Type of singularities of complex functions This problem is one of my qualifying exam problem, which I am still confused now.
Suppose $z_{0} \in \mathbb{C}$ is a pole of f. Prove that $z_{0}$ is an essential singularity of $g:=e^{f}$
There is also a hint: consider the image of g. We need to prove that it is neither a removable singularity nor a pole, but I cannot find the contradiction for those two cases.
Any hint is appreciated, thank you.
 A: WLOG assume $z_0 = 0$.  A local holomorphic change of coordinates transforms a pole into $\frac{1}{z^k}$ for $k > 0$.  The trick is to just show that $g(z) = e^{1/z^k}$ has an essential singularity at the origin.  This can be done in a bunch of ways.  To use the image of the function you note the following:  The image near the origin should be bounded for a removable singularity or the modulus should go to infinity as we approach the origin.  So take two sequences that both approach the origin $z_n = \sqrt[k]{1/n}$ and $w_n = \sqrt[k]{-1/n}$ (simply take some root for each $n$).  Then $g(z_n) = e^n \to \infty$ and $g(w_n) = e^{-n} \to 0$ as $n \to \infty$.  So arbitrarily close to the origin $g$ achieves both arbitrarily small and arbitrarily large values, and so the point is neither removable singularity nor a pole.
Another funky way (using some more heavy hitting theorems) using the image is this:  Suppose that $k > 0$. If we show $e^{1/z^k}$ has an essential singularity at 0, then so does $e^f$ as we said above.  The function $e^{1/z^k}$ now extends holomorphically to the Riemann sphere minus the origin (in particular it is holomorphic at infinity).  If it we had a pole or a removable singularity at the origin, then the function would be a rational function.  By fundamental theorem of algebra, it would then have a zero (as it is not constant because $k > 0$), but $e^{1/z^k}$ is never zero (even at infinity), so it cannot possibly be rational, hence the singularity at the origin is essential.
So remember that local questions can always be done at the origin and in one variable a meromorphic function at the origin is always locally biholomorphically equivalent to $z^k$ for some $k \in {\mathbb Z}$.  Many problems can be simplified this way: you expose the real underlying concept.
