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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!

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    $\begingroup$ The key is that you are not computing the entire Laurent series, only the $1/z$ term. $\endgroup$ – Simply Beautiful Art May 9 at 13:56
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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$$\operatorname{res}_{z=0}\left(\frac{z^{n-1}}{\sin^nz}\right)=1.$$

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  • $\begingroup$ 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 $\lim_{z\to 0}\frac{\sin(z)}{z}$? $\endgroup$ – Pedro Gomes May 9 at 14:14
  • $\begingroup$ 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 $\lim_{z\to0}\frac{z^n}{\sin^nz}=1$. $\endgroup$ – José Carlos Santos May 9 at 14:16

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