TITLE: Showing that $e^z$, $\sin(z)$, $\cos(z)$ have essential singularities at $\infty$
QUESTION [9 upvotes]: Problem: Show that $e^z$, $\sin(z)$, $\cos(z)$ have essential singularities at $\infty$.

Attempt for $e^z$:

We have that $e^z$ has an essential singularity at $\infty$ iff $e^{1/z}$ has an essential singularity at $0$ iff
$$
0 \ne \lim_{z \to 0} |z|^{\alpha} |e^{1/z}| \ne \infty
$$
for all values of $\alpha \in \mathbb{R}$.

My intuitions must be off here, because it seems to me like $|e^{1/z}|$ tends to infinity much quicker than $|z|^\alpha$ tends to zero.  That is, it seems to me that indeed
$$\lim_{z \to 0} |z|^{\alpha} |e^{1/z}| = \infty$$
which evidently is not the case.

REPLY [4 votes]: Since we have that
$$
\lim_{z \to 0+} \left|e^{1/z}\right| = \infty
$$
$$
\lim_{z \to 0-} \left|e^{1/z}\right| = 0
$$
it follows that
$$
0 \ne \lim_{z \to 0} |z|^{\alpha} |e^{1/z}| \ne \infty
$$
so that $e^z$ has an essential singularity at $\infty$ by definition.
Furthermore, the definitional dependencies of $\sin(z)$ and $\cos(z)$ on the exponential function means that
$$\cos(z) = {e^{iz} + e^{-iz} \over 2}$$
$$\sin(z) = {e^{iz} - e^{-iz} \over 2i}$$
have essential singularities at $\infty$ if and only if $\exp(z)$ has an essential singularity at $\infty$.  Therefore, both $\cos(z)$ and $\sin(z)$ have essential singularities at $\infty$ as well. This completes the proof.