Determining Laurent Series

Question

This question is from Complex Variables and Applications by Brown & Churchill, 8ed. Section 62, #2.

Determine the Laurent Expansion of:

$$\frac{\exp(z)}{(z+1)^2}$$

for the interval, $ 0 < |z + 1| < \infty$. How do I go about doing that?

Here's what I have so far. If I substitute $ w = (z + 1) \implies w^2 = (z+1)^2$ then I get

$$\begin{aligned} \exp(z) \sum_{k = 0}^{\infty} \frac{1}{w^2} &= \exp(z) \sum_{k=0}^{\infty}(-1+w)^n(-1)^n(1+n)\\ &= \exp(z)\sum_{k =0}^{\infty}(z^n)(-1)^n(1+n)\end{aligned}$$

But I don't know if this is the correct approach or if this will even yield a viable Laurent Series. Any ideas?

If you're looking for a Laurent series with $z$ centered at $-1$, then you would want the one with $w$ centered at $0$. Also, why not treat $\exp z$ the same way? — , Apr 10 '13 at 20:06
@Hurkyl This is related but for a different question, just so I understand. If I were to expand $\frac{z}{(z-1)(z-3)}$ about the region $0<|z-1|<2$, then I would convert: $\frac{3}{2}\frac{1}{z-3} - \frac{1}{2}\frac{1}{z-1} = \frac{3}{2}\frac{1}{(z-1)-2} - \frac{1}{2}\frac{1}{z-1}$ — franklin, Apr 10 '13 at 20:19

score 2 · Accepted Answer · edited Apr 21 '13 at 05:48

2

$$\frac{e^z}{(z+1)^2}=\frac{e^{z+1}}{e(z+1)^2}=\frac{1}{e(z+1)^2}\sum\limits_{n=0}^{\infty}\frac{(z+1)^n}{n!}=\frac{1}{e}\sum\limits_{n=0}^{\infty}\frac{(z+1)^{n-2}}{n!}$$

edited Apr 21 '13 at 05:48

franklin

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answered Apr 10 '13 at 19:57

M. Strochyk

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score 1 · Answer 2 · edited Apr 13 '17 at 12:21

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Determining Laurent Series

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