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Suppose $f(z)$ and $g(z)$ are entire functions and that $f(z)$ is not constant. If $|f(z)| < |g(z)|$ for all $z \in \mathbb C, $ prove that $f(z)$ can not be a polynomial.

I was thinking what I could do was using the fact $|f(z)| < |g(z)|$ , I can argue $ \frac {|f(z)|} {|g(z)|}< 1$ if $g$ not equal $0$. And, I use the Louiville's Theorem to conclude $ \frac {|f(z)|} {|g(z)|}$ is constant. Then I don't know where to go with that. I think I am not going in the right direction. Please help.

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    oops, there is one thing missing, that equal sign should be removed, let me edit that. Sorry about that.2012-12-30

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You are almost done.

Since $\frac{f(z)}{g(z)}=C$, you get that $f(z)=C g(z)$.

Now, if $f$ is a polynomial, since it is not constant it has some toot $z_0$. Then $0=C g(z_0)$ which implies that $g(z_0)=0$, contradiction.

P.S. You actually get something stronger: You can prove that $f(z)$ has no zeroes. The funny part is that the stronger version probably makes the problem easier, since it guides you towards the last step....

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    @BrettFrankel Since |f(z)| < |g(z)| it is impossible for $g$ to be zero. Note that with $\leq$, one can still prove that $\frac{f}{g}$ is entire, BUT then the rest of the problem is not true ;)2012-12-30