TITLE: Show that holomorphic $f_1, . . . , f_n $ are constant if $\sum_{k=1}^n \left| f_k(z) \right|$ is constant.
QUESTION [21 upvotes]: 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.

REPLY [4 votes]: 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.