$\frac{\sin(z)}{z}$ 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.
 A: 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 $x \to \pm \infty$.
A: Notice that it does have a limit at $z=0$. This means it has a removable singularity there. 
If it were bounded on $\mathbb{C}\setminus\{0\}$ it would be bounded everywhere. This is because it is also bounded near $z=0$.
Since it is holomorphic everywhere (notice at $z=0$, since it has a limit, it has a removable singularity), then by Liouville's theorem it would be constant. 
But it is not, because if it were then $\sin(z)=Cz$, for some constant $C$. But you know that $\sin$ vanishes in ($z=\pi k, k\in\mathbb{Z}$) more places than $Cz$ so they can't be equal.
A: No, it is not bounded. Since this function is analytic in its domain. if it is bounded then it would be constant by Liouvill's Theorem but function is no constant. 
