How to find the order of a pole How can I find the order of the pole $z = \frac{\pi}{2}$ for $f(z)=\frac{1}{(2\log(z))(1-\sin(z))}$? I know the answer should be 2, but I can't solve it, mostly due to poor understanding of the pole order theory. As far as I've understood, the order should be equal to the index of the first non-zero member of the expansion in a series. However, I don't know how to expand this function in the singularity for a Taylor series, and a geometrical series doesn't seem to do the trick
 A: 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.
A: 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.
A: Write $f(z)=\frac{ 1/\log z^2}{(1-\sin z)}=\frac{g(z)}{h(z)}$
Observe that $g(z)$ is analytic and non- zero at $z=\pi/2$ hence the order of zero of $h(z)$ there determines the order of pole for rational function $f(z)$.
$h(\pi/2)=h'(\pi/2)=0$ and $h''(\pi/2)\ne 0$ suggests $h(z)$ has zero of order $2$ at $z=\pi/2$ so $f(z)$ has pole of order $2$ there.
