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What is the simplest way of calculating the Residue of $$\frac{e^z-1}{1-\cos z}$$

I know this function has simple pole at $z=2n\pi$ but I just need $z=0$. I can find the residue by expanding in Laurent series. I have got that but , do we have any other simplest way?

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Try calculating $$\lim_{z\to0}zf(z)= \Bigl(\lim_{z\to0}\frac{z^2}{1-\cos z}\Bigr) \Bigl(\lim_{z\to0}\frac{e^z-1}{z}\Bigr)$$ by using L'Hopital's rule for each limit on the RHS.

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  • $\begingroup$ This idea works. Thanks $\endgroup$ – user178061 Oct 16 '14 at 5:19
  • $\begingroup$ You're welcome Santosh. BTW just noticed - if $n\ne0$, your function doesn't have a simple pole at $z=2n\pi$, it's a double pole. Perhaps the function was supposed to be $(e^{iz}-1)/(1-\cos z)$? $\endgroup$ – David Oct 16 '14 at 6:21

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