Determine singular point of $f(z)=\frac{e^z}{\sin\frac{1}{z}}$ at $z=0$ and $z=\infty$

I believe $z=0$ is an essential singular point because I cannot find the limit of this function and $z=\infty$ is removable singular point.

Am I right?

  • $\begingroup$ There are also simple pole singularities at $z=1/(n\pi)$ for all $n\ne 0$. $\endgroup$ – Mark Viola May 12 '17 at 20:38
  • $\begingroup$ @Dr.MV So I'm right when z=0 and z=infinity? $\endgroup$ – Parting May 12 '17 at 20:48
  • $\begingroup$ $z=\infty$ will not be a removable singularity. Think about $z=\frac{1}{\zeta}$ as $\zeta$ goes to $0$ in the exponential. $\endgroup$ – sharding4 May 12 '17 at 20:51
  • $\begingroup$ @sharding4 so what point it is? $\endgroup$ – Parting May 12 '17 at 21:23

Since $e^z$ is non singular at $z=0$ and $e^0=1$, you can just look at $$ \frac{1}{\sin(1/z)} $$ and this has no limit for $z\to0$: approach $0$ with $\frac{2}{(2n+1)\pi}$ or $-\frac{2}{(2n+1)\pi}$.

The given function has no limit for $z\to\infty$ either. Consider instead $$ g(w)=\frac{e^{1/w}}{\sin w} $$ and its behavior at $0$. In order for $0$ to be a removable singularity you need $\lim\limits_{w\to0}e^{1/w}=0$, which doesn't hold. Can $w=0$ be a pole?


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