# Does $z^{2}\sin \frac 1 z$ have a removable singularity at $0$ in the Riemann sphere?

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

I'm not sure is $f(z):\mathbb C \cup\{\infty\} \to \mathbb C \cup\{\infty\}$ has removable singularity, or has essential singularity at 0. I think it has isolated essential singularity at 0.
but my textbook says it has removable singularity at 0.

this is my thinking.
since laurent expansion of $z^2\sin(1/z)$ at $z=0$ is $$z^2(\frac 1 z -\frac 1 {3!z^3}+\frac 1 {5!z^5} - ....)$$ so by definition, $f(z)$ has essential at 0.
(in my book, $a$ is isolated essential singularity of $f$ if,
$f$ has laurent at $a$, (say, $f(z)=\sum_{n=-\infty} ^{\infty} c_n (z-a)^n$) and there is no $n \in \mathbb Z$ s.t. $c_m=0$ for all $m<n$)

is it correct?

or maybe something changed because this function is in reimann sphere? i'm very confused

No, the singularity at $0$ is essential. Consider for example what happens along the imaginary axis $z=iy$. The function explodes exponentially in that direction and therefore the singularity is not removable.
• Just looking at the real picture, there is a removable singularity at $0$ (at least with respect to continuity, if not analyitcity), but the moment you allow complex inputs, the singularity becomes essential. – Arthur Oct 23 '16 at 10:49