Order of the pole 
What is the order of the pole at $z=0$:
$$f(z)=\frac{1}{(2\cos(z)-2+z^2)^2}$$
and find and classify the isolated singularities of:
  $$\frac{1}{e^z-1}$$

My attempt:
If I let $f(z)=(2\cos(z)-2+z^2)^2$ then I get that:
$$f(z)=(2\cos(z)-2+z^2)^2 = [z^2-2+2\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} z^{2n}]^2=[2\sum^{\infty}_{n=2} \frac{(-1)^n}{(2n)!} z^{2n}]^2$$ but now I am having trouble finding th poles. I assumed that it is essential singularity at $0$, but the answer is $8$. I do not know why?
For the second part the answer is simple poles at $2n\pi i$. I get:
$$\frac{1}{e^z-1} = 
    \frac{1}{\sum^\infty_{n=0} {z^n\over n!} =1 + z + {z^2 \over 2!} + ... -1}=\frac{1}{z + {z^2 \over 2!} + ... }=\frac{1}{z}[\frac{1}{1+{z \over 2!}+... }]$$
but again I do not know why it is simple poles at $2n\pi i$?
 A: You don't need to go that far in the expansions.
You have $\cos z=1-\frac{z^2}{2}+\frac{z^4}{24}+O(z^6)$, hence $2\cos z-2+z^2=\frac{z^4}{12}(1+O(z^2))$. Finally
$$
f(z)=\frac{1}{\frac{z^8}{144}(1+O(z^2))^2}=\frac{144}{z^8}(1+O(z^2)).
$$
So the order of the pole is $8$ and the coefficient is $144$.
The set of zeros of $g(z)=e^z-1$ is indeed $2i\pi\mathbb{Z}$. Note that $g'(z)=e^z$ so $g'(2in\pi)=1$ at each zero. So
$$
\lim_{z\rightarrow 2in\pi}\frac{z-2in\pi}{g(z)}=\frac{1}{g'(2in\pi)}=1. 
$$
Therefore each zero of $g$ yields a pole of order $1$ (simple pole) with coefficient $1$ (also called the residue) for the function $\frac{1}{g(z)}=\frac{1}{e^z-1}$.
Both times, we used the following characterization: if $\lim_{z\rightarrow z_0}(z-z_0)^nf(z)=A\neq 0$, then $z_0$ is a pole of order $n$ of $f$ with coefficient $A$ (of course, the converse is true).
A: Hint: For the first part, you can factor a $z^4$ out of the denominator, and then since there is a square, you end up factoring a $z^{-8}$ out of $f(z)$. That is, you have $$f(z)=\left(2z^{-4}\sum_{n=2}^\infty\frac{(-1)^n}{(2n)!}z^{2n-4}\right)^{-2}=z^{-8}\cdot\left(2\sum_{n=2}^\infty\frac{(-1)^n}{(2n)!}z^{2n-4}\right)^{-2}.$$ Therefore,the order of the pole at $z=0$ is $8$.
For the second part, solve the equation $e^z-1=0$, perhaps by logarithms if you've learned those in your course. Otherwise, if $z=x+iy$, then you have $e^z=e^x(\cos y+i\sin y)=1$. Equating real and imaginary parts, we have the $y=n\pi$ for $n\in\Bbb Z$ (via the imaginary part since $e^x>0$). Now we can evaluate the real part and see that $y=2n\pi i$. 
