$\frac{\sin(z)}{z}$ Bounded on $\mathbb{C}-\{0\}$? Is $\frac{\sin(z)}{z}$ bounded on $\mathbb{C}-\{0\}$?
Based on real analysis intuition it is, and even though complex $\sin$ is not bounded the $\frac{1}{z}$ makes me think it is.

Consider the values that this function takes on the imaginary axis.
We have that 
$$
\sin z = \frac{e^{iz} - e^{-iz}}{2i},
$$
and so if $z=ix$, where $x \in \mathbb R$,
then 
$$
\sin ix = \frac{e^{-x}-e^x}{2i}.
$$
So we see that
$$
\left \vert 
\frac{\sin ix}{ix} 
\right \vert
=\left \vert
\frac{e^{-x} - e^x}{2x}
\right \vert
\to \infty,
$$
as $x \to \pm \infty$.
