To show each $f_i$ is bounded so for $\Omega$ a connected open subset in $\mathbb{C}$ each $f_I \in \mathcal{H}(\Omega)$ for all $j = 1,2,3...n $ such that $\sum |f_j|^{2}$  is constant on $\Omega$ then I need to show that each $f_j$ is a constant on $\Omega$ 
so from  $\sum |f_j|^{2}$  is constant we can say that it is bounded by some $M$ so each $|f_j|^{2} \leq M $ 
But i dont know what next ! 
 A: Let $\sum_{i=1}^{n}\lvert f_i\rvert^2=c$. If $c=0$, then we are done, so assume $c > 0$.
If $n=1$, then writing $f_1=u+iv$ with $u,\ v$ real, we note that $u^2+v^2=c$. Taking derivatives w.r.t. $x$ and $y$ we get:
$$u\frac{\partial u}{\partial x} + v\frac{\partial v}{\partial x} = 
u\frac{\partial u}{\partial y} + v\frac{\partial v}{\partial y} = 0 \tag{1}$$
Using the Cauchy-Riemann equations
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\ 
\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \tag{2}$$
we get the following:
$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} = 0 \tag{3}$$
Since $\Omega$ is connected, this implies, $u = v =$ constant, i.e. $f_1 =$ constant.
For $n > 1$, choose any $w\in\Omega$ and define the holomorphic function:
$$ f(z)=\sum_{i=1}^nf_i(z)\overline{f_i(w)} \tag{4} $$
Then by Cauchy-Schwarz,
$$ \lvert f(z)\rvert^2\leq\left(\sum_{i=1}^n\lvert f_i(z)\rvert^2\right) \left(\sum_{i=1}^n\lvert f_i(w)\rvert^2\right)\leq c^2 \tag{5}$$
so that $\lvert f(z)\rvert\leq c$. But $\lvert f(w)\rvert=c$ and $w$ is an internal point of $\Omega$ so by maximum modulus
$$\lvert f(z)\rvert=c\ \forall\ z\in\Omega \tag{6}$$
Then $f$ is a holomorphic function on connected $\Omega$ with $\lvert f\rvert^2=$ constant, so using the $n=1$ case, we conclude $f=$ constant. Thus $f(z)=f(w)=c$.
Equation $(6)$ also implies we actually have equalities in $(5)$, and using the condition of equality in Cauchy-Schwarz, we get for all $z\in\Omega$:
$$ (f_1(z),\ldots,f_n(z))=\alpha(z)(f_1(w),\ldots,f_n(w)) \tag{7}$$
Use this to evaluate $f(z)$
$$ f(z) = \sum_{i=1}^{n}f_i(z)\overline{f_i(w)} = \alpha(z)f(w) \tag{8}$$
Using $f(z)=f(w)=c\neq 0$, we conclude $\alpha(z)=1$ for all $z\in\Omega$, so that $(7)$ implies $f_i(z)=f_i(w)$ for all $z\in\Omega$ and $1\leq i\leq n$, i.e. $f_i$'s are constant.
NOTE: This is not my solution, at least not completely. The question appeared in my Complex Analysis midterm, and professor gave a hint towards the solution, I just filled in the blanks.
