Problem.src) Compute $$\oint_C \frac{5\sec 2z}{e^{-3z}-1}\,\mathrm{d}z,$$ where $C:|z|=10^{-4}$.
I solved the problem like this, but I don't know if this is the right way
$$\oint_C \frac{5\sec 2z}{e^{-3z}-1}\,\mathrm{d}z = 2πi\cdot\text{Res}\frac{5\sec 2z}{e^{-3z}-1}$$ $$= 2πi\cdot\text{Res}\frac{5}{cos2z\cdot\ e^{-3z}-1}$$ $$= 2πi\cdot\text{Res z=0 f(z)} = 2πi\cdot\frac{-5}{3}\, $$
$$ \text{Res}\frac{5\sec 2z}{e^{-3z}-1} = \frac{5}{-2sin2z\cdot\ (e^{-3z}-1) + cos2z\cdot\ (-3e^{-3z})} (z=0) =\frac{-5}{3}\ $$
Here's an easy formula/principle to remember.
The point is that you can avoid the tedius Maclaurin expansions and computations with a bit of cleverness. i.e. you don't need to expand all that is there, you only need to focus on the problematic bit. You are essentially doing the same, but you don't need to even bother with the $\sec(2z)$ part.
To see this, just note that in a neighbourhood of $z_{0}$, $q(z)=(z-z_{0})h(z)$ where $h(z)\neq 0$ for all $z$ in that neighbourhood.
So in fact, in that nbd, $\frac{p(z)}{h(z)}$ is an analytic function and hence has a power series of the form $\sum_{k=0}^{\infty}a_{k}(z-z_{0})^{k}$
Hence, in that neighbourhood, $$\frac{p(z)}{q(z)}=\frac{1}{z-z_{0}}\bigg(\sum_{k=0}^{\infty}a_{k}(z-z_{0})^{k}\bigg)$$
Which directly gives you the residue as $\displaystyle a_{0}=\frac{p(z_{0})}{h(z_{0})}$
But what is $h(z_{0})$? As $h(z)=\frac{q(z)}{z-z_{0}}$ it is simply the limit $$\lim_{z\to z_{0}}\frac{q(z)}{z-z_{0}}=q'(z_{0})$$ due to L'Hospital's rule.
So in this case, what you only need to do find the residue is given by $$\frac{\sec(2\cdot 0)}{\bigg(e^{-3z}-1\bigg)'(0)}=\frac{1}{-3}$$