Homework #10 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1
p. 220: 3, 6
1
3. FInd the Taylor series for the function
about the point z0 = 2. Then, by differentiating that
z
series term by term, show that
∞
1X
1
=
(−1)n (n + 1)
2
z
4
n=0
z−2
2
n
(|z − 2| < 2).
Solution:
1
1
1
1
=
= ·
z
2 + (z − 2)
2 1 + z−2
2
n
∞ X
z−2
1
−
=
2
2
n=0
n
∞
1X
n z−2
=
(−1)
.
2
2
n=0
n !
∞
d 1
d 1X
z
−
2
.
=
(−1)n
dz z
dz 2
2
n=0
∞
1
1X
z−2 n
n d
− 2 =
(−1)
z
2
dz
2
n=1
∞
1X
z − 2 n−1
n 1
=
(−1) n
2
2
2
n=1
∞
1
1X
z − 2 n−1
nn
(−1)
=−
z2
2
2
2
n=1
∞
1X
z − 2 n−1
n+1
=
(−1)
n
.
4
2
n=1
Shift the index to start from n = 0:
∞
1
1X
=
(−1)n (n + 1)
z2
4
n=0
Convergence: The series for
z−2
2
n
. (since (−1)n+2 = (−1)n )
1
|z − 2|
about the point z0 = 2 converges for
< 1, or |z − 2| < 2.
z
2
1
6. In the w plane, integrate the Taylor series expansion
∞
X
1
=
(−1)n (w − 1)n (|w − 1| < 1)
w
n=0
along a contour interior to the circle of convergence from w = 1 to w = z to obtain the representation
Log z =
∞
X
(−1)n+1
n
n=1
(z − 1)n (|z − 1| < 1).
Solution:
Z
C
1
dw =
w
=
∞
X
Z
C
∞
X
!
n
n
(−1) (w − 1)
dw
n=0
n
Z
(−1)
(w − 1)n dw.
C
n=0
1
is continuous and has an antiderivative that exists
w
n
inside the circle of convergence and (w − 1) is also continuous with an antiderivative that exists
inside the circle of convergence.
Z
Z z
1
1
=⇒
dw =
dw
w
w
C
1
= Log w|z1
The integrals are path-independent, because
= Log z − Log 1
= Log z,
and
Z
n
z
Z
(w − 1)n dw (n ≥ 0)
z
1
n+1 =
(w − 1)
n+1
1
1
n+1
=
(z − 1)
.
n+1
(w − 1) dw =
C
1
So,
Log z =
∞
X
(−1)n
n=0
1
(z − 1)n+1 ,
n+1
or
Log z =
∞
X
(−1)n+1
n=1
Convergence: Since the series expansion for
n
(z − 1)n ,
1
converges for |z − 1| < 1, so does its integral.
w
2
2
p. 225: 1, 2, 4
1. Use multiplication of series to show that
ez
1
1
5
= + 1 − z − z 2 + · · · (0 < |z| < 1).
2
z(z + 1)
z
2
6
Solution:
z2
1
1
=
+1
1 − (−z)2
∞
X
=
(−z 2 )n
=
n=0
∞
X
(−1)n z 2n , | − z 2 | < 1 =⇒ |z| < 1.
n=0
So,
ez
1
1
= ez 2
2
z(z + 1)
z z +1
z2 z3
1
1+z+
+
+ ···
1 − z2 + z4 − z6 + · · ·
=
z
2!
3!
1 2 5 3
1
1 + z − z − z + · · · , 0 < |z| < 1
=
z
2
6
1
1
5 2
= + 1 − z − z + · · · , 0 < |z| < 1.
z
2
6
2. By writing csc z =
1
and using division, show that
sin z
1
1
1 3
1
−
z + · · · (0 < |z| < π).
csc z = + z +
z 3!
(3!)2 5!
Solution: First, we know that
sin z =
∞
X
(−1)n
n=0
z 2n+1
z3 z5 z7
=z−
+
−
+ ··· .
(2n + 1)!
3!
5!
7!
So,
1
=
sin z
z−
1
z3
3!
+
z5
5!
−
z7
7!
−
1
5!
i
+ ···
,
or
1
z
+
1
3! z
+
h
1
(3!)2
z3 + · · ·
z − 3!1 z 3 + 5!1 z 5 − 7!1 z 7 + · · · 1
1 + 0 − 13 z 2 + 0 +
0+
0+
1 2
3! z
1 2
3! z
h
+0
+0
1
2
h (3!)
1
(3!)2
1 4
1 6
5! z − 7! z · · ·
− 5!1 z 4 + 0 − 7!1 z 6 + · · ·
1
1 6
4
− (3!)
0 + 3!5!
z + ···
2z +
−
−
i
1
4
5! i z
1
4
5! z
..
.
3
+
+
1
0 −
0 −
h
+
7!
1
−
(3!)2
1
z6 + · · ·
3!5!
i
1 1 6
5! 3! z + · · ·
Therefore,
1 3
1
1
1
−
z + ··· .
csc z = + z +
z 3!
(3!)2 5!
1
Convergence:
has singularities at z = nπ, n ∈ Z, so the series can only converge for 0 < |z| <
sin z
π.
4. Use the expansion
1
1 1
1
7
− · +
z + · · · (0 < |z| < π)
3
z
6 z 360
in Example 2, Sec. 67 and the method illustrated in Example 1, Sec. 62, to show that
Z
dz
πi
=− ,
2
3
C z sinh z
z 2 sinh z
=
where C is the positively oriented unit circle |z| = 1.
Solution: From Example 1, Sec. 62, we know that
Z
1
b1 =
f (z) dz,
2πi C
where C is a simple closed contour and b1 is the residue of f (z) at z = z0 , the only singularity
inside C. Then,
Z
f (z) dz = 2πib1 ,
C
and since
Res
z=0
1
z 2 sinh z
1
=− ,
6
we obtain
Z
C
dz
πi
1
= 2πi −
= − i.
2
z sinh z
6
3
4