Math 210B Homework 1
Edward Burkard
Problem 1. Prove Goursat’s Theorem using rectangles instead of triangles.
Proof.
Problem 2. Each of the following functions f has an isolated singularity at z = 0. Determine its nature; if it is
a removable singularity define f (0) so that f is analytic at z = 0; if it is a pole, find the singular part; if it is an
essential singularity, determine f ({z | 0 < |z| < δ} for arbitrarily small values of δ.
sin z
(a) f (z) =
z
cos z
(b) f (z) =
z
(c) f (z) =
cos z − 1
z
1
(d) f (z) = e z
(e) f (z) =
(f) f (z) =
log (z + 1)
z2
cos z1
1
z
(g) f (z) =
z2 + 1
z(z − 1)
(h) f (z) =
1
1 − ez
(i) f (z) = z sin
1
z
(j) f (z) = z n sin
1
z
Solution.
Problem 3. Let f (z) =
1
. Find the Laurent series of f (z) in
z(z − 1)(z − 2)
(a) ann0 (0, 1);
(b) ann0 (1, 2);
(c) ann0 (2, ∞).
Solution.
Problem 4. Prove that if f : G → C, G a region, is analytic except for poles, then the set of poles of f cannot have
a limit point in G.
Proof. Let P be the set of poles of f and suppose that P has a limit point in G. If the limit point is a pole, then there
is no punctured disk around the limit point on which f is analytic since each punctured disk will contain infinitely
many poles, contradicting the fact that the limit point is an isolated singularity. If the limit point is not a pole, then
f should be analytic at that point, but it cannot be since any neighborhood of that point contains infinitely many
poles, contradicting the fact that f is analytic. Therefore P cannot have any limit points in G.
1
2
Problem 5.
1
about the origin.
(z − 1)(3 − z)
1
about z = i.
(b) Find all possible Laurent series of f (z) =
1+z
1
(c) Find all possible Laurent series of f (z) =
about
z(1 − 2z)
(i) the origin,
1
(ii) z = .
2
Solution.
(a) Find all possible Laurent series of f (z) =
Problem 6. Find the Laurent series of the function f (z) = log
z−1
z+1
around the origin of the disk.
Solution.
Problem 7.
(a) Prove that an entire function has a removable singularity at infinity iff it is a constant.
(b) Prove that an entire function has a pole of order m at infinity iff it is a polynomial of degree m.
Proof.
(a)
∞
X
an z n . Since f has a removable
(=⇒) Let f : C → C be entire. Then f has a power series expansion f (z) =
n=0
1
singularity at infinity, we have that f
has a removable singularity at z = 0. Thus by Corollary 1.18
z
1
of Chapter 5, the coefficients in the Laurent series of f
are zero for all negative powers of z. The
z
X
∞
0
X
1
1
is f
=
an z −n =
Laurent series of f
a−m z m . Hence a−m = an = 0 for all m < 0
z
z
m=−∞
n=0
(or n > 0). Therefore f must be a constant function, namely f (z) = a0 .
(⇐=) Let f (z) = c be a constant function. Then
itis entire and is its own power series expansion. f has a
1
removable singularity at infinity if lim zf
= 0. Observe:
z→0
z
1
= lim zc = 0,
lim zf
z→0
z→0
z
hence f (z) has a removable singularity at infinity.
(b)
∞
X
(=⇒) Let f : C → C be an entire function. Then f has a power series expansion f (z) =
an z n . If f has
n=0
1
a pole of order m at infintity, then f
has a pole of order m at z = 0. Since the Laurent series
z
0
X
1
1
of f
is f
=
a−k z k , by Corollary 1.18 of Chapter 5, it must be that a−k = 0 for all
z
z
k=−∞
−k < −m or k > m. Hence f (z) = a0 + a1 z + · · · + am z m .
(⇐=) Let f (z) = a0 + a1 z + · · · + am z m be a polynomial
of degree m. Then it is entire and is its own power
1
series expansion. Since the Laurent series of f
is f z1 = am z −m +am−1 z −(m−1) +· · ·+a1 z −1 +a0 ,
z
1
by Corollary 1.18 of Chapter 5, f
has a pole of order m at z = 0, hence f (z) has a pole of order
z
m at infinity.