The order of pole of $\frac{1}{(2\cos z -2 +z^2)^2} $at $z=0$ What is the order of the pole of : 
$$\frac{1}{(2\cos z -2 +z^2)^2}$$
at $z=0$
This is what I did : 
$$\cos z = \frac{e^{iz}-e^{-iz}}{2} $$
then : 
$$ \implies \frac{1}{(e^{iz}-e^{-iz}+z^2-2)^2} $$
How do I continue?

Hint : Let , $f(z)=2\cos z-2+z^2$. Find the order of zero of $f$.
$f'(z)=-2\sin z+2z$.
$f''(z)=-2\cos z+2$
$f'''(z)=2\sin z$
$f^{iv}(z)=2\cos z$.
Clearly , $f(0)=f'(0)=f''(0)=f'''(0)=0$ , but $f^{iv}(z)\not= 0$. So $f$ has a zero of order $4$.

 So , $(2\cos z-2+z^2)^2$ has a zero of order $8$ at $z=0$.

