1
$\begingroup$

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Oh Yess! Thanks. This gives it directly ! $\endgroup$
    – Rusty
    Commented Mar 30, 2015 at 14:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .