Suppose $f$ is analytic on the ring $D(0,1;2)$ Through the Laurent series of $f$ prove that there exist $f_1$ analytic on $D(0;2)$ and $f_2$ analytic on the ring $D(0;1,+\infty)$ such that $f=f_1+f_2$ on the initial ring $D(0,1;2)$ .
I dont know even how to start .I know that having the laurent series the function will look like 2 polynomials one with negative powers of $z$ kai one with positive powers of $z$(taylor expansion) those 2 polynomials might do the work since they are analytic inside the radius of convergence but i have no idea if its the correct way to go .And i dont have any more info. and i dont know how to write those things down.
You are on the right path.
If $f$ is analytic on the ring $D(0;1,2)$, there exists a Laurent series expansion by the Laurent's theorem: $$f(z)=\sum_{n\in \mathbb{Z}}a_nz^n$$ for some $a_n\in \mathbb{C}$.
Let $f_1$ and $f_2$ be the following function: $$f_1(z)=\sum_{n=0}^\infty a_nz^n$$ $$f_2(z)=\sum_{n=1}^\infty a_{-n}z^{-n}$$
Then $f=f_1+f_2$ and $f_1$ is analytic on the disc $|z|<2$. The function $f$ is analytic for $1<|z|<2$, so the radius of convergence of the power series $f_1$ is greater or equal to $2$, otherwise the function $f$ would not be defined in the ring. One of the properties of the radius of convergence is that a power series converges if $|z|<\rho_1$, where $\rho_1$ is the radius of convergence of the power series. This proves the convergence on the disk $D(0;2)$ of the power series $f_1$.
For $f_2$, substitute $w=z^{-1}$. The power series $$\sum_{n=1}^\infty a_{-n}w^{n}$$ have a radius of convergence greater or equal to $1$, otherwise the function $f$ would not be analytic on the ring. For the same reason as the construction of $f_1$, the power series is convergent if $|w|<1$ (or $|z|>1$).