About the singularity of $z^{-2}\sin z$ at $\infty$ On Priestly's Introduction to Complex Analysis, second edition, at page 206 it says that $z^{-2}\sin z$ has removable singularity at $\infty$. If this is not a typo, what's the reasoning behind it?
 A: $\sin(z)$ has an essential singularity at infinity (it is periodic so it takes lots of value (in fact all of $\Bbb C$) infinitely often)
Since $z^2$ doesn't have an essential singularity there, $z^{-2}\sin(z)$ also has an essential singularity at $\infty$.
Saying what is the typo and what he meant to be writing is hard to tell without more context.
A: It is not a typo. We refer to points at infinite as singularity points on complex analysis, because their substance revolves around a lot of calculations and crucial stuff. 
For your specific example, we have the function : 
$$f(z) = \frac{1}{z^2}\sin(z)$$
The function $f(z)$ has an essential singularity because of $\sin(z)$ which can take infinitely many values at some point $\infty$ due to its periodic nature.
So, the point at $\infty$ acts like a singularity point to your function $z$. More specifically, you'll see about singularities at $\infty$ when you'll be learning about residues (shortly if not already, judging on your "residue-calculus" tag), which are important there. For example, here is a formal statement for a residue at $\infty$ and the condition that shall be met (the singularity at infinity) : 

Residue at infinity :
  $$\text{Res}(f(z),\infty) = -\text{Res}\bigg(\frac{1}{z^2}f\bigg(\frac{1}{z}\bigg),0\bigg)$$
  If the condition :
  $$\lim_{|z| \to \infty}f(z) = 0$$
  is met, then the residue at infinity can be computed using the following formula :
  $$\text{Res}(f,\infty)= -\lim_{|z|\to \infty}zf(z)$$
  If instead :
  $$\lim_{|z| \to \infty}f(z) = c \neq 0$$
  then the residue at infinity is :
  $$\text{Res}(f,\infty)= -\lim_{|z|\to \infty}z^2f'(z)$$

It's also interesting to note, that every non-constant entire function $f:\mathbb C \to \mathbb C$ has a singularity at infinity. Technically, $f$ has a singularity at $\infty$ by virtue of not being defined there.  What is really meant is that a non-constant entire function $f$ has a non-removable singularity at $\infty$, and this follows directly from Liouville's theorem: if the singularity at $\infty$ was removable, $f$ would be bounded in a neighbourhood of $\infty$, say $\{z: |z| \ge r\}$, and since $f$ is also bounded on $\{z: |z| \le r\}$ (because a continuous function is bounded on a compact set) $f$ would be bounded on $\mathbb C$, therefore constant by Liouville's theorem.
