# What sort of singularity is this?

I have the following function:

$f(z) = \frac{\cos z}{\left(z-\frac{\pi}{2}\right)^3}$

Where I have to classify the singularity $z_o=\frac{\pi}{2}$

I've taken the limit to see if it's a removable singularity since it is indeterminate when plugging in $z_o$ but the limit does not exist.

I'm conflicted here -- I know it's not removable because the limit diverges, but then what type of singularity is it? Is it just a pole of order 3?

It is a pole of order $2$$\lim_{z\to\dfrac{\pi}{2}}(z-\dfrac{\pi}{2})^2f(z)=\lim_{z\to\dfrac{\pi}{2}}(z-\dfrac{\pi}{2})^2\frac{\cos z}{\left(z-\frac{\pi}{2}\right)^3}=\lim_{z\to\dfrac{\pi}{2}}\frac{\cos z}{z-\frac{\pi}{2}}=\lim_{z\to\dfrac{\pi}{2}}\frac{-\sin z}{1}=-1$$ • Thanks! But what about it tells you it's a pole of order 2? Mar 20, 2018 at 0:19 • In fact a pole$x_0$of order$m$of function$f(x)$is such that $$\lim_{x\to x_0}(x-x_0)^mf(x)\in\Bbb R-\{0\}$$ Mar 20, 2018 at 7:34 Hint:$\cos(\pi/2) = 0\$ and the order of this zero is...