Complex Analysis - Maths KU

The coefficients defined by (8) are
seldom used. See (ii) inthe
Remarks at the end of this section.
y
y
0 1
(a)
0 1
(c)
x
x
0
y
y
x
0 1
(b)
1
(d)
Figure 6.8 Annular domains for
Example 2
x
☞
6.3 Laurent Series 329
The integral formula in (8) for the coefficients of a Laurent series are
rarely used in actual practice.As a consequence, finding the Laurent series of
a function in a specified annular domain is generally not an easy task.But this
is not as disheartening as it might seem.In many instances we can obtain a
desired Laurent series either by employing a known power series expansion of
a function (as we did in Example 1) or by creative manipulation of geometric
series (as we did in Example 2 of Section 6.2). The next example once again
illustrates the use of geometric series.
EXAMPLE 2 Four Laurent Expansions
Expand f(z) =
domains.
1
in a Laurent series valid for the following annular
z(z − 1)
(a) 0< |z| < 1 (b) 1< |z| (c) 0< |z − 1| < 1 (d) 1< |z − 1|
Solution The four specified annular domains are shown in Figure 6.8. The
black dots in each figure represent the two isolated singularities, z = 0 and
z =1,off.In parts (a) and (b) we want to represent f in a series involving
only negative and nonnegative integer powers of z, whereas in parts (c) and
(d) we want to represent f in a series involving negative and nonnegative
integer powers of z − 1.
(a) By writing
f(z) =− 1
z
1
1 − z ,
we can use (6) of Section 6.1 to write 1/(1 − z) as a series:
f(z) =− 1 � � 2 3
1+z + z + z + ··· .
z
The infinite series in the brackets converges for |z| < 1, but after we
multiply this expression by 1/z, the resulting series
f(z) =− 1
z − 1 − z − z2 − z 3 −···
converges for 0 < |z| < 1.
(b) To obtain a series that converges for 1 < |z|, we start by constructing a
series that converges for |1/z| < 1. To this end we write the given function
f as
f(z) = 1
z2 1
1 − 1
z
and again use (6) of Section 6.1 with z replaced by 1/z:
f(z) = 1
z2 �
1+ 1
�
1 1
+ + +... .
z z2 z3