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lets consider $f:\mathbb C\setminus \{0\}\rightarrow \mathbb C, f(z)=\frac{1}{1-e^z}$ and I want to determine the type of the singularity $0$.

I am pretty sure that this singularity is an essential singularity but I think it does hold

$\lim_{|z|\rightarrow \infty}\frac{1}{1-e^z}=\infty$ so this points must be a pole, right? If this is really true how one can determine the order?

Another thing: I know that $e^{\frac{1}{z}}$ has an essential singularity in $0$, but its also true that $\lim_{|z|\rightarrow \infty}e^{\frac{1}{z}}=\infty$, what am I missing?

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    $\begingroup$ For the first, $0$ is a pole since $\lim_{z\rightarrow 0}\frac{1}{1-e^z}=\infty$ and $\lim_{z\rightarrow 0}\frac{z}{1-e^z}=1$ (why are you taking the limit at $\infty$?). For the second I think you mean to take $\lim_{z\rightarrow 0}e^{1/z}$ which does not exist (therefore $0$ is essential singularity). $\endgroup$ – Nikolaos Skout Mar 5 '16 at 2:42
  • $\begingroup$ and $\lim_{|z| \to \infty} e^{1/z} = 0$ while $\lim_{z \to 0} e^{1/z}$ doesn't exists (it depends on the path along which $z$ tends to $0$, which together with the fact that $e^{1/z}$ is holomorphic on $\mathbb{C}^*$, proves that $e^{1/z}$ has an essential singularity at $0$) $\endgroup$ – reuns Mar 5 '16 at 3:15
  • $\begingroup$ as you know, $e^z = \sum_{k=0}^\infty \frac{z^k}{k!} = 1 + z + \mathcal{O}(z^2)$, i.e. $e^z - 1 \sim z$ when $z \to 0$. hence $\frac{1}{1-e^{z}} \sim -\frac{1}{z}$ and $f(z) = \frac{1}{1-e^z}$ has a pole or order $1$ and residue $-1$ at $z=0$. its other poles are at $z = 2 i n \pi, n \in \mathbb{Z}$, again of order $1$ and residue $-1$ since $f(z+2i \pi) = f(z)$. $\endgroup$ – reuns Mar 5 '16 at 3:18
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$f$ is the reciprocal of the entire function $1-e^z$, which has series expansion $$ 1 - e^z = -\sum_{j=1}^\infty \frac{z^j}{j!} $$ From the series, $1 - e^z$ has a simple zero at $0$.(divide by $z$ to see this) Therefore, its reciprocal $f$ has a simple pole at $0$. I think that is all you are asked to do.

For $e^{1/z}$, Riemann's Theorem indicates that $z_0$ is a pole of $f$ if and only if $\lim_{z \to z_0} |f(z)| = \infty$ for every path of $z$ approaching $0$. But $\lim_{z \nearrow 0} e^{1/z} = 0$.

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You can think of the pole in $\frac{1}{1-e^z}$ at $z=0$ as just being a manifestation of the simple pole of $\frac{1}{1-z}$ at $z=1$. The chain rule would explain this (since $e^z$ is holomorphic everywhere).

As for $e^{1/z}$, you are not correct. In particular, $\lim_{r \to 0} e^{1/(r e^{i \theta})}$ depends very strongly on $\theta$: for $\theta=0$ you have $+\infty$, for $\theta=\pi$ you have $0$, and for $\theta=\pi/2$ the limit doesn't exist at all.

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