Let $f$ be an entire function s.t. $\forall z \in \mathbb{C}, f(z)\text{ is real}\iff z\text{ is real}$. Prove that $\exists a,b \in \mathbb{R}$ s.t. $a\neq 0$ and $\forall z \in \mathbb{C}, f(z)=az+b$.

I think I have a proof for this using some advanced theorems, but I want to find one that can be done by elementary means (by a student in undergraduate complex analysis).


A sketch of the not so simple proof:

Let $f=u+iv$. Assume $f$ is not constant.

All the points above the real line must have the imaginary part of $f$ with the same sign. Similarly for the points below.

If $u$ is constant on the real line then $f$ is constant (contradiction).

$v_y=u_x\neq 0$ somewhere on the real line.

So points above the real line must all be of different sign from the points below.

Hence $u_x$ must stay the same sign throughout the real line, and hence $u$ is monotonic on the real line. If $u$ is constant in some interval, then $f$ is constant (contradiction). So $u$ is strictly monotone on the real line.

If $f$ is a polynomial of degree at least 2, then $f$ has at least 2 roots. These roots must all be the same, so we can write $f(z)=c(z-a)^n$ and then either $(z-a)^n=-c^{-1}$ or $(z-a)^n=c^{-1}$ has imaginary solutions (contradiction).

Otherwise, $f(1/z)$ has an essential singularity at 0 and by the great Picard's theorem there are at least two occurrences of some real number not equal to the (possibly) one excluded by Picard's theorem (contradiction).

  • $\begingroup$ $f$ is equal to $u$ on the real line! $\endgroup$ Nov 30 '14 at 12:26
  • $\begingroup$ yeah you are right, sry. $\endgroup$
    – sranthrop
    Nov 30 '14 at 12:30
  • $\begingroup$ But there was a flaw, which I just corrected. $\endgroup$ Nov 30 '14 at 13:04

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