5

Let $f : \mathbb C \to \mathbb C$ be a holomorphic function. Suppose that for some $a<b<c \in \mathbb R$ we have $$ |e^{f(z)}| = 1 $$ for all $z = x+iy, x \in \mathbb R, y \in \{ a,b,c \}$. That is, $|e^f|$ is constant one for such $x$ and $y$. Can we conclude that $e^f$ is globally constant, i.e. $f(z)=iu$ for some real $u$? I was thinking to apply a Phragmen Lindelöf argument, but failed.

Marius S.L.
  • 2,245
user975628
  • 125
  • Use identity theorem. – MathFail Jul 03 '23 at 19:35
  • 2
    Not clear how the identity theorem can help here. – copper.hat Jul 03 '23 at 19:47
  • 2
    I think that schwarz reflection principle implies there are infinitely many such parallel lines and $e^{f}$ would be periodic – Keplerto Jul 03 '23 at 20:03
  • @Keplerto Thanks for your comment! If e.g. $a=0$ and $b=1$ then the function $|e^f|$ would satisfy $|e^{f(\cdot + i)}| = e^f$ using schwarz reflection principle? Is this what you mean with periodicity? So $|e^f|$ is one-periodic in direction of the imaginary axis? – user975628 Jul 06 '23 at 17:32
  • @user975628 I think in fact you can get that $f$ is periodic (it is in the given answer). If $a=0, b=1$ then from the assumption, $$\Im z = 0, 1 \implies \Re f(z) = 0,$$ so by Schwarz principle you get $f(\overline z) = \overline{f(z)}$ and $f(\overline z) = \overline{f(z+2i)}.$ Thus $$f(z+2i) = f(z).$$ So yes, $f$ is periodic in the imaginary direction with period $2i$, and not just $e^f$ – Keplerto Jul 06 '23 at 18:03

1 Answers1

8

Write $f(x+yi) = u(x,y) + v(x,y)i$. Suppose $a=-\pi, b=0, c=\pi$. Consider $u(x,y)=\exp(x)\sin(y)$. One can check that $u$ is harmonic, and thus one can find a $v$ such that $f$ is holomorphic. Furthermore, on the prescribed lines one can check that $|e^{f(z)}|=1$. So no, $f$ need not be constant.

G. Bellaard
  • 757