Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2. 
Let $$f(z) = \frac{\pi^2}{\sin^2(\pi z)}.$$ 
  For some $z=n$, show it has a pole of order 2.

Aside from appear elsewhere, this function is discussed on p188 of Ahlfors.
I can answer the question I posed above by expanding $\sin(\pi z)$ into its power series, taking the square of the series in the denominator, and then using polynomial long division to get the actual Laurent series.
What would be some other, faster ways to achieve the answer? For finding out the order of the pole, I tried to use the definition of a pole to put $f(z)$ in the form
$$
f(z) = \frac{g(z)}{(z-n)^n}
$$
and solve for $n$. Aside from using the knowledge that $1/sin(z)$ is has a pole of order one, how else can we find the order?
 A: 
For finding out the order of the pole, I tried to use the definition of a pole to put $f(z)$ in the form
  $$
f(z) = \frac{g(z)}{(z-n)^n}
$$
  and solve for $n$.

Good idea, but if you tried exactly what you wrote, you've made a mistake : you use $n$ for two independant things !!!
You want to solve 
$$f(z) = \frac{g(z)}{(z-n)^k}$$
Or in other words, to find the smallest $k \in \mathbb{N}$ such that $\lim_{z\to n} (z-n)^kf(z)$ exist and is finite
And this is immediate to answer if you remember that $\sin(z) \sim z  $
A: Note that $ \sin(x+iy) = \sin(x)\cos(iy) + \cos(x)\sin(iy) = \sin(x)\cosh(y) - i\cos(x)\sinh(y) $, then this function is complex differentiable at $ x + iy = 0$ as it satisfies the Cauchy-Riemann equations there. Then, the complex derivative must agree with the real derivative at that point, which is $ 1 $. Therefore, we have
$$\sin'(0) = \lim_{z \to 0} \frac{\sin(z) - \sin(0)}{z - 0} = \lim_{z\to 0} \frac{\sin(z)}{z} = 1 $$
which shows that the pole at $ z=0 $ is of order $ 1 $.
