# Why does $\sin\left(\frac{\pi}{z^2}\right)$ not have a Laurent series at $z=0$?

I tried to compute the Laurent series of $$\sin\left(\frac{\pi}{z^2}\right)$$ at $$z=0$$, then I realized something was off altogether and there simply doesn't exist a Laurent espansion centered at $$z=0$$ for that function (Mathematica says "no series expansion available"). I thought of a possible explanation, could you tell me if it makes sense? If $$C_{r,R}$$ is the annulus of internal radius $$r$$ and external radius $$R$$ centered at $$z=0$$.

$$\sin{\frac{\pi}{z^2}}=\sum_{k=0}^{\infty}{\frac{(-1)^k \pi^{2k+1}}{(2k+1)!}(z-0)^{-4k-2}},\forall z\in C_{r,R},\space where\space \frac{1}{R}=\lim_{k\rightarrow\infty}{\sup\left|\frac{{(-1)^k \pi^{2k+1}}}{(2k+1)!}\right|^{1/k}=0}\space and\space r=\lim_{k\rightarrow\infty}{\sup\left|\frac{{(-1)^-k \pi^{-2k+1}}}{(-2k+1)!}\right|^{1/k}=0}$$ $$\iff$$ $$f\space holomorphic\space in\space C_{r,R}=C_{0,\infty}\space and\space\forall \space closed\space path\space \gamma\in C_{0,\infty},\space c_{k}=\frac{(-1)^k\pi^{2k+1}}{(2k+1)!}=\frac{1}{2\pi i}\oint_{\gamma}{\frac{\sin\left({\frac{\pi}{z^2}}\right)}{(z-0)^{k+1}}dz}$$ though if I compute the coefficients $$c_k$$ on a $$\gamma$$ unitary circle centered at $$z=0$$: $$\frac{1}{2\pi i}\oint_{\gamma}{\frac{\sin\left({\frac{\pi}{z^2}}\right)}{(z-0)^{k+1}}dz}=\frac{1}{2\pi}\int_{0}^{2\pi}{\frac{\sin\left(\pi e^{-2 i \theta}\right)}{e^{i\theta k}}d\theta}=0\neq\frac{(-1)^k\pi^{2k+1}}{(2k+1)!}, \forall k\in\mathbb{Z}$$, so the proposition on the left of the $$\iff$$ can't be true.

Or said in another way, if the expansion exists it is unique, but then it would mean $$f$$ is zero everywhere which is obviously not true, so the expansion doesn't exist.

P.S. I'm sorry about poor formatting, if someone improves it I promise I'll look at the corrections and learn form them :)

It does have a Laurent series. Just render the usual Maclaurin series for $$\sin u$$ and then put $$u:=\pi/z^2$$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $$z=0$$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
• sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $\frac{\sin\left(\frac{\pi}{z^2}\right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists. – EugenioDiPaola Jan 19 at 21:18
Ok, found the mistakes, obviously I was wrong, $$\sin\left({\frac{\pi}{z^2}}\right)$$ does indeed have a Laurent expansion at $$z=0$$, namely $$\frac{\pi}{z^2}-\frac{\pi^3}{z^6}+\frac{\pi^5}{z^10}-\frac{\pi^7}{z^14}+…$$ First, the implications are slightly incorrect, then while computing $$c_{n}=\frac{1}{2\pi i} \oint_{\gamma}{\frac{\sin\left({\frac{\pi}{z^2}}\right)}{z^{k+1}}dz}$$, in order to check if I was getting the right coefficients I plugged in only positive $$n$$ in the integral on Wolphram Alpha obtaining always $$0$$ (and obviously for positive $$n$$ they are $$0$$), and so I assumed every coefficient was $$0$$. For a generic circle of radius $$R$$ I have:
$$c_{n}=\frac{1}{2\pi i} \oint_{\gamma}{\frac{\sin\left({\frac{\pi}{z^2}}\right)}{z^{k+1}}dz} =\frac{1}{2\pi i} \int_{0}^{2\pi}{\frac{\sin\left({\frac{\pi}{R^2}e^{-i2\theta}}\right)}{R^n e^{i\theta n}}d\theta}$$
Which gives the right coefficients for $$n=-4k-2,k\in\mathbb{N}$$ and $$0$$ for all other $$n$$. My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $$some\space terms+O(something),\space z\rightarrow z_0$$. For example, it says no series expansion available also for $$\sin{z}$$ at $$\infty$$.