# Computing the $\operatorname{res}_{z=0}\frac{z^{n-1}}{\sin^n z}$ via an aproximation

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?

• The key is that you are not computing the entire Laurent series, only the $1/z$ term. – Simply Beautiful Art May 9 at 13:56
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$$\operatorname{res}_{z=0}\left(\frac{z^{n-1}}{\sin^nz}\right)=1.$$
• Thanks for your answer! $\lim_{z\to0}\frac z{\sin z}=1$ using the Hopital's rule, since in complex analysis I cannot use that rule. How would I prove $\lim_{z\to0}\frac z{\sin z}=1$? Could I derive it as the inversion of $\lim_{z\to 0}\frac{\sin(z)}{z}$? – Pedro Gomes May 9 at 14:14
• You have $\lim_{z\to0}\frac{\sin z}z=\sin'(0)=\cos(0)=1$. Therefore, $\lim_{z\to0}\left(\frac{\sin z}z\right)^n=1^n=1$, and so $\lim_{z\to0}\frac{z^n}{\sin^nz}=1$. – José Carlos Santos May 9 at 14:16