Computing the residues of $\dfrac{1}{\sin^2 z}$.

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 ... – kiwi Mar 17 '12 at 23:50

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 Harald, this is indeed much easier. Is there a way to salvage the work I've done? Even if it's less elegant, that's the way I first begin to see it. – Hana Bailey Mar 18 '12 at 0:04
@Cowbell: if the answer is useful, which it is, upvote (+1). :-) – robjohn Mar 18 '12 at 0:25
@Cowbell: Sure, in your calculation, you may notice that $f_2$ must be an odd function, by considering the symmetry properties of the other terms in the equation defining $f_2$. In particular, $f_(0)=0$. – Harald Hanche-Olsen Mar 18 '12 at 0:35
From the defining equation $\sin z=\cos k\pi(z-k\pi)+f_2(z)(z-k\pi)^2$, I find that $f(z)+f(-z)=2k\pi\cos k\pi$, not zero. Is this the incorrect way to check if a complex valued function is odd? – Hana Bailey Mar 18 '12 at 0:58
@Cowbell: That gives you the right answer for $k=0$, anyhow. And for other values of $k$, you should consider even and odd symmetry around $z=k\pi$. So you should get $f_2(k\pi-z)=-f_2(k\pi+z)$. (But it is easier to just do it for $k=0$ and then let periodicity take care of the other cases.) – Harald Hanche-Olsen Mar 18 '12 at 10:15

$\sin(z)=z-z^3/6+\dots$ 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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Thanks, this was a nice simple way to look at it. – Hana Bailey Mar 18 '12 at 1:28
How do you lose the exponent $-2$ in the second row? – vladimirm Sep 16 '14 at 17:49
@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. – robjohn Sep 16 '14 at 18:11