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$ r\in\mathbb{R}_{>0} $. Let $ f,g:\overline{U_r(0)}\rightarrow\mathbb{C} $ be continuous and without zero. Let $ f,g $ be holomorphic on $ U_r(0) $ and $ |f(z)|=|g(z)| $ for all $ z\in\partial U_r(0) $. Show that there is a $ \lambda\in\mathbb{C} $ with $ |\lambda|=1 $, such that $ f=\lambda g $.

I am trying to solve this and I have seen a hint that applying the maximum modulus principle to $f/g$ implies $f/g$ constant and thereby solves the exercise. But I fail to apply. I guess it's about this version of the maximum modulus principle:

If $D$ is a bounded domain and $f$ is holomorphic on $D$ and continuous on its closure $\bar D$, then $|f|$ attains its maximum on the boundary $∂D := D \backslash \bar D$.

This tells us only that $f/g$ attains its maximum on $\partial U_r(0)$, which we know to be $1$. So $|(f/g)(z)| \leq 1$ for all $z \in U_r(0)$. But how do we know that $f/g$ is constant?

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    $\begingroup$ Note that you can apply the MMP to both $f/g$ and to $g/f$ ... $\endgroup$
    – Martin R
    Jul 11, 2023 at 9:19
  • $\begingroup$ Do you also know that a nonconstant holomorphic functions attains its maximum only on the boundary? $\endgroup$
    – Martin R
    Jul 11, 2023 at 9:23
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    $\begingroup$ @MartinR thanks! I think I got it: Suppose $|(f/g)(z)| = c <1$ for some $z \in U_r(0)$. But then $(g/f)(z)=1/c>1$ which is a contradiction since we can apply MMP to $g/f$ as well. So we get $(f/g)(z)=1$ for all $z\in U_r(0)$. So it takes its maximum in $U_r(0)$, so by applying MMP again (the "normal form" this time), we get that $f/g$ is constant, which solves the exercise. Is this correct? $\endgroup$ Jul 11, 2023 at 9:39
  • $\begingroup$ Almost. It should be $|(f/g)(z)|=1$ etc. $\endgroup$
    – Martin R
    Jul 11, 2023 at 10:57

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You have already solved the exercise in the comments, I present a solution which is quite similar.

Let $f_1 = f/g$, $f_2 = g/f$ both of these functions are holomorphic and thus will attain maxima at the boundary. However both of them are $=1$ on the boundary thus.

$$|f_1| \le 1 \implies |f| \le |g|$$ $$|f_2| \le 1 \implies |g| \le |f|$$

Thus we have $|f/g| = 1$

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