Not so clear singular points 
Find the singular points and determine the type:
  
  
*
  
*$g(z)=\frac{z^2-3z+2}{z^2-2z+1},z_0=1$
  
*$f(z)=\frac{1}{1-\sin(z)}$
  

For $g$, we have that $$ \frac{z^2-3z+2}{z^2-2z+1}=\frac{z-2}{z-1}=1-\frac1{z-1}=\frac1{(z-1)^0}-\frac1{z-1}$$
And according to the criteria to detrmine what type of singular point us $1$ we have to look at the principal part of the Laurent series, wich in this case I think is zero hence (?) it is a removable point, but what's bugging me a little is that there are only two terms ib the sum, is this correct?
Now, for $f$ we hace that $1-\sin z=0$ iff $z=\pi/2(4n+1), n\in\Bbb Z$, I took one in particular $\pi/2$, but I'm having a lit of trouble findind its Laurent series, I tried using cosecant but it turned in to very complucated calculations, how do I find it? And also is it correct to just take $\pi/2$?
 A: Your work for g is correct, showing only one pole - a single pole at z=1. Here's the approach for f:
f has poles when $\sin(z)=1$, i.e. $z=\pi(2n+1/2)$ and these are all similar by sin's periodicity.
Consider $z=\pi/2$, i.e. $w=0$ where $w=z-\pi/2$
$$
f(z)=\frac{1}{1-\sin(w+\pi/2)}
$$
$$
\sin(w+\pi/2)=\cos(w)
$$
$$
\cos(w)=1-w^{2}/2+O(w^4)
$$
$$
\therefore f(z)=\frac{1}{1-\cos(w)}=\frac{1}{1-1+w^{2}/2 + O(w^4)}=\frac{1}{w^{2}}\frac{1}{1/2 + O(w^2)}=\frac{h(w)}{w^{2}}
$$
Because h is holomorphic at $w=0$, this pole at $w=0$ is a double pole and so all of f's poles are double.
A: In the first case the singularity is a simple pole (i.e. a pole of multiplicity $1$), while in the second case all poles are double (i.e. of multiplicity $2$). 
The principal part of the Laurent series of $g(z)$ around $z=1$ is $\frac{-1}{z-1}$, not $0$.
Since $f(z)$ is periodic and all its (potential) singularities differ by multiples of the period $2\pi$, it suffices to examine just one of them, say $z=\pi/2$, as you suggest. Consider the reciprocal function $F(z) = 1-\sin z$. Observe that $F'(z) = -\cos z$ vanishes at the zero $z = \frac{\pi}{2}$ of $F(z)$, while $F''(z) = \sin z$ does not. This means that in a neighbourhood of $z=\frac{\pi}{2}$ we can write $F(z) = \left(z-\frac{\pi}{2}\right)^2G(z)$ with $G(z)$ holomorphic and $G(\frac{\pi}{2}) \neq 0$. The non-vanishing of $G(z)$ at $\frac{\pi}{2}$ implies that $1/G(z)$ is holomorphic around $\frac{\pi}{2}$ and thus, we can write 
$$ \frac{1}{G(z)} = a_0 + a_1\left(z-\frac{\pi}{2}\right) + a_2\left(z-\frac{\pi}{2}\right)^2 + \cdots$$
with $a_0, a_1, a_2, \ldots \in \mathbb{C}$ and $a_0 \neq 0$. Therefore,
$$ f(z) = \frac{1}{\left(z-\frac{\pi}{2}\right)^2}\cdot \frac{1}{G(z)} = \frac{a_0}{\left(z-\frac{\pi}{2}\right)^2} + \frac{a_1}{\left(z-\frac{\pi}{2}\right)} + a_2 + \cdots$$
This shows that $z=\frac{\pi}{2}$ is a double pole of $f(z)$. 
