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Suppose $f$ and $g$ are entire functions with $|f(z)|\leq|g(z)|$ for all $z$.

How can we show that $f=cg$ for some complex constant $c$?

Thanks for any help :)

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    ok, thanks for the advice2011-07-18

1 Answers 1

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Assume $g(z) \neq 0$. Consider the quotient $\nu(z)=\displaystyle\small\frac{f(z)}{g(z)}$. Then the singularities of $\nu$ are isolated since the zeros of $g$ are isolated. Clearly $\nu(z)$ is bounded in each deleted neighborhood of each zero of $g$. By Riemann's theorem, $\nu$ extends, uniquely to an entire function and using continuity we have $|\nu(z)| \leq 1$ for all $z \in \mathbb{C}$. Now use Liouville's theorem.

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    In fact, I do not think it is referred to as "Riemann's theorem on removable singularities" in Walter Rudin's *Real and Complex Analysis*. Of course, this does not imply that the name "Riemann's theorem on removable singularities" is uncommon ...2011-07-23