
=
2
z=πi z sinh z
π
Solution:
(a) Res
z − sinh z
p(z)
=
=⇒ p, q are entire.
2
z sinh z
q(z)
p(πi) = πi − sinh(πi) = πi − i sin(π) = πi 6= 0.
q(πi) = (πi)2 sinh(πi) = 0.
q 0 (z) = 2z sinh z + z 2 cosh z
=⇒ q 0 (πi) = 2πi sinh(πi) + (πi)2 cosh(πi)
= −π 2 cos(π)
= π 2 6= 0.
p(πi)
z − sinh z
= 0
=⇒ Res 2
z=πi z sinh z
q (πi)
πi
i
= 2 = .X
π
π
ezt
ezt
+ Res
= −2 cos(πt)
z=−πi sinh z
z=πi sinh z
Solution: First, we write
(b) Res
ezt
p(z)
=
,
sinh z
q(z)
so we see that p and q are entire. Then
z0 = πi:
p(πi) = eπit = cos(πt) + i sin(πt) 6= 0.
q(πi) = sinh(πi) = 0.
q 0 (z) = cosh z =⇒ q 0 (πi) = −1 6= 0.
Therefore,
ezt
p(πi)
=
= − cos(πt) − i sin(πt).
z=πi sinh z
q(πi)
Res
z0 = −πi:
p(−πi) = e−πit = cos(πt) − i sin(πt) 6= 0.
q(−πi) = sinh(−πi) = 0.
q 0 (z) = cosh z =⇒ q 0 (−πi) = −1 6= 0.
Therefore,
p(−πi)
ezt
=
= − cos(πt) + i sin(πt).
z=−πi sinh z
q(−πi)
Res
Therefore,
ezt
ezt
+ Res
= −2 cos(πt). X
z=πi sinh z
z=−πi sinh z
Res
4. Let C denote the positively oriented circle |z| = 2 and evaluate
14
Z
tan z dz
(a)
C
Solution:
f (z) = tan z =
sin z
p(z)
=
cos z
q(z)
=⇒ p and q are entire.
1
n+
2
π, n ∈ Z. The only singularities inside
The singularities are where cos z = 0, or z =
π
π
C are z0 = and z0 = − .
2
2
π
z0 = :
2
π π = sin
= 1 6= 0.
p
2
2
π
q
= 0.
2
π q 0 (z) = − sin z =⇒ q 0
= −1.
2
Therefore,
Resπ tan z = −1.
z= 2
π
z0 = − :
2
π
π
p −
= sin −
= −1 6= 0.
2
2
π
= 0.
q −
2
π
q 0 (z) = − sin z =⇒ q 0 −
= 1.
2
Therefore,
Res tan z = −1.
z=− π2
So,
Z
tan z dz = 2πi(−1 + (−1)) = −4πi .
C
Z
(b)
C
dz
sinh(2z)
1
nπ
occur where sinh(2z) = 0 =⇒ z =
i, n ∈
sinh(2z)
2
π
Z. The only singularities inside C are z0 = 0, z0 = ± i. Since
2
Solution: The singularities of f (z) =
f (z) =
1
p(z)
=
,
sinh(2z)
q(z)
we see that both p and q are entire.
15
z0 = 0:
p(0) = 1 6= 0.
q(0) = 0.
q 0 (z) = 2 cosh(2z) =⇒ q 0 (0) = 2.
Therefore,
Res f (z) =
z=0
z0 =
1
p(0)
= .
0
q (0)
2
π
i:
2
π i = 1 6= 0.
2 π
q
i = 0.
2
p
q 0 (z) = 2 cosh z =⇒ q 0
π i = 2 cosh(πi) = −2 6= 0.
2
Therefore,
1
Res
tan z = − .
π
2
z= 2 i
π
z0 = − i:
2
π p − i = 1 6= 0.
2 π
q − i = 0.
2
π q 0 (z) = 2 cosh z =⇒ q 0 − i = 2 cosh(−πi) = −2 6= 0.
2
Therefore,
1
Resπ tan z = − .
2
z=− 2 i
So,
Z
C
dz
= 2πi
sinh(2z)
16
1
1
1
+ −
+ −
= −πi .
2
2
2