Let $C_1$ and $C_2$ be simple closed curves in $\mathbb C$ and assume that $C_2$ is in the interior of $C_1$. Let $U$ be the region bounded by $C_1$ and $C_2$.
Prove that an analytic function $f(z)$ on $U$ can be decomposed as $$ f(z)=f_1(z)+f_2(z) ,$$ where $f_1(z)$ is analytic in the interior of $C_1$ and $f_2(z)$ is analytic in the exterior of $C_2$(including $\infty$). Moreover the decomposition is unique up to an additive constant.
My Attempt:
If $C_1$ and $C_2$ are circles then we know that $f_1(z)$ are the terms in the Laurent series with non-negative power, and $f_2(z)$ are those with negative power. Otherwise, we can imitate the important technique to prove the Laurent series locally so that we can still treat it as is the circle(since within a small angle, it preserves the inequalities and we can write $\frac{1}{\zeta-z}=\frac{1}{\zeta-z_0+z_0-z}=\frac{1}{1+\frac{z_0-z}{\zeta-z_0}}\cdot\frac{1}{\zeta-z_0}$ and use the Talor series of $\frac 1{1-z}$). Then since the closed curve is also compact, we can get finitely many expansions of $f(z)$ at different parts of the curve. Finally, since the Laurent series is unique locally, then we can glue each part to get a Laurent series of $f(z)$ in $U$. Does this work?
Shift $C_1,C_2$ slightly so that $f$ is analytic on an open containing $U$ and them.
From the Cauchy integral formula (the one for simply connected domains, adding an edge to make $U$ simply connected, edge which disappears since traversed in two opposite directions)
For $z$ in between the two simple closed curves $$f(z) = \frac{1}{2i\pi}\int_{C_1} \frac{f(s)}{s-z}ds-\frac{1}{2i\pi}\int_{C_2} \frac{f(s)}{s-z}ds$$ It is immediate that this is your decomposition.
When $C_1,C_2$ are circles expanding $\frac{1}{s-z}$ in geometric series in $s/z,z/s$ is how you get the Laurent series.