Andrew Monnot
210B Homework
(Winter 2013)
Homework 1
5.1.1 Classify the isolated singularity at z = 0 for the following functions: (a) f (z) = sinz z ,
1
z 1 +1
(b) f (z) = cosz z , (d) f (z) = e1/z , (e) f (z) = ln(z+1)
, (h) f (z) = 1−e
, (g) f (z) = z(z−1)
z , (i)
z2
f (z) = z sin(1/z).
Proof. For this problem, we will find the Laurent series’ and apply Corollary 5.1.18.
(a) Here we have
∞
∞
1 X (−1)n z 2n+1 X (−1)n z 2n
sin z
=
=
.
z
z n=0 (2n + 1)!
(2n
+
1)!
n=0
Thus it has a removable singularity, and we may define f (0) = 1 to make it analytic (this
can also be seen by limiting z → 0 in the power series).
(b) Similarly we have
∞
∞
∞
X
1 X (−1)n z 2n X (−1)n z 2n−1
(−1)n+1 z 2n+1
cos z
=
=
=
.
z
z n=0 (2n)!
(2n)!
(2n
+
2)!
n=0
n=−1
So this has a pole of order 1, and its singular part is 1/z.
(d)
∞
0
X
X
1
zn
e1/z =
=
.
n n!
z
|n|!
n=0
n=−∞
So this one has an essential singularity at 0.
(e)
∞
∞
∞
X
(−1)n+1 z n
1 X (−1)n+1 z n X (−1)n+1 z n−2
ln(z + 1)
=
=
=
.
z2
z 2 n=1
n
n
n
+
2
n=1
n=−1
This has a pole of order 1 whose singular part is 1/z.
(g) The Laurent series about 0 is given by
∞
∞
∞
∞
X
X
X
z2 + 1 1
z2 + 1 X n
n+1
n−1
n+2
n
2
−
=−
z =−
z +z
=−
z +z = −(z +1)
zn,
z 1−z
z n=0
n=0
n=−1
n=−1
whose singular part is − z1 .
(h) Here we have a pole since limz→0 |1/(1 − ez )| = ∞, and it’s of order one (see super
duper online Taylor series calculator, which hopefully gave right answer) with singular
part −1/z.
(i) This series is
∞
∞
0
X
X
1
(−1)n (1/z)2n+1 X
(−1)n
(−1)n z 2n
z sin
=z
=
=
.
z
(2n + 1)!
z 2n (2n + 1)! n=−∞ (1 − 2n)!
n=0
n=0
1
Hence the singularity is essential.
1
. Find the Laurent series expansions on the annuli (0; 0, 1),
5.1.4 Let f (z) = z(z−1)(z−2)
(0; 1, 2), and (0; 2, ∞).
Proof. We get a partial fraction decomposition of
1
1
1
1
=
−
+
z(z − 1)(z − 2)
2z z − 1 2(z − 2)
1
1
1
=
+
−
2z 1 − z 2(2 − z)
1
1
1
=
+
−
2z 1 − z 4 1 − z2
For the first annulus, we have |z| < 1 and hence
1
1
1
+
−
2z 1 − z 4 1 −
∞
∞
X
1
1 X zn
n
=
+
z
−
z
2z n=0
4 n=0 2n
2
∞
X
1
zn
n
+
=
z − n+2
2z n=0
2
∞
X
1
1
=
+
1 − n+2 z n
2z n=0
2
∞ X
1
=
1 − n+2 z n .
2
n=−1
For the other two annuli...
5.1.5 Show that f (z) = tan z is analytic in C except at the simple poles z = π2 + nπ.
Find the singular part of f at these poles.
Proof. Since tan z = sin z/ cos z, it follows that this function is defined provided
π
cos z 6= 0. Thus the singularities are S = 2 + nπ . Moreover, on any annulus centered
on these singularities and not including another singularity, or on any other open set in
C − S, tan z is differentiable and hence analytic by Goursat’s theorem. The union of
all such annuli together with other open sets in C − S clearly covers C − S, so tan z is
analytic on C − S.
−1
Since the poles are simple, the singular part is a−1 z − π2 + nπ
. We use the rule
π
on pg. 113 to compute a−1 . So if a = 2 + nπ for some n, we have
(z − a) sin z
sin z + (z − a) cos z
= lim
= −1.
z→a
z→a
cos z
− sin z
a−1 = lim (z − a) tan z = lim
z→a
Thus the singular part is −(z − a)−1 .
5.1.6 If f : G → C is analytic except for poles, show that the poles of f cannot have a
limit point in G.
Proof. By definition, a is a pole iff limz→a |f (z)| = ∞. Since f is otherwise analytic,
it follows that the function g(z) = 1/f (z) is analytic everywhere in G except at the zeros
of f, and has zeros equal to the poles of f. But note by Theorem 4.3.7 that the set of zeros
2
of g has a limit point in G iff g = 0 everywhere. But this would imply that f is undefined
everywhere, which is a contradiction. Thus the set of zeros of g, and equivalently the set
of poles of f, has no limit point in G.
5.1.11 Find the Laurent series of f (z) = e1/z . Can this be generalized?
Proof. This function is analytic outside
0, so we give the Laurent series at 0,
P of z =
zn
. Can we generalize? What ?
as we did in 5.1.1(d) (see above): e1/z = 0n=−∞ |n|!
5.1.13ab Let R > 0 and G = {z : |z| > R}. f : G → C is said to have a singularity of
type X at z = ∞ iff f (z −1 ) has a singularity of type X at z = 0.
(a) An entire function has a removable singularity at ∞ iff f is constant.
(b) An entire function has a pole at ∞ of order m iff it is a polynomial of degree m.
Proof. (a) Suppose f is entire with removable singularity at z = ∞. Then f (z −1 )
has a removable singularity at z = 0, which means
lim zf (z −1 ) = 0.
z→0
Let f (z) =
P∞
n=0
an z n be the Taylor series of f centered at z = 0. Then
−1
0 = lim zf (z ) = lim z
z→0
z→0
∞
X
an
n=0
= lim a0 z + a1 +
zn
z→0
a2
+ ··· .
z
Thus an = 0 for n ≥ 1. So f (z) = a0 is constant.
Conversely, suppose f (z) = c. Then
lim zf (z −1 ) = lim zc = 0,
z→0
z→0
so f has a removable singularity at z = ∞.
(b) Suppose f is entire and has a poleP
of order m at z = ∞. Then f (z −1 ) has a pole
n
of order m at z = 0. As before, if f (z) = ∞
n=0 an z is the Taylor series of f centered at
z = 0, then we have
−1
f (z ) =
∞
X
an
n=0
zn
0
X
=
n
a−n z =
n=−∞
0
X
a−n z n
n=−m
where the last equality
from the fact that f (z −1 ) has a pole at 0 (say, of order
Pm follows
n
m). Thus f (z) = n=0 aP
n z and is hence a polynomial.
m
Conversely if f (z) = n=0 an z n , then obviously f (z −1 ) has a pole of order m at z = 0
as desired.
5.1.14 Let G = {z : 0 < |z| < 1} and f : G → C be analytic. Suppose
γ is a closed
R
rectifiable curve in G such that n(γ, a) = 0 for all a ∈ C − G. What is γ f ?
Proof. It’s analytic in the annulus G and thus admits a Laurent series expansion
centered at 0:
∞
X
f (z) =
an z n .
n=−∞
Hence
Z
Z
f (z) dz =
γ
∞
X
γ n=−∞
n
an z dz =
−2
X
n=−∞
Z
n
an
z dz +
γ
3
∞
X
n=−1
Z
an
n
z dz =
γ
∞
X
n=−1
Z
an
γ
z n dz
since
R
γ
z n dz = 0 for n ≤ −2 (problem 4.5.5, pg. 87). Now we have
∞ Z
X
n=−1
Z
n
an z dz =
γ
γ
a−1 dz
+
z
Z X
∞
an z n .
γ n=0
However since 0 ∈ C − G, we have n(γ, 0) = 0 and hence
Z
a−1 dz
= 2πia−1 n(γ, 0) = 0.
z
γ
And lastly, the positive part of f ’s Laurent series clearly has a primitive in G, so by
Theorem 4.1.18 (pg. 65),
Z X
∞
an z n dz = 0.
γ n=0
Thus
R
γ
f (z) dz = 0.
4