Finding Laurent's series of a function I am trying express the function $$f(z)=\frac{z^3+2}{(z-1)(z-2)}$$ like a Laurent's series in each ring centering in $0$, but I do not now how could I express it, in first I said that $$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
Ok, now, I see two posibilities: $A\equiv open\ ring=A(C;r,R)$ where $$C\equiv center\\ r<R$$
so, I see: 
$a)$ $A(0;1;2)$
$b)$ $D(0,1)\cup\{\mathbb{C}-\overline{D}(0,2)$
It is correct? Ok, then I suposse that $a)$ I think that I have to try express like series of potence with $\frac{1}{z}$, and in the second case, $b)$, in $\overline{D}(0,1)$ like potences of $z$ and in $\mathbb{C}-\overline{D}(0,2)$ like potences of $\frac{1}{z}$ but I can not, I need someone help, please.
All I know is that $\sum_{0}^{\infty} z^n=\frac{1}{1-z}$ when $|z|<1$
I do not know if it is so, I need someone clarify my doubts

You have to find possible Laurent series expansion for each case !
Case I : $0<|z|<1$
Case II : $1<|z|<2$
Case III : $2<|z|$
Case I : $0<|z|<1$
$$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
$$\Rightarrow f(z)=(z^3+2)\left[\frac{1}{1-z} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
For all $|z|<1$ , $$\displaystyle \frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$ 
And $\displaystyle  |z|<1 \Rightarrow \frac{|z|}{2}<1$. 
So $$\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^{\infty}\frac{z^n}{2^n}$$
Therefore for each $|z|<1$ , $$f(z)=(z^3+2)\left[ \sum_{n=0}^{\infty}z^n-\frac{1}{2}\cdot\sum_{n=0}^{\infty}\frac{z^n}{2^n} \right]$$
$$f(z)=\left[ \sum_{n=0}^{\infty}z^{n+3}-\frac{1}{2}\cdot\sum_{n=0}^{\infty}\frac{z^{n+3}}{2^n} +2\sum_{n=0}^{\infty}z^{n}-\cdot\sum_{n=0}^{\infty}\frac{z^{n}}{2^n}\right]$$
Case II : $1<|z|<2$
Now consider $1<|z|<2$ , 
Then $$f(z)=(z^3+2)\left[\frac{1}{1-z} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
$$\Rightarrow f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
Observe that $1<|z| \Rightarrow \frac{1}{|z|}<1$
So $$\frac{1}{1-\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{1}{z^n}$$
And clearly for $|z|<2$ , $$\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^{\infty}\frac{z^n}{2^n}$$
Thus for each $1<|z|<2$
$$f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
$$f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\sum_{n=0}^{\infty}\frac{1}{z^n} -\frac{1}{2}\cdot \sum_{n=0}^{\infty}\frac{z^n}{2^n}\right] $$
Case III : $2<|z|$
Now consider $2<|z|$.
Clearly $2<|z|\Rightarrow \frac{2}{|z|}<1 \Rightarrow \frac{1}{|z|}<1$ 
Now change $f(z)$ .
$$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
$$f(z)=(z^3+2)\left[\frac{1}{z}\cdot \frac{1}{1-\frac{2}{z}}-\frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}} \right] $$
Clearly for each $2<|z|$ , 
$$\frac{1}{1-\frac{2}{z}}=\sum_{n=0}^{\infty}\frac{2^n}{z^n}$$
So
$$f(z)=(z^3+2)\left[\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{2^n}{z^n}-\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{1}{z^n}  \right] $$
