$\frac{\sin(z)}{z}$ 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.

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.
