0
$\begingroup$

My thoughts so far: since the zeros of $f$ are isolated, $f(z)$ never assumes the value zero. It follows that $f(z) = e^{g(z)}$ for some entire function $g$, and that $Re(g(z))$ depends only on $Re(z)$ only. The problem then reduces to showing that $g(z) = uz+v$, with $u$ real. It may or may not be easier to show this than to show that $f$ has the given form directly. Intuitively I doubt that it is possible for $g$ to be a polynomial of higher degree, although I have yet to prove this. Therefore I'd be satisfied with a proof that assumes that $g$ is not a polynomial and reaches a contradiction.

1 Answers 1

0

Nevermind, I figured out a proof as soon as I posted this. Note that $g(z + ih) = g(z) + g'(z)ih + o(h)$. If $h$ is real and $|h|$ is sufficiently small, then it follows that $Re(g(z+ih)) = Re(g(z))$ iff $g'(z)$ is real. We therefore conclude $g'(z)$ is a purely real entire function, which is constant by the Cauchy-Riemann equations. The result follows.

Alternatively you could try the same approach with $f$ directly. We have $f(z + ih) = f(z) + f'(z)ih + o(h)$, where $h$ is real. As $h$ goes to zero, the only way that $|f(z+ih)| = |f(z)|$ is if $arg(f'(z)) = \pm arg(f(z))$ for all $z$. By the polar Cauchy-Riemann equations $f'(z) = af(z)$ for some real $a$, and of course you can conclude that $f(z) = ce^{az}$ for some constant $c$ by noting the uniqueness of solutions to that differential equation.

  • 0
    Yes. I would look at $g(z) = \log f(z)$ on a domain where $f(z)$ doesn't vanish, so that $Re(i g'(x+iy)) = \frac{\partial }{\partial y}Re( g(x+i y)) = 0$ which means that $g'(z)$ is real, so it is constant and $g(z) = az+b, f(z) = c e^{az}$ – 2017-01-30
  • 0
    I established in the beginning that $f(z)$ is never zero, so this is essentially exactly what I did by writing $f(z) = e^{g(z)}$ for an entire function $g$ and then looking at properties of $g$. – 2017-01-30