Singularities of $f(z)=z /\ (\cos(z)-1)$

Question

I'm having difficulties with this function's singularities. As far as I understand $f(z)$ has order $2$ poles in $z=2\pi k$ (where $k$ is integer), but I'm not sure about $z=0$ since it turns numerator to zero as well. Could you help me identify what type of singularity $z=0$ is?

Answer 1

The order of $z$ at $0$ is $1$ and the order of $\cos(z)-1$ at $0$ is $2$. Therefore, the order of $f$ at $0$ is $-1(=1-2)$. In other words, $f$ has a simple pole at $0$.

Answer 2

We say that a function has a pole of order $n$ if $1/f$ has a zero of order $n$. Since $$\cos(z) = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} +O(z^6)$$ it implies that $$g(z)=1/f(z)=\frac{\cos(z)-1}{z} =- \frac{z}{2!} + \frac{z^3}{4!} +O(z^5)$$ So $g(0)=0$, but $g'(0)=-\frac{1}{2}$, so $1/f$ as a zero of order $1$ in $z=0$, it implies that $f$ has a pole of order $1$ in $z=0$

Now to calculate the residue is jut: $$Res(f,0)=\lim_{z\to 0}z\frac{z}{\cos z-1}=_{L'H}\lim_{z\to 0}\frac{2z}{-\sin z}=_{L'H}\lim_{z\to 0}\frac{2}{-\cos z}=-2$$