How to find the order of a pole

Solution 1:

First, note that $2\log(z)$ does not vanish at $z=\frac\pi2$, so we can ignore it; it does not contribute to the pole. So consider simply $\dfrac1{1-\sin(z)}$. We want to write things in terms of $w = z-\frac\pi2$, so let $\sin(z)=\sin(w+\frac\pi2) = \cos(w)$, where at the end I have used a simple trig formula. Then we can expand $\cos(w) = 1-\frac{w^2}2+\frac{w^4}{24}-...$, so $\dfrac{1}{1-\cos(w)} = \dfrac{1}{\frac{w^2}2-\frac{w^4}{24}+...} = \dfrac1{w^2(\frac12-\frac{w^2}{24}+...)}$. Thus, we can see that the order of the pole is 2.

Solution 2:

you have the function $h(z)=\log(z)(1-\sin z)$ then

$h(\pi/2)=0$,

$h'(z)= \frac{1-\sin z}{z}- \log z \cos z$ then $h'(\pi/2)=0$

$h''(z)= \frac{-z\cos z-(1-\sin z)}{z^2}-\frac{\cos z}{z}+\log z\sin z$ then $h''(\pi/2)\ne 0$ hence $\pi/2$ is a zero of order two for $h$, therefore is a pole for $\frac{1}{h(z)}$ of order two.