Can you find a residue by using Taylor's formula for a single coefficient instead of expanding into a Laurent series?

Question

For example, say I was trying to determine the residue of the functions $\frac{e^z}{\sin z}$ and $-\cot(z)$ at $-\pi$. Clearly there is a singularity in both of these functions at this point, but the Laurent series are difficult to determine. My instinct is that you can just use the formulas $a_{-1}=\frac{1}{2\pi i} \int_c \frac{-\cot(z)}{(z + \pi)^0} dz = \frac{1}{2\pi i} \int_c -\cot(z)$ to directly find the residue for $-\cot z$ at $-\pi$, for example, but seemingly you run into the same problem because you need the residue to determine that integral if you're trying to use Cauchy's Integral Formula.

Answer 1

Simply use that fact that if $g(z_0)\ne0$, $h(z_0)=0$, and $h'(z_0)\ne0$, then$$\operatorname{res}\left(z_0,\frac{g(z)}{h(z)}\right)=\frac{g(z_0)}{h'(z_0)}.$$This will work for both of your residues.