$z=0$ is a Pole of order $2$

Question

I was trying to find order of $z=0$ for $$f(z)=\frac{\cos z}{z\sin z}$$

We have $$g(z)=z\sin z$$ Evidently $g(0)=0$

Also $$g'(z)=z\cos z+\sin z$$

So $g'(0)=0$ Now $$g''(0)=\cos z-z\sin z+\cos z$$ So $g''(0)\ne 0$

So since $g(0)=g'(0)=0$ and $g''(0)\ne 0$ $z=0$ is Pole of order $2$

Is this method valid for Non polynomial functions?

That is, we know that if $f(x)=0$ is a polynomial with root $\alpha$ repeated $r$ times, then $f'(x)=0$ has root $\alpha$ repeated $r-1$ times.

Is this concept valid for non polynomial functions too?

Answer 1

Yes. If an analytic function has a zero of order $n$ at some point $a$ in it's domain, then we can find another function $g$ analytic in a neighborhood of $a$ such that $$f(z)=(z-a)^ng(z)$$ and $g(a)\neq 0.$