Solutions for Math 311 Assignment #9
(1) Obtain the Taylor series representation
∞
X
z 4n+1
z cosh(z 2 ) =
(2n)!
n=0
at z = 0 and show that it holds for all z, i.e., its radius of
convergence is ∞.
Solution. Since
ez =
∞
X
zn
n!
n=0
for |z| < ∞, we have
z2
e
=
∞
X
(z 2 )n
n=0
and
2
e−z =
n!
∞
X
(−z 2 )n
n=0
n!
=
∞
X
z 2n
n=0
=
n!
∞
X
(−1)n
n=0
n!
z 2n
for |z| < ∞. Therefore,
2
2
∞
X
ez + e−z
1 1 + (−1)n
z cosh(z ) = z
=z
z 2n
2
2
n!
n=0
2
for |z| < ∞. Note that
(
2 if n = 2k
1 + (−1) =
0 if n = 2k + 1
n
Therefore,
∞
∞
X
X
z 4n+1
z 4k
=
z cosh(z ) = z
(2k)! n=0 (2n)!
k=0
2
for |z| < ∞.
(2) Obtain the Taylor series
∞
X
(z − 1)n
ez = e
for |z − 1| < ∞
n!
n=0
of f (z) = ez at z = 1 by
(a) using f (n) (1);
(b) writing ez = ez−1 e.
1
2
Solution. (a) Let f (z) = ez . Then f (n) (z) = ez and f (n) (1) =
e. Therefore,
∞
∞
∞
X
X
X
f (n) (1)
e
(z − 1)n
z
n
n
e =
(z − 1) =
(z − 1) = e
n!
n!
n!
n=0
n=0
n=0
for |z − 1| < ∞. The radius of convergence is ∞ since f (z) is
entire.
(b) Since
∞
X
zn
ez =
n!
n=0
for |z| < ∞, we see that
z−1
e
=
∞
X
(z − 1)n
n=0
n!
for |z − 1| < ∞ by substituting z − 1 for z. Therefore,
∞
X
(z − 1)n
ez = eez−1 = e
n!
n=0
for |z − 1| < ∞.
(3) Find the Taylor series of f (z) = sin(z 2 ) at z = 0 and use it to
show that
f (4n) (0) = f (2n+1) (0) = 0
for n = 0, 1, 2, ....
Solution. Since
∞
X
(−1)n 2n+1
sin(z) =
z
(2n + 1)!
n=0
for |z| < ∞, we see that
∞
X
(−1)n 4n+2
2
sin(z ) =
z
(2n + 1)!
n=0
by substituting z 2 for z. Therefore,
∞
∞
X
f (m) (0) m X (−1)n 4n+2
z =
z
m!
(2n + 1)!
m=0
n=0
It follows that f (m) (0) 6= 0 if and only if m = 4n + 2 for some
integer n ≥ 0. Therefore,
f (2n+1) (0) = f (4n) (0) = 0
3
for all integers n ≥ 0.
(4) Derive the Taylor series representation
∞
X (z − i)n
1
=
1−z
(1 − i)n+1
n=0
of
√ f (z) = 1/(1 − z) at z = i and show√that it holds for |z − i| <
2, i.e., its radius of convergence is 2.
Solution. Since
∞
X
1
=
wn
1 − w n=0
for all |w| < 1, we have
1
1
1
1
=
=
1−z
(1 − i) − (z − i)
1 − i 1 − (z − i)/(1 − i)
∞
1 X (z − i)n
=
1 − i n=0 (1 − i)n
∞
X
(z − i)n
=
(1 − i)n+1
n=0
for all z satisfying
√
z − i
< 1 ⇔ |z − i| < 2.
1 − i
√
Another way to show that √the radius of convergence is 2 is
by noticing
√ that |i − 1| = 2. Hence 1/(1 − z) is analytic√in
|z − i| < 2 and not√analytic in |z − i| < R for all R > 2.
In other words, R = 2 is the largest R such that 1/(1 − z) is
analytic in |z − i| < R. So it is the radius of convergence.
(5) Find the Laurent series that represents the function
1
2
f (z) = z sin
z2
in the domain 0 < |z| < ∞.
Solution. Since
∞
X
(−1)n 2n+1
sin(z) =
z
(2n
+
1)!
n=0
4
for |z| < ∞, we have
∞
∞
X
1
(−1)n −4n−2 X (−1)n 1
2
2
z sin
=z
z
=
z2
(2n + 1)!
(2n + 1)! z 4n
n=0
n=0
for 0 < |z| < ∞ by substituting z −2 for z.
(6) Derive the Laurent series representation
!
∞
ez
1
1
1 X (z + 1)n
+
+
for 0 < |z + 1| < ∞.
=
(z + 1)2
e n=0 (n + 2)! z + 1 (z + 1)2
Solution. Since
z
e =
∞
X
zn
n=0
n!
for |z| < ∞, we have
z+1
e
=
∞
X
z + 1n
n=0
n!
for |z + 1| < ∞ by substituting z + 1 for z. Therefore,
∞
X (z + 1)n−2
ez
ez+1
=
=
(z + 1)2
e(z + 1)2
e(n!)
n=0
1
=
e
=
1
e
∞
X (z + 1)n−2
1
1
+
+
(z + 1)2 z + 1 n=2
n!
!
∞
n
X
1
1
(z + 1)
+
+
2
(z + 1)
z + 1 n=0 (n + 2)!
!
for 0 < |z + 1| < ∞.
(7) Give two Laurent Series expansions in powers of z for the function
1
f (z) = 2
z (1 − z)
and specify the regions in which those expansions are valid.
5
Solution. We observe that f (z) is analytic in {z 6= 0, 1}. So it
is analytic in 0 < |z| < 1 and 1 < |z|. When 0 < |z| < 1,
∞
∞
1
1 X n X n−2
= 2
z =
z
z 2 (1 − z)
z n=0
n=0
∞
=
∞
1
1
1 X n−2
1 X n
+
z
=
+
z
+
+
z 2 z n=2
z 2 z n=0
When 1 < |z| < ∞,
∞
1
1
1
1 X −n
=
−
=
−
z
z 2 (1 − z)
z 3 1 − 1/z
z 3 n=0
∞
∞
X
X
1
1
=−
=
n+3
z
zn
n=0
n=3
(8) Show that when 0 < |z − 1| < 2,
∞
X
(z − 1)n
1
z
= −3
−
.
n+2
(z − 1)(z − 3)
2
2(z − 1)
n=0
Solution. We first write z/((z − 1)(z − 3)) as a sum of partial
fractions:
1
3
1
z
=
+
−
(z − 1)(z − 3)
2
z−1 z−3
When 0 < |z − 1| < 2, |(z − 1)/2| < 1 and hence
1
1
1
1
=−
=−
z−3
2 − (z − 1)
2 1 − (z − 1)/2
∞
∞
X
1 X (z − 1)n
(z − 1)n
=−
=
−
2 n=0
2n
2n+1
n=0
Therefore,
1
3
−
+
z−1 z−3
∞
3 X (z − 1)n
1
=−
−
2 n=0 2n+1
2(z − 1)
1
z
=
(z − 1)(z − 3)
2
= −3
∞
X
(z − 1)n
n=0
2n+2
−
1
2(z − 1)
6
(9) Write the two Laurent series in powers of z that represent the
function
1
f (z) =
z(4 + z 2 )
in certain domains and specify these domains.
Solution. We observe that f (z) is analytic in {z 6= 0, ±2i}.
Therefore, it is analytic in 0 < |z| < 2 and 2 < |z|. When
0 < |z| < 2,
n
∞ 1
1
1
1 X
z2
=
=
−
z(4 + z 2 )
4z 1 − (−z 2 /4)
4z n=0
4
∞
∞
1 X (−1)n 2n X (−1)n 2n−1
=
z =
z
4z n=0 4n
4n+1
n=0
∞
∞
X (−1)n
1 X (−1)n+1 2n+1
1
2n−1
=
+
z
+
z
=
4z n=1 4n+1
z n=0 4n+2
When 2 < |z| < ∞,
∞
1
1
1X
1
=
=
(−4z −2 )n
z(4 + z 2 )
z 3 1 − (−4/z 2 )
z n=0
∞
∞
X
1 X
n n −2n
(−1)n 4n z −2n−3
(−1) 4 z
=
= 3
z n=0
n=0
=
∞
X
(−1)n 4n
n=0
z 2n+3