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
 A: 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
$$f(z)=(z^3+2)\left[\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{2^n}{z^n}-\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{1}{z^n}  \right] $$
A: Executing the polynomial division we have
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1+\frac{3z+1}{(z-1)(z-2)},
$$
and with partial fractions
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1-\frac{4}{z-1}+\frac{7}{z-2}.
$$
This can be rewritten as
$$
f(z)=1+4\cdot\frac{1}{1-z}-\frac{7}{2}\cdot\frac{1}{1-\frac{z}{2}}.
$$
Using the geometric series known expansion 
$$
\frac{1}{1-z}=\sum_{n=0}^\infty z^n,\quad|z|<1
$$
we have
$$
f(z)=1+4\cdot\sum_{n=0}^\infty z^n-\frac{7}{2}\cdot\sum_{n=0}^\infty \left(\frac{z}{2}\right)^n=1+\sum_{n=0}^\infty\left(4-\frac{7}{2^{n+1}}\right)z^n.
$$
This expansion is only valid in $|z|<1$.
For $1<|z|<2$ we have to use the Laurent series, whose coefficients are given by the formula
$$
a_n=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z^{n+1}}dz=\sum_k\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},z_k\right],
$$
where the curve can be taken as a circle centered in $z=0$ and with radius $1<r<2$, and the sum is extended to the residues of the function inside the circle.
The same formula, applied to the circle $|z|<1$, where we already have found the coefficients, provides
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},0\right]=\begin{cases}
\displaystyle{1+4-\frac{7}{2}}, & n=0, \\
\\
\displaystyle{4-\frac{7}{2^{n+1}}}, & n > 0.
\end{cases}
$$
Furthermore
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},1\right]=\lim_{z\to1}(z-1)\frac{f(z)}{z^{n+1}}=\left.\frac{z^2+3}{z^{n+1}(z-2)}\right|_{z=1}=-4,
$$
So, for $1<|z|<2$
$$
a_n=\begin{cases}
-4, & n < 0,\\
\\
\displaystyle{1-\frac{7}{2}}, & n=0, \\
\\
\displaystyle{-\frac{7}{2^{n+1}}}, & n > 0,
\end{cases}
$$
i.e.
$$
f(z)=-4\sum_{n=-1}^{-\infty}z^n+1-\sum_{n=0}^\infty\frac{7}{2^{n+1}}z^n=-4\sum_{n=1}^{\infty}\frac{1}{z^n}+1-\sum_{n=0}^\infty\frac{7}{2^{n+1}}z^n.
$$
Finally, for $|z|>2$ we have to add the residue in $z=2$, given by
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},2\right]=\lim_{z\to2}(z-2)\frac{f(z)}{z^{n+1}}=\left.\frac{z^2+3}{z^{n+1}(z-1)}\right|_{z=2}=\frac{7}{2^{n+1}},
$$
so that
$$
f(z)=-\sum_{n=-1}^{-\infty}\left(4-\frac{7}{2^{n+1}}\right)z^n+1=1-\sum_{n=1}^{\infty}\left(4-7\cdot 2^{n-1}\right)\frac{1}{z^n}.
$$
