Residue of $f(z)=\frac{1}{\sin(z)-1}$

Question

I'm trying to find $$\text{Res}_{z=\frac{\pi}{2}}\frac{1}{\sin(z)-1}$$

Since I don't know where to begin writing out the Laurent series expansion for $f$ I assume I need to know the order of the pole at $z=\frac{\pi}{2}$. I don't know how this is found, but Wolfram Alpha tells me it's a double pole. Using this info I've tried the formula:

$$\text{Res}\left(f(z),\frac{\pi}{2}\right)=\lim_{z\to \frac{\pi}{2}}\frac{d}{dz}\left(z-\frac{\pi}{2}\right)^2\frac{1}{\sin(z)-1}$$

When I try to differentiate this things go from bad to worse. Any ideas?

Answer 1

You can remark that $\sin$, hence also $f(z) = \frac1{\sin z - 1}$, is symmetric around $z = \pi/2$.

It follows that its Laurent series can not have odd terms, and the residue has to be $0$.

Answer 2

Notice that $\sin(\pi/2-z)=\cos(z)$, hence substituting $z=\pi/2-w$ it is: $$\lim_{z \to \pi/2} \frac{(\pi/2-z)^2}{\sin(z)-1}=\lim_{w \to 0} \frac{w^2}{\cos(w)-1}=-2$$ So the pole is of order $2$. Hence, it is: $$\text{Res}(f(z),z=\pi/2)=\lim_{z \to \pi/2} \frac{\text{d}}{\text{d}z} \frac{1}{\sin(z)-1}(\pi/2-z)^2=\lim_{z \to \pi/2}\left[-\frac{\cos z}{(\sin z-1)^2}(\pi/2-z)^2-\frac{2(\pi-z)}{\sin z-1}\right]=0$$