Prove that $ f(z)=\sum_{j=1}^n f_j(z) \quad \text{for all } z\in\Bbb C\setminus\bigcup_{j=1}^n K_j. $

Question

Let $K_1,K_2,\ldots,K_n$ be closed and disjoint discs in $\Bbb C$. Prove that if $f\in H(\Bbb C\setminus\bigcup_{j=1}^n K_j)$, then there are functions $f_j\in H(\Bbb C\setminus K_j)$, $j=1,2,\ldots,n$ such that $$ f(z)=\sum_{j=1}^n f_j(z) \quad \text{for all } z\in\Bbb C\setminus\bigcup_{j=1}^n K_j. $$

My try: I can define functions $f_j\in H(\Bbb C\setminus K_j)$ by Laurent series in the annulus but how to merge it up to show that the series equals $f$. Please help me from the scratch.

Answer 1

Let $i=1$ and $f_{i-1}=f$, $K_i = \{|z-a_i|\le r_i\}$.

Expand $f_{i-1}$ in Laurent series on the annulus $r_i<|z-a_i|<r_i+\epsilon$ you'll get $f_{i-1}(z) = \sum_n c_{i,n} (z-a_i)^n$, let $g_i(z)=\sum_{n\le 0} c_{i,n} (z-a_i)^n$ which converges on $\Bbb{C}-K_i$.

$f_i=f_{i-1}-g_i$ is analytic on $\Bbb C\setminus\bigcup_{l=i+1}^j K_l$. This is because on $|z-a_i|<r_i+\epsilon$ it is given by $\sum_{n>0} c_{i,n}(z-a_i)^n$.

Then repeat with $i=2,3,\ldots, j$.

$f_j$ will be entire and $f= f_j+\sum_{i=1}^j g_i$