Compute the $$\operatorname{res}_{z=0}\frac{z^{n-1}}{\sin^n z}\:\:\text{for}\:\:n\in\mathbb{N}$$
I was given the hint that $\frac{z^{n-1}}{\sin^n z}\sim\frac{1}{z}$ as $z\to 0$. However I am not understanding this approximation.
I tried $\frac{z^{n-1}}{\sin^n z}\sim\frac{1}{z}=\frac{1}{z}\frac{z^{n}}{\sin^n z}\sim\frac{1}{z}$. However I have no idea on how to compute the Laurent series of $\frac{z^{n}}{\sin^n z}\sim\frac{1}{z}$ because of $\sin^n z$
I have no idea how I could adapt $\sin(z)=\sum_\limits{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}$.
Question:
Can someone explain me this approximation? How do I derive it?
Thanks in advance!
Note that$$\lim_{z\to0}\frac{\frac{z^{n-1}}{\sin^nz}}{\frac1z}=\lim_{z\to0}\frac{z^n}{\sin^n z}=\lim_{z\to0}\left(\frac z{\sin z}\right)^n=1.$$Therefore, $0$ is a removable singularity of$$\frac{\frac{z^{n-1}}{\sin^nz}}{\frac1z}$$and, near $0$, you can write it as$$1+a_z+a_2z^2+\cdots$$So, near $0$,$$\frac{z^{n-1}}{\sin^nz}=\frac1z+a_1+a_2z+\cdots$$from which youn deduce that$$\operatorname{res}_{z=0}\left(\frac{z^{n-1}}{\sin^nz}\right)=1.$$