Residue at $z=0$ where $f(z)=\frac{e^{4z}-1}{\sin^2(z)}$ Im struggling a bit with finding residue of $f(z)=\frac{e^{4z}-1}{\sin^2(z)}$ at $z=0$
My solution so far is not long.
Clearly there is a pole of order $m=2$ from $\sin^2(z)$.
By experience I want to do Taylor expansion since it's trigonometry, but the squared $\sin$-term makes it difficult to continue.
Is there any other theorem that is applicable in situations like this?
Thanks!
 A: $f(z)=\frac{e^{4z}-1}{\sin^{2}(z)}$ .
Well the fastest way to do it would be to write :-
$$f(z)=\frac{e^{4z}-1}{z^{2}}\frac{z^{2}}{\sin^{2}(z)}=\frac{\frac{e^{4z}-1}{z}\cdot\frac{z^{2}}{\sin^{2}(z)}}{z}.$$
And notice that the numerator has a removable singularity at $0$ and hence can be extended holomorphically  and hence $f$ has a pole of order $1$ at $0$.
So it only come's down to evaluating the value of the numerator(holomorphic extension) at $0$ which is given by $\lim_{z\to 0}{\frac{e^{4z}-1}{z}\cdot\frac{z^{2}}{\sin^{2}(z)}}=4$ and that's your residue . But these require clear justifications .
Otherwise you can always go for Taylor Expansion(Technically it is called Laurent Expansion). You don't lose any rigor and it only requires clever manipulation of the terms. For people comfortable with taylor series manipulations, this is perhaps easier as it avoids having to justify each step and invoke other theorems like the Riemann Removable Singularity theorem.
$$\frac{1}{\sin^{2}(z)}=\frac{1}{(z-\frac{{z}^{3}}{3!}+...)^{2}}=\frac{1}{z^{2}}\frac{1}{(1-\frac{z^{2}}{3!}+...)^{2}}$$
Now $\frac{1}{1-z}=\sum_{r=0}^{\infty}z^{r}$
So $\frac{1}{(1-z)^{2}}=\sum_{r=1}^{\infty}rz^{r-1}$
So you get $\frac{1}{z^{2}}\bigg(1+2\cdot\bigg(\frac{z^{2}}{3!}-\frac{z^{4}}{5!}+...\bigg)+3\cdot \bigg(\frac{z^{2}}{3!}-\frac{z^{4}}{5!}+...\bigg)^{2}+....\bigg)\\=\frac{1}{z^{2}}+\{\text{terms containing}\, z^{n}\, \text{where} \,n\geq 0\}$
So $$\frac{e^{4z}-1}{\sin^{2}(z)}=(\frac{4z}{1!}+\frac{(4z)^{2}}{2!}+...)\bigg(\frac{1}{z^{2}}+\{\text{terms containing}\, z^{n}\, \text{where} \,n\geq 0\}\bigg)$$
Which at once gives that the coefficeint of $\frac{1}{z}$ is $4$ .
