How to find the Laurent series for $\frac1{\sin(z)}$ in the regions $0<|z|<\pi$ and $\pi<|z|<2\pi$?

Question

I know that

$$ \begin{align} \frac1{\sin(z)} &=\frac1z\frac{z}{\sin(z)}\\ &=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\ &=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\ &=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots \end{align} $$

but I am not sure how to continue to find the Laurent series for $\frac1{\sin(z)}$ in the regions $0<|z|<\pi$ and $\pi<|z|<2\pi$?

Answer 1

Since the power series for $\;\sin z\;$ around zero has infinite convergence radius we can use everywhere, thus: for $\;z\;$ close to zero, we have

$$\frac1{\sin z}=\frac1{z-\frac{z^3}6+\frac{z^5}{120}-\ldots}=\frac1{z\left(1-\frac{z^2}6+\mathcal O(z^4)\right)}\stackrel{\text{geometric series}}=$$

$$=\frac1z\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=\frac1z+\frac z6+\ldots$$

The above is the beginning of the Laurent series, just as you got. We usually are interested only in the first few terms: enough to know the multiplicity of the pole and the residue of the function there.

For $\;z=\pi\;$ you can do as above around $\;\pi\;$ using $\;\sin(z-\pi)=-\sin z\;$ (observe that then we get residue $\;-1\;$ , as expected...)

Answer 2

In the Laurent series for $1/\sin(z)$ valid in $0<|z|<\pi$, the formula for the coefficients involves Bernoulli numbers (LINK). The first few terms were found by the OP.
$$ \frac{1}{\sin z} = z^{-1}+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots\qquad 0<|z|<\pi . $$

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What about $\pi < |z| < 2\pi$? Jack D'Aurizio commented the method to use.

The residue of $1/\sin(z)$ at $z=\pi$ is $-1$. So $$ f_1(z) := \frac{1}{\sin z}+\frac{1}{z-\pi} $$ is analytic in $0 < |z| < 2\pi$. Its Laurent series, valid in $0<|z|<2\pi$ begins $$ f_1(z) = z^{-1} - \frac{1}{\pi} + \left(\frac{1}{6}-\frac{1}{\pi^2}\right) z -\frac{1}{\pi^3} z^2 + \cdots\qquad 0<|z|<2\pi . \tag1$$ Now note that $1/\sin(z) = f_1(z)-f_2(z)$ where $$ f_2(z) := \frac{1}{z-\pi} $$ A geometric series, valid in $|z|>\pi$ is $$ f_2(z) = z^{-1}+\pi z^{-2} + \pi^2 z^{-3} +\cdots\qquad |z|>\pi. \tag2$$ So subtract using $(1)$ and $(2)$ to get the series valid in $\pi < |z| < 2\pi$. $$ \frac{1}{\sin z} = \cdots + \pi^2 z^{-3} + \pi z^{-2} - \frac{1}{\pi} + \left(\frac{1}{6}-\frac{1}{\pi^2}\right) z -\frac{1}{\pi^3} z^2 + \cdots \qquad \pi < |z| < 2\pi . $$ It seems the $z^{-1}$ term is $0$.