Residue of $\cot(z)/(z-\frac{\pi}{2})^2$ at $\frac{\pi}{2}$

Question

I want to know what type of singularity has $f(z)=\cot(z)/(z-\frac{\pi}{2})^2$ at $\frac{\pi}{2}$ and what is the residue of $f(z)$ at $\frac{\pi}{2}$. I thought that $f$ has a pole of order $2$ at $\frac{\pi}{2}$, but the problem is that $\cot(\pi/2)=0$. Can you help me, please?

Answer 1

Since $\cot$ has a simple zero at $\frac\pi2$, $f$ has a simple pole there. Besides,$$f(z)=\frac{-\left(z-\frac\pi2\right)-\frac13\left(z-\frac\pi2\right)^3+\cdots}{\left(z-\frac\pi2\right)^2}=-\left(z-\frac\pi2\right)^{-1}-\frac13\left(z-\frac\pi2\right)+\cdots$$and therefore $\operatorname{res}_{z=\frac\pi2}\bigl(f(z)\bigr)=-1$.

Answer 2

Since $ \cot$ hat a simple zero at $ \pi/2$, the function $f$ has a pole of order $1$ at $ \pi/2$ !

Answer 3

Write $\cos z$ as $(z-\frac {\pi} 2) g(z)$ near $\frac {\pi} 2$ and check (by looking at the derivative) that $g(\frac {\pi} 2)\neq 0$. Hence the function has a pole of order $1$ at $\pi /2$. Can you use $g$ to find the residue now? (The answer is $-1$).

Answer 4

The residue is the coefficient of $(z - z_0)^{-1}$ in the Laurent series for expr. $$f(z)=\cot(z)/(z-\frac{\pi}{2})^2= -\frac{1}{z-\frac{{\pi} }{2}}-\frac{z-\frac{{\pi} }{2}}{3}-\frac{2 {{\left( z-\frac{{\pi} }{2}\right) }^{3}}}{15} -\mbox{...}$$

Then $$\mbox{residue}(f(z),z,\pi/2)=-1$$