MA30056: Complex Analysis

Exercise Sheet 9: Laurent series & Singularities

Please hand solutions in at the lecture on Monday 27th April.
1.) Let p be a non-constant polynomial. Show: |p(z)| as |z| .
Optional and not examinable: Regarding p as a function from the extended com-
plex plane

C = C{} to C, what are we showing here in terms of singularities?
Do other entire functions like exp or sin also have this property?
Solution: We have actually already shown this in the proof of Gauss Theorem
(see Exercise sheet 6 Question 4 or Exercise sheet 7 Question 1). Let p(z) =

n
k=0
a
k
z
k
with a
n
= 0. Then, since
p(z)
a
n
z
n
= 1 +
a
n1
a
n
1
z
+ +
a
0
a
n
1
z
n
1 as
|z| , there is R > 0 so that
|z| > R

p(z)
a
n
z
n

1
2
.
Thus, |p(z)|
1
2
|a
n
| |z|
n
for |z| > R and the claim follows.
Alternative: Using the reverse triangle inequality, one obtains
|p(z)| =

k=0
a
k
z
k

|a
n
| |z|
n

n1

k=0
a
k
z
k

.
Setting M = max{|a
0
|, |a
1
|, . . . , |a
n1
|}, one obtains for the second term (using the
triangle inequality) for |z| > 1:

n1

k=0
a
k
z
k

n1

k=0
|a
k
| |z|
k
M
n1

k=0
|z|
k
M n |z|
n1
.
Thus, whenever |z| > max {1, M n/|a
n
|}, we have
|p(z)| |a
n
|
_
|z|
M n
|a
n
|
_
|z|
n1
and |p(z)| as |z| .
Remark: We can also uses the characterisation of singularities at the point of
innity (i.e., in the extended complex plane

C = C{}): A function f(z) has
a removable singularity/pole/essential singularity at i f(1/z) has a removable
singularity/pole/essential singularity at 0.
Please turn over!
Using this or the characterisation in Theorem V.1.1, it is easy to see that a non-
constant polynomial has a pole at . In fact, one can also show the converse: An
entire function is a polynomial i it has a pole (or, if it is constant, a removable
singularity) at . Entire functions such as the exponential function, sine, cosine
are also called (entire) transcendental function. They have an essential singularity
at (compare Theorem V.1.1), see the Question 1 on Exercise sheet 10 or the
examples in the lecture.
2.) Find the Laurent series expansions of f(z) =
1
z(1z)(2z)
for the annuli
(i) 0 < |z| < 1, (ii) 1 < |z| < 2, (iii) 2 < |z|.
Hint: Do not compute integrals.
Solution: We write f(z) =
1
z
_
1
1z

1
2(1
z
2
)
_
and use geometric series.
0 < |z| < 1: In this case |z| < 1 and
|z|
2
< 1 so that
f(z) =
1
z
_

k=0
z
k

1
2

k=0
_
z
2
_
k
_
=

k=1
(1 2
(k+2)
) z
k
.
1 < |z| < 2: In this case
1
|z|
< 1 and
|z|
2
< 1 so that
f(z) =
1
z
2
1
1
1
z

1
2z
1
1
z
2
=
1
z
2

k=0
1
z
k

1
2z

k=0
_
z
2
_
k
=
2

k=
z
k

k=1
2
(k+2)
z
k
.
2 < |z|: Now
1
|z|
< 1 and
2
|z|
< 1 and we get
f(z) =
1
z
2
_
1
1
1
z

1
1
2
z
_
=
1
z
2
_

k=0
_
2
z
_
k

k=0
1
z
k
_
=
2

k=
(2
(k+2)
1) z
k
.
Note how the Laurent series expansions of the very same function are dierent for
dierent annuli!
Note: you can also use the partial fraction expansion
f(z) =
1
2z
+
1
1 z
+
1
2(z 2)
.
Please turn over!
More precisely, you may use
f(z) =
1
2
1
z
+
1
1 z

1
4
1
(1
z
2
)
=
1
2
1
z

1
z
1
1
1
z

1
4
1
(1
z
2
)
=
1
2
1
z

1
z
1
1
1
z
+
1
2z
1
(1
2
z
)
for parts (i), (ii), (iii), respectively.
3.) Find the Laurent series expansions of f(z) =
1
z
+
1
1z
+
1
2z
for the annuli
(i) 0 < |z| < 1, (ii) 0 < |z 1| < 1, (iii) 0 < |z 2| < 1.
Hint: Do not compute integrals.
Solution: We use geometric series throughout, rewriting the function suitably.
0 < |z| < 1: Write f(z) =
1
z
+
1
1z
+
1
2
1
1
z
2
; since |z|, |
z
2
| < 1 we get
f(z) =
1
z
+

k=0
z
k
+
1
2

k=0
z
k
2
k
=
1
z
+

k=0
_
1 +
1
2
k+1
_
z
k
.
0 < |z1| < 1: Write f(z) =
1
1(1z)

1
z1
+
1
1(z1)
; since |z1| = |1z| < 1
we get
f(z) =

k=0
(1 z)
k

1
z 1
+

k=0
(z 1)
k
=
1
z 1
+

j=0
2(z 1)
2j
.
0 < |z 2| < 1: Write f(z) =
1
2
1
1
(2z)
2

1
1(2z)

1
z2
; since |z 2|, |
z2
2
| < 1
we get
f(z) =
1
2

k=0
(2 z)
k
2
k

k=0
(2z)
k

1
z 2
=
1
z 2

k=0
(1)
k
_
1
1
2
k+1
_
(z2)
k
.
What are the (nonzero) residues of f?
4.) Let f : D C be holomorphic. We say that z
0
D is a zero of order m N of f
if the Taylor series expansion of f at z
0
f(z) =

k=m
a
k
(z z
0
)
k
, where a
m
= 0.
Prove that z
0
D is a zero of order m N i there is a holomorphic function g
with g(z
0
) = 0 so that f(z) = (z z
0
)
m
g(z).
Conclude that the zeros of a nonzero holomorphic function are isolated.
Solution: This is pretty much a book keeping exercise.
First assume that z
0
D is a zero of order m of f, that is, its Taylor expansion
starts with the m-th term:
f(z) =

k=m
a
k
(z z
0
)
k
.
The radius of convergence R of the Taylor series satises R > 0, where
B

(z
0
) D (Cauchy-Taylor Theorem). Thus
g(z) =
1
(z z
0
)
m

k=m
a
k
(z z
0
)
k
=

k=0
a
k+m
(z z
0
)
k
denes, as a power series with radius of convergence R > 0, a holomorphic function
on some B
R
(z
0
) (by Theorem IV.2.2).
Conversely, if g is holomorphic in some B

(z
0
), > 0, with g(z
0
) = 0 then
g(z) =

k=0
b
k
(z z
0
)
k
on B

(z
0
), where b
0
= 0,
by the Cauchy-Taylor Theorem. Now, if f(z) = (z z
0
)
m
g(z), then the Taylor
expansion of f at z
0

k=0
a
k
(z z
0
)
k
= f(z) = (z z
0
)
m

k=0
b
k
(z z
0
)
k
so that a
k
= 0 for k = 0, . . . , m1 and a
m
= b
0
= 0 since the Taylor expansion is
unique (see the Identity Theorem for power series).
Finally (to show that z
0
is isolated), if f(z) = (z z
0
)
m
g(z) with g(z
0
) = 0, g
holomorphic around z
0
, then there is R > 0 so that g(z) = 0 for z B
R
(z
0
) (why
can we assume this?). Thus f(z) = 0 on B
R
(z
0
) \ {z
0
}. (The last part can also be
proven from the Identity Theorem for power series by contraposition: Suppose z
0
is not isolated, then show that the function f is the zero-function; so z
0
is not a
zero of order m).
Optional question:
5.) Convince yourself that z
1
sin z
is meromorphic in C, i.e., it is holomorphic in C
except for poles.
Solution: Clearly
1
sin z
is holomorphic on
D = {z C| sin z = 0} = {n | n Z},
and all singularities n of
1
sin z
are isolated. Since sin is periodic,
sin(z + n) = (1)
n
sin z,
it suces to investigate the function at z = 0. Since
lim
z0
z
sin z
= lim
z0
1
1
1
6
z
2
+
1
120
z
4
. . .
= 1
the function B

(0) z g(z) =
z
sin z
extends holomorphically to B

(0) (e.g., by
Theorem V.1.1(i) and hence
1
sin z
=
1
z
g(z) =

k=1
a
k+1
z
k
on B

(0), where g(z) =

k=0
a
k
z
k
is the Taylor expansion of g(z) on B

(0).
Consequently,
1
sin z
has a simple pole at z = 0 and therefore, at z = n, n Z.