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

Determine the Laurent Expansion of:

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

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

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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