# Property of Entire Functions

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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You may want to check out math.stackexchange.com/questions/50421/… – algebra_fan Jul 18 '11 at 8:51
Hint: Use Riemann's theorem of removable singularities. – Hendrik Vogt Jul 18 '11 at 8:52
@all: algebra_fan's proposed link is better than mine, so if you vote for closing this question, please use his. – t.b. Jul 18 '11 at 8:58
@Hendrik: yes, sure. now we have a good answer by Chandru, I see no reason for closure. – t.b. Jul 18 '11 at 9:07

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.
Actually its a neat (and very elementary) result from the fact that entire functions are analytic that $v$ can be extended to the roots of $g$. – Listing Jul 18 '11 at 9:13