While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated.
Let $U ⊂ \mathbb{C}$ be a domain and $f_1, . . . , f_n : U \rightarrow \mathbb{C}$ holomorphic functions, such that $\sum_{k=1}^n \left| f_k(z) \right|$ is constant on $U$. Show that $f_1, . . . , f_n$ are constant.
For simplicity, let's consider $n=2$. In $U$, we have $|f|+|g|=C$ for some constant. Fix a point $z_0$ in $U$. Then, for appropriately chosen unimodular constants
$$|\alpha f(z_0) +\beta g(z_0)|=|f(z_0)|+|g(z_0)|=C.$$
This means the holomorphic function $\alpha f(z) +\beta g(z)$ attains its maximum in $U$ (since the supremum in $U$ is at most $C$ by the triangle inequality, and it attains $C$), so it is a constant. So for all $z$, we have
$$|\alpha f(z) +\beta g(z)|=|f(z)|+|g(z)|.$$
Equality is only possible in the triangle inequality all the vectors point in the same direction. So $c(z)\alpha f(z) = \beta g(z)$ for some real holomorphic $c(z)$. But since real holomorphic functions are constant, $c$ is constant. Then $\alpha f(z) +\beta g(z)=(1+c)(\alpha f)$, and the latter is a holomorphic function that attains its maximum on $U$, so it is constant. So $f$ is constant, and it follows that $g$ is constant.
The result would be easier to show if we would have as assumption that $\sum_{j=1}^n|f_j|^{\color{red}2}$ is constant (then we would take the derivative, and use Cauchy-Riemann equations, or as Julien suggest, the Laplacian).
To deal with this case, consider $z_0\in U$ and $r>0$ such that $B(z_0,r)$ contains none of the zeroes of the $f_k$ (if one is $0$, remove it, and in a ball there are finitely many zeroes of $f_k$). Then on this ball, using a determination of logarithm, we can find for each $k$, $g_k$ such that $f_k=g_k^2$ on $B(z_0,r)$ for some $g\in\mathcal O(B(z_0,r))$.
We get that each $g_k$ is constant on $B(z_0,r)$ (hence so is each $f_k$) and we conclude invoking again principle of isolated zeroes (and connectedness of $U$).
To complete @Potato's solution, where he/she did the $n=2$ case, here's how the induction step can be done (I would have added this as a comment, but it's too long to post it as a comment):
Let $n \geq 3$ and $|f_1(z)| + \dots + |f_n(z)|=C$. Then we can choose an arbitrary $z_0 \in U$ and choose $w_1,\dots,w_n$ such that $|f_i(z_0)| = w_i f_i (z_0)$, $|w_i| = 1$.
Then by the same reason as above (in Potato's answer), $g(z) = w_1 f_1(z) + \dots + w_n f_n(z)$ is constant on $U$ (by the maximum principle because $|g(z)|$ attains a maximum at $z_0$.) In fact $g(z)=C$ for all $z \in U$. So $$C = |g(z)| \leq |w_1 f_1(z) + w_2 f_2(z)| + |f_3(z)| + \dots + |f_n(z)| \leq |f_1(z)| + \dots + |f_n(z)| = C$$ So by the induction hypothesis, the $n-1$ functions $(w_1 f_1 + w_2 f_2) , f_3 , \dots , f_n$ are all constant and by considering that $(w_1 f_1(z_0) + w_2 f_2(z_0) ) = |f_1(z_0)| + |f_2(z_0)|$ we can apply @Potato's reasoning again to conclude that $f_1$ and $f_2$ are also constant.
Here is a different approach at the solution: Let $c = \sum_{i=1}^n |f_i| $ where c is a constant. For any holomorphic function $f$ we have the mean value theorem as $$ f(z_0) = \frac{1}{2\pi} \int_0 ^{2\pi} f(z_0 + re^{i\theta}) d\theta \\ \implies |f(z_0)| \leq \frac{1}{2\pi} \int_0 ^{2\pi} |f(z_0 + re^{i\theta})| d\theta$$ Hence $|f|$ is subharmonic and also $-|f|$ is superharmonic. Now we know that, $ |f_1| = c - \sum_{i=2}^n |f_i|$. This means that $|f_1|$ is both subharmonic as well as superharmonic or in other words $|f_1|$ is harmonic.
Thus the problem reduces to showing that if the modulus of a holomorphic function is harmonic then it is a constant. This can be shown in several ways.
1) One way would be to note that $|f|$ would follow the mean value theorem itself. Hence $$ |f(z_0)| = \frac{1}{2\pi} \int_0 ^{2\pi} |f(z_0 + re^{i\theta})| d\theta $$ Since this is true for any $r>0$ in $D(z_0,r)$, hence $ |f(z_0 + re^{i\theta})| = |f|$ is independent of $r = |z-z_0|$. This signifies that on all rays emanating from $z_0$, $|f|$ would be constant. This would violate the maximum value theorem for $f$ unless it is a constant.
2) A second approach is this: If $f$ is non-constant, then the derivative $f'$ has isolated zeroes. On a small disk away from the zeroes of $f'$, the function $f$ has a holomorphic inverse. The composition of a harmonic function with a holomorphic function is again harmonic, so composing with $f^{-1}$ reduces the problem to showing that the function $|z|$ is not harmonic in a neighborhood of the origin. But this conclusion is evident, since $|z|$ attains a global minimum at the origin.
Now that we have shown $|f_1|$ is a constant, we can keep repeating the process for any $f_i$ from $\tilde{c} = \sum_{i=2}^n |f_i|$ to show that each $f_i$ is a constant.
