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The following question came up in a past year paper.

Question. Consider the function on $\mathbb{C} \setminus \pi\mathbb{Z}$

$$f(z)=\frac{1}{\sin z}$$

Let the Laurent series of $f$ on $A=\{z \mid 0 < |z| < \pi\}$ be

$$c_{-1}z^{-1}+\sum_{n=0}^{\infty}c_nz^n$$

and let the Laurent series of $f$ on $B=\{z \mid \pi < |z| < 2\pi\}$ be

$$\sum_{n=-\infty}^{\infty}d_nz^n.$$

By considering expressions for $c_n$ and $d_n$ as contour integrals around circles with radii $0 < R_1< \pi < R_2 < 2\pi$, express the coefficient $d_n$ in terms of $c_n$ for $n \geq -1$, and obtain an explicit formula for $d_n$ for $n \leq 1$.

My attempt.

$$c_n = \frac{1}{2 \pi i}\oint_{\partial B(0,R_1)}\frac{f(w)}{w^{n+1}}dw,$$

$$d_n = \frac{1}{2 \pi i}\oint_{\partial B(0,R_2)}\frac{f(w)}{w^{n+1}}dw$$

and I am stuck. I don't think that the two integrals have the same limit as $R_1 \rightarrow 1^-$ and as $R_2 \rightarrow 1^+$ because of the pole at $\pm \pi$.

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Here is my solution after a week of thinking.

Let $r=R_1, R = R_2$, make a "cut" in the annulus with radii $0<r<R$, then we see by Cauchy Residue Formula that

\begin{eqnarray} d_n - c_n &=& \frac{1}{2\pi i} \oint_{\partial B(0,R)} \frac{f(w)dw}{w^{n+1}} -\oint_{\partial B(0,r)} \frac{f(w)dw}{w^{n+1}} \\ &=& \sum_{s \text{ singularity : } R_1 < |s| < R_2}\text{Res}(\frac{f(w)}{w^{n+1}},s) \\ &=& \text{Res}(\frac{1}{\sin(w)w^{n+1}},\pi) + \text{Res}(\frac{1}{\sin(w)w^{n+1}},-\pi) \\ &=& \frac{1}{\pi^{n+1}}(-1+(-1)^n). \end{eqnarray}

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