0

Prove or give a counterexample to this:

$f(z)$ is an analytic function on domain D containing some circle $\lvert z \rvert = r$. If $\lvert f(z) \rvert$ remains constant along $\lvert z \rvert = r$, then $f(z)$ is of this form $f(z) = c\cdot z^n$ where $c$ is a complex constant.

Intuitively I think it's right, but I don't know how to prove it.

Community
  • 1
Hongyan
  • 283
  • The question as stated isn't correct, e.g. $f(z) \equiv 1$. Also, what is $D$? What if it doesn't contain any circle $|z| = r$? You probably need to assume that $0 \in D$. – John Samples Oct 03 '17 at 07:00
  • $\mathbb{D}$ is a fairly standard way to represent the unit ball, usually open. –  Oct 03 '17 at 07:05
  • for $f(z) \equiv 1$, just let c = 1 and n = 0. – Hongyan Oct 03 '17 at 07:06

1 Answers1

2

Hint for a counterexample. Consider the map $$B(z)=e^{i\theta}\prod_{k=1}^n\frac{z-a_k}{1-\overline{a}_k z}\quad \mbox{with $|a_k|<1$, for $k=1,\dots,n$ }$$ ($B$ is called finite Blaschke product).

Show that $|B(z/r)|=1$ when $|z|=r$. It suffices to verify that for $|z|=1$, $$\left|z-a_k|=|{1-\overline{a}_k z}\right|.$$

P.S. As remarked by mrf, if $D=\mathbb{C}$ then the property holds. For a proof see Characterizing nonconstant entire functions with modulus 1 on the unit circle

Robert Z
  • 145,942
  • sorry, haven't learned Blaschke product. is there another way easier? – Hongyan Oct 03 '17 at 06:53
  • Forget the name "Blaschke product". Consider the map $M(z)=\frac{z-a}{1-\overline{a} z}$ and show that $|M(z)|=1$ along $|z|=1$. – Robert Z Oct 03 '17 at 06:58
  • another related question: https://math.stackexchange.com/questions/2455505/lvert-c-n-rvert-frac-max-lvert-z-rvert-r-vert-fz-rvert-rn – Hongyan Oct 03 '17 at 08:15
  • Just for completeness, if $D=\mathbb{C}$, i.e. if $f$ is entire, the conclusion actually holds. – mrf Oct 03 '17 at 08:46
  • @mrf How to prove it? – Hongyan Oct 03 '17 at 09:25
  • @Hongyan See https://math.stackexchange.com/questions/3408/characterizing-nonconstant-entire-functions-with-modulus-1-on-the-unit-circle – Robert Z Oct 03 '17 at 09:31
  • If I reinforce the condition: along every circle $\lvert z \rvert = r$ in $\Bbb D$, $\lvert f(z) \rvert$ remains constant. Is there any chance of the statement being true? – Hongyan Oct 03 '17 at 12:51
  • See https://math.stackexchange.com/questions/95570/does-constant-modulus-on-boundary-of-annulus-imply-constant-function?rq=1 So the answer is yes as soon as the function is constant on $|z|=r_1>0$ and on $|z|=r_2>r_1$ and it does not vanish in the annulus. – Robert Z Oct 03 '17 at 13:15