Analytic Functions Prove or give a counter-example:
If $f_j(j=1,2,...,n)$ is analytic on the domain $D$ such that $\sum_{j=1}^n |f_j(z)|^2$ is constant on $D$. Then each $f_j$ is a constant function.
Inputs:
We know that, if $f$ is analytic on a domain D and if $|f|$ is a constant then $f$ must also be a constant.
I tried to prove the problem, but I am having a real difficult time. Starting with 
$\sum_{j=1}^n |f_j(z)|^2=k $, where $k\in\Bbb C$  is a constant
$\sum_{j=1}^n (u_j^2 +v_j^2)=k$, (As $f_j=u_j+iv_j$)
Then I differentiated the above equation partially w.r.t $x$ as well as $y$ and tried a lot of manipulations and used CR equations in order to prove it. But I have had no success.
Now I am wondering whether the above result is at all true? Is there a counter-example for the problem?
 A: We may suppose wlog that $0\in D$, and $B(0,R)\subset D$. Put $f_j(z)=\sum_{n\geq 0} a_{n,j}z^n$, that is convergent in $B(0,R)$. Let $0<r<R$. We have
$$\int_0^{2\pi}|f_j(r\exp(i\theta)|^2d\theta=\int_0^{2\pi}\sum_{n,m}a_{n,j}\overline{a_{m,j}}r^{n+m}\exp(i(n-m)\theta)d\theta=2\pi\sum_{n\geq 0}|a_{n,j}|^2 r^{2n}$$
Now if $\sum_{j=1}^m|f_j(z)|^2=c$ is constant, we get that
$$2\pi c=2\pi\sum_{n\geq 0}(\sum_{j=1}^m|a_{n,j}|^2)r^{2n}$$
As this is true for all $r$, $0\leq r<R$, we get that $\sum_{j=1}^m|a_{n,j}|^2=0$, and hence $a_{n,j}=0$, for all $j$ and $n\geq 1$. So each $f_j$ is constant on $B(0,R)$, and it is easy to finish. 
A: Here's another proof of the more general statement mentioned by @MartinR. The domain $D$ is assumed to be connected and the constant equal to $1$. Take a fixed $z_0\in D$ such that $$\sum_{k=1}^n\lvert f_k(z_0) \rvert^2 =1$$ and define the function $f(z)$ by
$$f(z)=\sum_{k=1}^nf_k(z)\overline{f_k(z_0)}.$$
Then


*

*$f$ is holomorphic on $D$

*$f(z_0) = 1$

*$\lvert f(z) \rvert \leq 1$ by Cauchy-Schwarz.


Then $f(z) = 1$ for all $z\in D$ by the maximum modulus principle. Again by Cauchy-Schwarz this means that $(f_1(z), \ldots, f_n(z))$ is a scalar multiple of $(f_1(z_0), \ldots, f_n(z_0))$ for all $z\in D$. Since $f(z)=1$ it follows that this scalar must be $1$ and therefore $f_k(z)=f_k(z_0)$ for $k\in\{1, \ldots, n\}$.
Alternatively, a bit more verbose (using $f(z)=1$ in the second equality):
$$\begin{eqnarray}
1 \geq \sum_{k=1}^n\lvert f_k(z) \rvert^2 &=& \sum_{k=1}^n\lvert f_k(z) - f_k(z_0) + f_k(z_0) \rvert^2 \\
&=& \sum_{k=1}^n\lvert f_k(z) - f_k(z_0)\rvert^2 + \sum_{k=1}^n \lvert f_k(z_0) \rvert^2\\
&=& 1+\sum_{k=1}^n\lvert f_k(z) - f_k(z_0)\rvert^2
\end{eqnarray}$$
and so $f_k(z) - f_k(z_0)=0$ for $k\in\{1, \ldots, n\}$.
