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The function $f$ is constant

Posted: Mon Nov 09, 2015 1:00 pm
by Tolaso J Kos
Let \( f \) be an entire function across the complex plane. If \( \mathfrak{Im}(f(z))>\mathfrak{Re}^2 (f(z))-2 \) holds, then prove that \( f \) is constant.

Re: The function $f$ is constant

Posted: Mon Nov 09, 2015 2:13 pm
by Demetres
This is immediate by the little Picard theorem which says that the range of every non-constant entire function is either the whole of the complex plane, or the complex plane minus one point.

Here however the range of $f$ does not contain any point with imaginary value less than or equal to $-2$. Since $f$ is entire, the only possibility left is that $f$ is constant.

Re: The function $f$ is constant

Posted: Thu Sep 01, 2016 2:24 pm
by S.F.Papadopoulos
Is easy to prove:
The range of entire no constant function is dense.