Laurent series of $\dfrac{1}{\sin(\frac{1}{z})}$

Question

What is the Laurent series for the function $f(z)=\dfrac{1}{\sin(\frac{1}{z})}$ at the point $z=\dfrac{1}{\pi}$ and $z=0$?

Answer 1

The Laurent series of $f(z)=\dfrac{1}{\sin(\frac{1}{z})}$ at $z=\dfrac{1}{\pi}$. \begin{align} f(z)&=-\dfrac{1}{\sin\left(\dfrac{1}{z}-\pi\right)}=\frac{1}{\sin\left(\dfrac{\pi^2\zeta}{\pi\zeta+1}\right)}\quad (\text{where}\quad \zeta=z-\frac{1}{\pi})\\ &=\dfrac{\pi\zeta+1}{\pi^2\zeta}+\frac{1}{6}\cdot \dfrac{\pi^2\zeta}{\pi\zeta+1}+\frac{7}{360}\cdot\left(\dfrac{\pi^2\zeta}{\pi\zeta+1}\right)^3+O(\zeta^5), \end{align} since $$ \frac{1}{\sin x}=\frac{1}{x}+\frac{x}{6}+\frac{7}{360}x^3+O(x^5)$$ around $x=0$.
Since$$ \dfrac{\pi^2\zeta}{\pi\zeta+1}=\pi^2\zeta\left(1-\pi\zeta+\pi^2\zeta^2-\pi^3\zeta^3+O(\zeta^4)\right) $$ and $$ \left(\dfrac{\pi^2\zeta}{\pi\zeta+1}\right)^3=\pi^6\zeta^3\left(1-\pi\zeta+o(\zeta^2) \right)^3=\pi^6\zeta^3\left(1-3\pi\zeta+O(\zeta^2)\right), $$ we have \begin{align} f(z)&=\frac{1}{\pi^2\zeta}+\frac{1}{\pi}+\frac{\pi^2}{6}\zeta-\frac{\pi^3}{6}\zeta^2+\frac{\pi^4(60+7\pi^2}{360}\zeta^3-\frac{\pi^5(20+7\pi^2}{120}\zeta^4+O(\zeta^5)\\ &=\frac{1}{\pi^2\left(z-\frac{1}{\pi}\right)}+\frac{1}{\pi}+\frac{\pi^2}{6}\left(z-\frac{1}{\pi}\right)-\frac{\pi^3}{6}\left(z-\frac{1}{\pi}\right)^2\\ &\quad +\frac{\pi^4(60+7\pi^2)}{360}\left(z-\frac{1}{\pi}\right)^3-\frac{\pi^5(20+7\pi^2)}{120}\left(z-\frac{1}{\pi}\right)^4+O\left(\left(z-\frac{1}{\pi}\right)^5\right). \end{align}

At $z=0$.

For $z=\frac{1}{n\pi} \,( n \in \mathbb{Z},n\ne 0)$, $\sin \left(\frac{1}{z}\right)=0.$ Therefore $f(z)$ has a pole at $z= \frac{1}{n\pi}.$
Since these points accumulate at $z=0$, $z=0$ is not an isolated singularity. This means that $f(z)$ does NOT have it's Laurent expansion at $z=0.$

Answer 2

About $z=0$:

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

Answer 3

At $z=\frac{1}{π}$, the expansion is:

$\frac{1}{π^2(z-\frac{1}{π})}+\frac{1}{π}+\frac{1}{6}π^2(z-\frac{1}{π})-\frac{1}{6}π^3(z-\frac{1}{π})^2+\frac{1}{360}π^4(60+7π^2)(z-\frac{1}{π})^3-\frac{1}{120}(π^5(20+7π^2))(z-\frac{1}{π})^4+O((z-\frac{1}{π})^5)$

At $z=0$ we note that $\frac{1}{\sin(z)}=\csc(z)$ and take the expansion...never mind, see Ron Gordon's answer.

Answer 4

There are many Laurent expansions around $z=0$, one for $|z|>\frac1{\pi}$ and one for each annulus $$\frac1{(n+1)\pi}<|z|<\frac1{n\pi}.$$

Start with the identity $$\frac1{\sin(z)}=\frac1z+\sum_{k=1}^{\infty}\frac{(-1)^k\,2z}{z^2-k^2\pi^2}$$ and substitute $z\leftarrow \frac1z$ to get $$\frac1{\sin(\frac1z)}=z+\sum_{k=1}^{\infty}\frac{(-1)^k\,2z}{1-k^2\pi^2z^2}.$$ Now determine the Laurent expansion of each summand for any of the annuli indicated above. For example if $|z|>\frac1{\pi}$ then for all $k \geq 1$ $$\frac{2z}{1-k^2\pi^2z^2}=\frac{-2}{k^2\pi^2 z}\cdot\frac1{1-\frac1{k^2\pi^2z^2}}=-2z\sum_{m=1}^{\infty}\left(k \pi z\right)^{-2m}.$$

For $|z|>\frac1{\pi}$ this gives the Laurent series $$\frac1{\sin(\frac1z)}=z+\sum_{m=1}^{\infty}\frac{2-4^{1-m}}{\pi^{2m}}\zeta(2m)z^{1-2m}.$$