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

Question

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?

In the title, is it $|z-\frac{1}{3}|<\frac{1}{12}$? – Mhenni Benghorbal Mar 31 '13 at 12:43 — Mhenni Benghorbal, Mar 31 '13 at 12:43

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

2

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

$$ \lim_{z \to \frac{1}{3}} \frac{\left(z-\frac{1}{3}\right)^{2}}{\sin^2\left(\frac{\pi}{z}\right)}=\frac{1}{81\pi^2} . $$

L'hopital's rule can be used twice to evaluate the above limit.

edited Apr 13 '17 at 12:20

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answered Mar 31 '13 at 12:55

Mhenni Benghorbal

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

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