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