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I'm having a hard time correctly computing the residues of $\dfrac{1}{\sin^2 z}$. I know the poles occur at $k\pi$, with order $2$.

By a Taylor expansion I can rewrite $\sin z=\cos k\pi(z-k\pi)+f_2(z)(z-k\pi)^2$, and so $ \sin^2 z=(z-k\pi)^2(\cos k\pi+f_2(z)(z-k\pi))^2. $ I want to calculate the residue with Cauchy's Integral Theorem, so $ \text{Res}(f,k\pi)=\frac{1}{2\pi i}\int_{|z-k\pi|=1}\frac{dz}{(z-k\pi)^2[\cos k\pi +f_2(z)(z-k\pi)]^2}. $ This should equal the derivative of $(\cos k\pi+f_2(z)(z-k\pi))^{-2}$ evaluated at $k \pi$. The derivative comes out to be -2(\cos k\pi+f_2(z)(z-k\pi))^{-3}(f'_2(z)(z-k\pi)+f_2(z)) and evaluates to $\dfrac{-2f_2(k\pi)}{(\cos k\pi)^3}$. Apparently the residue should just be $0$, but I don't see how to conclude this. What am I missing to know $f_2(k\pi)=0$?

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    Since $\sin^2z$ has period $\pi$ the residue at $k\pi$ is the same for all $k$. But the function is even so the residue at $z=0$ is ...2012-03-17

2 Answers 2

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You're working too hard. Do the residue at $z=0$ first: $1/\sin^2 z$ is an even function, so its series expansion involves only even power of $z$. Or, if you wish, you can see that $\int_C \frac1{\sin^2z}\,dz=0$ for a small circle $C$ centered at the origin, by noticing that the integrand is the same on opposite points of the circle, while $dz$ on opposite sides are each other's negatives.

The residues at the other poles follow by periodicity.

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    Thanks, this has been very helpful.2012-03-19
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$\sin(z)=z-z^3/6+\dots$ so $ \begin{align} \frac{1}{\sin^2(z)} &=\frac{1}{z^2}\left(1-z^2/6+\dots\right)^{-2}\\ &=\frac{1}{z^2}\left(1+z^2/3+\dots\right)\\ &=\frac{1}{z^2}+\frac13+\dots \end{align} $ Thus, the residue at $z=0$ is $0$ since there is no $\dfrac{1}{z}$ term. The others follow by periodicity.

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    @vladimirm: There are several ways to show that $\left(1+x+o(x)\right)^{-2}=1-2x+o(x)$ one of the easiest is the [generalized binomial theorem](http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem).2014-09-16