Consider
the function: f(z) = 1/ (z+ 1) ((z + 3)

a)
Find a Laurent series for:
1 <

**÷**z**÷****< 3**
Solution: We first resolve the function f(z) by
partial fractions, so:

f(z)
= 1/ (z+ 1) ((z + 3) = ½ [1/ z + 1]
- ½ [1/ z + 3]

**÷**z

**÷**

**> 1:**

1/ 2

**÷**z + 1**÷****= 1/ 2z (1 + 1/z)**
= 1/ 2z [ 1 – 1/z + 1/ z

^{2}+ …….]
This
is the principal part of the series.

Next,
consider:

**÷**z**÷****< 3:**
1/
2 (z + 3) = 1/ 6(1 + z/3) = 1/6 -
z/ 18 + z

^{2}/54 + ……
This
is the analytic part of the series. So we just combine the two parts to get:

f(z)
= 1/ 2z [ 1 – 1/z + 1/ z

^{2}+ …….] + 1/6 - z/ 18 + z^{2}/54 + ……
b) Laurent series for 0 <

**÷**z + 1**÷****< 2**
Consider
first:

**÷**z + 1**÷****> 0**
We
let (z + 1) = u then write:

1/
(z+ 1) ((z + 3) = 1/ u (u + 2) – 1/ 2u (1 + u/2)

=
1/ 2u (1 – u/2 + u

^{2}/4 - u^{3 }/8 + ……)
Replace
u with z above:

1
/ 2(z + 1) - ¼ + (z – 1)/ 8 - (z
+ 1)

^{ 2}/ 16 + ……
Now
take:

**÷**z + 1

**÷**< 2 or

**÷**u

**÷**< 2 (letting z + 1 = u)

Then
for the same series above, since

**÷**z + 1**÷****< 2 we require: z ¹ -1**
WHY?

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