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Let $f(z)$, and $g(z)$ be two analytic functions defined on a region $D\subset\mathbb{C}$. Suppose there exist a constant $r>0$ such that $|f(z)|^2+|g(z)|^2=r$ for every $z$ in $D$. Show that both functions must be constant.

How do I solve this problem? I just need some hints to solve this.

BCLC
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KnobbyWan
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  • Also a special case of https://math.stackexchange.com/questions/289114/show-that-holomorphic-f-1-f-n-are-constant-if-sum-k-1n-left-f. – Martin R Jul 23 '18 at 09:43
  • @MartinR i think it is slightly different since that question only involves functions defined on unit disc, but of course the approach is perfectly identical.. – KnobbyWan Jul 23 '18 at 11:31
  • For generalizations of this see https://math.stackexchange.com/questions/2861422/vector-valued-holomorphic-functions-of-constant-norm . – David C. Ullrich Jul 24 '18 at 18:15

2 Answers2

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If you apply the Wirtinger derivative $\frac{\partial}{\partial \overline{z}}$ on both sides of $r=|f|^2+|g|^2=f\overline{f}+g\overline{g}$ you get $$f\overline{f'}+g\overline{g'}=0$$

where you need to take into account the product rule and Cauchy-Riemann equations: $\frac{\partial}{\partial \overline{z}}f=\frac{\partial}{\partial \overline{z}}g=0$

Therefore, $-\frac{f}{g}=\frac{\overline{g'}}{\overline{f'}}$

The left hand side is analytic, where $g\neq0$, and the right hand side is anti-analytic, when $f'\neq0$. Therefore, at all those points both of those quotients are constant.

If $f$ is a constant multiple of $g$, then $|f|^2$ is constant.

  • Nice solution. but is there a more elementary way of showing this without involving Wirtinger derivative? – KnobbyWan Jul 23 '18 at 05:44
  • @KnobbyWan Wirtinger derivative is just notation. All that has been done here is using Cauchy-Riemann equations, but using a notation that reduces the writing in half. –  Jul 23 '18 at 09:32
  • @KnobbyWan There's another answer below this one, with what might be regarded as a more elementary solution... – David C. Ullrich Jul 23 '18 at 13:11
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Assume wlog $\overline{D(0,1)}\subset D$, just to simplify the notation. Say $f(z)=\sum a_n z^n$ and $g=\sum b_n z^n$.

Then $$|a_0|^2+|b_0|^2=r$$and $$\sum|a_n|^2+\sum |b_n|^2=\frac1{2\pi}\int_0^{2\pi}(|f(e^{it})|^2+|g(e^{it})|^2)\,dt=r,$$ so $a_n=b_n=0$ for $n>0$, hence $f$ and $g$ are constant (in the unit disk, and hence constant in $D$).

David C. Ullrich
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