To show that $f = f_1 − f_2$ for $f \in \mathcal{H}(A(0; r_1 ; r_2 ))$

Question

I need to prove the foloowing.

Let $A(0; r_1 ; r_2 ) = \lbrace z \in \mathbb{C} : r_1 < |z_1| < r_2 \rbrace $ where $r_1 < r_2$

Show that if $f \in \mathcal{H}(A(0; r_1 ; r_2 ))$ such that $r_1 , r_2 \in \mathbb{R} \cup \infty $ then $f = f_1 − f_2$ where $f_1 \in \mathcal{H}(A(0; r_1; \infty))$ and $f_2 \in \mathcal{H}(D(0; r_2)$.

My try : I thought of using Maximum modulus on the boundary and then Rouche theorem. I define $g = f - f_1 + f_2$ Now let $\vert f_2 \vert \leq M_2$ for $|z| = r_2$ similarly $\vert f \vert \leq M$ for $|z| = r_2$ for some $M$ and $M_2$

But I dont know what next ...

Please help

Answer 1

If you know Laurent series, we can write $$ f(z) = \sum_{k=-\infty}^\infty a_k z^k = \underbrace{\sum_{k=-\infty}^{-1} a_k z^k}_{{}=f_1(z)} + \underbrace{\sum_{k=-0}^\infty a_k z^k}_{{}=-f_2(z)}, $$ and it's straight-forward to check that $f_1$ and $f_2$ have the desired properties.