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/ z2 + …….]
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 + z2 /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/ z2 + …….] + 1/6 -
z/ 18 + z2 /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 + u2/4 - u3
/8 + ……)
Replace
u with z above:
1
/ 2(z + 1) - ¼ + (z – 1)/ 8 - (z
+ 1) 2 / 16 + ……
Now
take:
Then
for the same series above, since ÷ z + 1÷ <
2 we require: z ¹ -1
WHY?
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