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

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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Arthur Oct 23 '16 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.