Essential singularity This is an exercise from Gamelin.

If $f(z)$ is a complex function with a not removable singularity in $ z_{0} \ $, then $e^{f(z)} \ $ has an essential singularity in $z_{0} $.

Any hint?
If your function $f$ has a pole at $z_0$ write $(z-z_0)^m f(z) = p(z) + (z-z_0)^m h(z)$ where $h$ is holomorphic and $p$ is a polynomial of degree $&lt;m$.  Then $f(z)= \displaystyle \frac{p(z)}{(z-z_0)^m} + h(z)$.  Now take $e$ to both sides, note that $e^{h(z)}$ is holomorphic, and use the power series expansion of the exponential to show that $e^{\frac{p(z)}{(z-z_0)^m}}$ has infinitely many negative powers of $z-z_0$.
I haven't looked at this stuff in a while, but Casorati-Weierstrass might help. 

In particular there is a sequence $(z_k)$ converging to $z_0$ such that $(f(z_k))$ converges to $w_0$ (this is page 175 of your book). Since $w_0$ is arbitrary can you think of a particular value that helps solve this problem?