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

Question

I don't know what type of singularity has $f(z)=\cot(z)/(z-\frac{\pi}{2})$ at $\frac{\pi}{2}$ and how can I calculate the residue of $f(z)$ at $\frac{\pi}{2}$. Can you help me?

Thanks in advance.

Answer 1

You have the complex function :

$$f(z) = \frac{\cot(z)}{\bigg(z-\frac{\pi}{2}\bigg)}$$

For $z=\pi/2$ you can see that the denominator is $0$ and thus $f(z)$ is not defined but also take note that $\cot(\pi/2) =0$, this is a zero point of order $1$, thus at $z=\pi/2$ the function $f$ has a removable singularity, which means that $\text{Res}(f,\pi/2) = 0$.

Answer 2

Since $\cot$ has at $\pi /2$ a zero of order $1$, the function $f$ has at $ \pi/2$ a removable singularity. Hence the residue at $ \pi /2$ is $0$.