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Find all singularities of $f(z)={z\over \sin({1\over z^2})}$, define them and compute the residues of $f$ in each.

This was pretty meticulous for me. I found $z=0$ to be an essential singularity and ,$z=\pm{1\over \sqrt{\pi z}},\pm{i\over \sqrt{\pi z}},k\in \Bbb{Z}_+\setminus \{0\}$ to be poles.

Although I tried computing the residues at the poles and it was a mass, what is more disturbing me is not being able to compute the residue at $z=0$.

I tried integrating $f(z)$ along a circle as small as desired around $z=0$, but it also seems too tedious, and also, there are infinitely many poles in every circle I would choose, which makes it pointless. I am completely lost. Can you help me? If I got the approach wrong and I'd better look at it all differently, it would help me even more than guiding me with the essential singularity.

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    Can you explain? Because $z=0$ is one point for which $f$ is not defined, What do I do with $z=0$? – 2017-02-09
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    $0$ is a non-isolated singularity. A residue only exists at isolated singularities. – 2017-02-09

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Let $f(z)=\frac{z}{\sin(1/z^2)}$.

The singularities of $f$ are comprised of simple poles at $z=\pm \frac{1}{\sqrt{n\pi}}$ and $z=\pm \frac{i}{\sqrt{n\pi}}$, for $n\ge 1$, a non-isolated singularity at $0$.

RESIDUES AT THE POLES

The residues along the real axis are given by

$$\begin{align} \text{Res}\left(f(z),z=\pm \frac{1}{\sqrt{n\pi}}\right)&=\lim_{z\to \pm 1/\sqrt{n\pi}}\frac{(z\mp 1/\sqrt{n\pi})z}{\sin(1/z^2)}\\\\ &=\lim_{z\to \pm 1/\sqrt{n\pi}}\frac{z+(z\mp 1/\sqrt{n\pi})}{\cos(1/z^2)(-2/z^3)}\\\\ &=\frac{(-1)^{n-1}}{2\pi^2 n^2} \end{align}$$

The residues along the imaginary axis are identical since $i^4=1$

THE NON-ISOLATED SINGULARITY

The singularity of $f(z)$ at $z=0$ is not an isolated singularity, hence is not an essential singularity. Hence, $f(z)$ has no Laurent series around $z=0$ and no definable residue at $z=0$.

RESIUDE AT INFINITY

The function $f(z)$ has a residue at infinity, which is given by

$$\text{Res}\left(\frac{z}{\sin(1/z^2)},z=\infty\right)=-\text{Res}\left(\frac {1}{z^3\sin(z^2)},z=0\right)=-\frac16$$

SUM OF THE RESIDUES

The sum of the residues from the poles is given by

$$\sum_{n =1}^\infty \frac{(-1)^{n-1}2}{\pi^2 n^2}=\frac{1}{6}$$

INTEGRATION AROUND THE POLES AND NON-ISOLATED SINGULARITIES

Let $R>1/\sqrt{\pi}$. Then, we have

$$\begin{align} \oint_{|z|=R} f(z)\,dz&=\int_0^{2\pi} \frac{iR^2e^{i2\phi}}{\sin\left(\frac{1}{R^2e^{i2\phi}}\right)}\,d\phi\\\\ &=\int_0^{2\pi}\frac{iR^4e^{i4\phi}}{1-\frac16 \left(\frac{1}{R^2e^{i2\phi}}\right)^2+O\left(\frac{1}{R^8}\right)}\,d\phi\\\\ &=2\pi i\frac{1}{6}+O\left(\frac{1}{R^4}\right)\\\\ &\to 2\pi i\frac{1}{6}\,\,\text{as}\,\,R\to \infty \end{align}$$

which agrees with the residue at infinity of $-\frac16$!

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    It seems that $f(z)$ does not tend to $0$ as $|z|\to \infty$, which is the condition for using the identity you used above, according to the Wikipedia criteria.Can you show me where I am wrong? – 2017-02-09
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    As for the criterion for the residue at $\infty$, we need only that $f$ be holomorphic on an annulus with inner radius $R$ and infinite outer radius. So just pick $R$ large enough to contain all poles and you're good to go. – 2017-02-10
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    @DanielFischer Inasmuch as the residue theorem is technically inapplicable here, I believe, due to the non-isolated singularity at the origin, it is curious to see that the integration around the poles seems to yield $2\pi i \sum_{n}\text{Res}(f(z),z=z_n)$ where $z_n$ is a pole of $f$. I'm interested in hearing your thoughts. -Mark – 2017-02-10
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    Or perhaps it isn't so curious inasmuch at the residue at infinity is $0$. – 2017-02-10