If $f,g$ analytic and $|f| = |g|$, prove that $f(z) = ag(z)$

Question

If $f,g$ are a pair of analytic functions on $U$, such that $|f| = |g|$, prove that $f(z) = ag(z)$. Where $a$ is a constant such that $|a| = 1$.

My attempt: I know I need to define an $h(z)$ as the quotient of the two fucntions but I am stuck there. Do I use the zeroes of the functions as was done in this problem here. If so, why can I assume that they have a zero in the domain $U$. Help needed.

When you say analytic, do you mean holomorphic ($f,g:\mathbb{C}\to\mathbb{C}$)? And what properties does $U$ satisfy? Is it open? Closed? ... — Daniel Robert-Nicoud, Nov 06 '15 at 13:44
If $g \equiv 0$ on some component of $U$, it follows immediately that also $f \equiv 0$ there. Otherwise, the zeros of $g$ are removable singularities of $f/g$. — Daniel Fischer, Nov 06 '15 at 13:46
I believe based on the question statement, $f,g: \Bbb C \to \Bbb C$ and we can only assume they are holomorphic on $U$ which is open. — Meecolm, Nov 06 '15 at 13:47

score 2 · Accepted Answer · answered Nov 06 '15 at 13:45

2

Hint: Use Liouville's theorem on $\tfrac{f}{g}$.

answered Nov 06 '15 at 13:45

Daniel Robert-Nicoud

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Oh wow, I cannot believe I did not see that. It follows immediately from L.T. then by defining $h =f/g$, $h$ bounded by $1$ $\implies h = a$. Then $f = ag$, correct? – Meecolm Nov 06 '15 at 13:53
@User52525 Yes, but you have some work to do to show that the quotient is well defined (see Daniel FIscher's comment to your original question). – Daniel Robert-Nicoud Nov 06 '15 at 13:55
Yes, and this is very similar to the argument as in the linked post I believe. – Meecolm Nov 06 '15 at 14:00
$f$ and $g$ are analytic functions defined in a region $U$. How can Liouville's theorem be applied? – Umberto P. Nov 06 '15 at 14:35
@UmbertoP. The quotient is bounded by $1$, so it must be constant by Liouville's theorem. This immediately gives the result. – Daniel Robert-Nicoud Nov 07 '15 at 02:13
What if $U \not= \mathbb C$? – Umberto P. Nov 07 '15 at 05:02
@UmbertoP. Yeah, ok, I admit that technically the theorem you should apply is the maximum modulus principle. The fact is that I also call that one Liouville's theorem... – Daniel Robert-Nicoud Nov 07 '15 at 11:19
@DanielRobert-Nicoud Don't we need to look into the multiplicity of the zeros as well? – Kashif Sep 18 '21 at 06:09

If $f,g$ analytic and $|f| = |g|$, prove that $f(z) = ag(z)$

1 Answers1