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

Question

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 :)

Answer 1

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.

Answer 2

My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value... You should be able to follow it on logical from there.

Answer 3

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 $\infty$.