If $f = u + iv$ is entire and $|u(z)| > |v(z)|$ for all $z$, is $f$ constant?

Question

What's a good way to approach this problem?

Maybe I want to use Liouville's theorem, so I need to show that $f(z)$ is bounded.

However I don't see how to prove this from the assumption.

What can you say about $\exp(i \log (i f(z)))$? ($if(z)$ avoids the real line, so the standard analytic branch of $\log$ is defined on its range) — ronno, May 18 '15 at 06:01

score 4 · Accepted Answer · answered May 18 '15 at 06:29

4

$f=u+iv$ maps $\mathbb {C}$ into $\{x+iy: |y|< |x|\}.$ The latter set doesn't intersect the $y$-axis. Because $f(\mathbb {C})$ is connected, it must be a subset of either the open right or left half plane. There is a holomorphic bijection $g$ from that half plane onto to the open unit disc. Thus $g\circ f$ is entire and bounded, hence must be constant. Therefore $f$ is constant.

answered May 18 '15 at 06:29

zhw.

105,693

Beautiful! Thanks! – Marcus Emilsson May 18 '15 at 06:37

If $f = u + iv$ is entire and $|u(z)| > |v(z)|$ for all $z$, is $f$ constant?

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