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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?

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1  
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? –  Hurkyl 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

2 Answers 2

up vote 1 down vote accepted

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

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Related problems (I), (II) .

Hint:

$$ \frac{e^{z}}{(z+1)^2}= \frac{e^{(z+1)-1}}{(z+1)^2}=e^{-1}\frac{e^{z+1}}{(z+1)^2}. $$

Now, put $w=z+1$ and expand the exponential function to get the Laurent series.

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