# Laurent expansion of $\csc^2(\frac{\pi}{z})$ about $\frac{1}{3}$ for $|z-\frac{1}{3}| \lt \frac{1}{12}$

The question is: Find the Laurent expansion of $\csc^2(\frac{\pi}{z})$ about $\frac{1}{3}$ for $|z-\frac{1}{3}| \lt \frac{1}{12}$. In particular what is the coefficient of $(z-\frac{1}{3})^{-2}$.

I am thinking on the following lines. Do the Taylor expansion of $\sin^2(\frac{\pi}{z})$, and then do the polynomial division $\frac{1}{\sin^2(\frac{\pi}{z})}$. Is this the right way? Is this the only way?

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In the title, is it $|z-\frac{1}{3}|<\frac{1}{12}$? –  Mhenni Benghorbal Mar 31 '13 at 12:43
Yeah! Thanks. I have edited it. –  user23238 Mar 31 '13 at 12:55

A related problem. To find the coefficient of $\left(z-\frac{1}{3}\right)^{-2}$, just evaluate the following limit
$$\lim_{z \to \frac{1}{3}} \frac{\left(z-\frac{1}{3}\right)^{2}}{\sin^2\left(\frac{\pi}{z}\right)}=\frac{1}{81\pi^2} .$$