f and g are entire and $ \left|f(z)\right| \leq \left|g(z)\right| $ then $\exists \alpha\in \mathbb{C}$ such that $f(z)=\alpha g(z)$

Asked Nov 27 '16 at 17:42

Active Nov 27 '16 at 17:42

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If f and g are entire and $\left|f(z)\right| \leq \left|g(z)\right| $ in $\mathbb{C}$ then $\exists \alpha\in \mathbb{C}$ such that $f(z)=\alpha g(z)$ $ \forall z \in \mathbb{C}$ with $\left|\alpha\right| \leq 1$

Is it possible to use without residues and so? I haven't learned neither Laurent series nor residues. This question is from a final exam so I don't know if im supposed to be able to prove it right now or i have to wait to learn those other tools before i can prove it. Any tips are appreciated.

asked Nov 27 '16 at 17:42

Rafa

1

The usual proof uses Riemann's theorem on removable singularities. – Robert Israel Nov 27 '16 at 17:49
Okay, thanks, and do you know where can i find that proof? – Rafa Nov 27 '16 at 18:04
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Whoops, didn't found that, it is a duplicate indeed. – Rafa Nov 27 '16 at 18:09

f and g are entire and $ \left|f(z)\right| \leq \left|g(z)\right| $ then $\exists \alpha\in \mathbb{C}$ such that $f(z)=\alpha g(z)$

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