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I'm trying to find the residue at $0$ of the function

$$f(z)=\frac{1+iz-e^{iz}}{z^3}$$

on $\mathbb{C} - \{0\}$.

I think it's a double pole at the origin, but I'm not entirely sure. I'm wondering if it's best to try and find the Laurent expansion of the function on a suitable annulus, or whether there's a neater trick to find the residue.

Thanks in advance.

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$e^{i z} = 1+i z-\tfrac{z^2}{2} + O(z^3)$. –  WimC May 27 '12 at 18:16
    
Thanks! I see it now. –  Mathmo May 27 '12 at 18:22
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1 Answer 1

up vote 1 down vote accepted

Wim's hint is huge: you can use MacClaurin's series for the exponential function and write $$\frac{1+iz-e^{iz}}{z^3}=\frac{1}{z^3}+\frac{i}{z^2}-\frac{1}{z^3}\left(1+iz+\frac{(iz)^2}{2!}+...\right)=\frac{1}{2z}+...$$

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