# Entire function real and imaginary part product

Let $f$ be an entire function such that $\Re(f(z))*\Im(f(z))\ge0$ for all $z$ then $f$ is constant. Prove or give contradicting example. I know about Louisville's theorem and Cauchy–Riemann equations but I don't see how to use them in this situation. I've tried to bound $|f(z)|$ but I could only show that $|f(z)|\le\Re(f(z))+\Im(f(z))$. I would like a hint.

• I don't suppose you know Picard's Little Theorem. – Michael Albanese Jun 14 '14 at 16:41
• I do know about that but I cannot use it for this. – Gyt Jun 14 '14 at 16:42
• @Gyt Is it specifically forbidden? Is this a homework problem? – user122283 Jun 14 '14 at 16:42
• No, I study for an exam and I can't use this theorem. – Gyt Jun 14 '14 at 16:43
• Can you find an entire function whose imaginary part is $\Re f(z) \cdot \Im f(z)$? – Daniel Fischer Jun 14 '14 at 16:47

Write $f(z) = u(z) + iv(z)$, so that $u=\Re(f)$ and $v = \Im(f)$. The trick to many of these problems is to find a suitable auxiliary function: as Daniel Fischer mentioned, a good candidate would be a function whose real or imaginary part is $uv$. How about $g(z) = f(z)^2 = u(z)^2-v(z)^2 + 2i\,u(z)v(z)$? By hypothesis, $\Im(g(z))\geq 0$ for all $z$ . . .
• Do we know that if $f(z)^2$ is constant, then $f(z)$ has to be constant, too? – el_tenedor Mar 2 '17 at 7:43
• @el_tenedor Since $f(z)^2$ is constant, $f(z)$ takes on at most two values. But since $f$ is continuous, it takes on exactly one of these values -- that is, $f$ is constant. Does that make sense? – user134824 Mar 4 '17 at 16:01
• This argument should work as long as $f$ is defined on a domain, since if the space is connected, every locally constant continuous function is globally constant. $f$ is entire, so there should be no problems, thanks! – el_tenedor Mar 4 '17 at 16:10
how to show image of a non constant entire function is dense in $\mathbb{C}$?
• Is it possible to prove this for the special case where $\mathbf{Re}(f(z))*\mathbf{Im}(f(z)) = 0$ using the Cauchy-Riemann equations only? – ignoramus Aug 25 '15 at 5:05