If $|f|+|g|$ is constant on $D,$ prove that holomorphic functions $f,~g$ are constant on $D$. 
Let $D\subseteq \mathbb{C}$ be open and connected and 
  $f,~g:D \rightarrow \mathbb{C}$ holomorphic functions such that 
  $|f|+|g|$ is constant on $D.$ Prove that $f,~g$ are constant on $D.$

Attempt. I noticed first that this problem is already set before, and is considered  to be duplicate (If $|f|+|g|$ is constant then each of $f, g$ is constant) and we are sent directly to problem sum of holomorphic functions. However, the last reference deals with the sum $|f|^2+|g|^2$, which I do not see how is connected to the sum $|f|+|g|$, as stated in our title (for example, the equality $|f|^2+|g|^2=(|f|+|g|)^2-2|f||g|$ is not helpful here).
Thanks in advance!
 A: Here is a proof that relies heavily on the open mapping theorem.
Since the zeros of $f,g,f',g'$ are isolated, we can find some $z_0$ such that $f,g,f',g'$
are non zero in a neighbourhood of $z_0$.
Suppose $z$ lies in this neighbourhood of $z_0$.
If we let $\phi(t) = {1 \over 2}|f(z+th)|$, it is straightforward to show that
$\phi'(0) = { \operatorname{re}(\overline{f(z)} f'(z) h ) \over |f(z)|}$, hence since $|f(z)| + |g(z)|$ is constant
we obtain
\begin{eqnarray}
{ \operatorname{re}(\overline{f(z)} f'(z) h ) \over |f(z)|} + { \operatorname{re}(\overline{g(z)} g'(z) h ) \over |g(z)|} &=& \operatorname{re} \left[ \biggl( { \overline{f(z)} f'(z)  \over |f(z)|} + { \overline{g(z)} g'(z) \over |g(z)|} \biggr) h\right] \\
&=& 0
\end{eqnarray}
for all $h$.
Hence, we have ${ \overline{f(z)} f'(z)  \over |f(z)|} + { \overline{g(z)} g'(z) \over |g(z)|} = 0$ and so
${f'(z) \over g'(z)} = -{|f(z)| \over \overline{f(z)} } {\overline{g(z)} \over |g(z)|}$, in particular, $| {f'(z) \over g'(z)}| = 1$. Hence
${f'(z) \over g'(z)} = c$ for some constant and so
${g(z) \over f(z)} = - \overline{c} | {g(z) \over f(z)} | $. Hence the values $z \mapsto {g(z) \over f(z)}$ takes lie on the line $\{t \overline{c} \}_{t \in \mathbb{R}}$ and hence we must have ${g(z) \over f(z)} = d$, another constant. Hence $(1+|d|)|f(z)|$ is a constant and so $f$ is a constant.
A: Here is an answer that will help you to solve the problem the way that you already tried to do it.
First let's enumerate our equation
\begin{eqnarray}
\big|f\big| + \big|g\big| = c. \quad(1)
\end{eqnarray}
It's easy to see that f,g can't be zero on $D$, otherwise if for example there was $z_0\in D$ such as $f(z_0)= 0$ then you have that
$$\big|g(z_0)\big| = c$$
but you also have that
$$\big|g(z)\big|\leq c \quad \forall z\in D$$
so from maximum modulus principal you have that $g=c$ on $D$ so (1) gives us that $f = 0$ on $D$.
So we have that there exists some holomorphic functions  on $D$, $f_1,g_1$  such us
\begin{eqnarray}
&f_1^2=f \\
&g_1^2=g
\end{eqnarray}
(Theorem 13.11 page 274 from Real and Complex Analysis by Rudin).
Now you have what you were looking for, you have that
$$\big|f_1\big|^2+\big|g_1|^2 = c \quad (2)$$
and now all you need is to prove is that $f_1,g_1$ are constant.
Now you can continue as that link you mention, but there is one other way to do it.
We have that $f_1,g_1$ are holomorphic so they have a powerseries expansion
\begin{eqnarray}
&f_1(z) = \sum_{n=0}^{\infty}a_nz^n \\
&g_1(z) = \sum_{n=0}^{\infty}b_nz^n
\end{eqnarray}
so if we integrate on some circle $\gamma$ with centre  $0$ and radious $r$
(I suppose that $\;D = D(0,R)\;$ and $\;0<r<R\;$)
equation (2), Parseval's formula gives us that
\begin{eqnarray}
&\frac{1}{2\pi }\int_{\gamma}|f_1|^2dz + \frac{1}{2\pi }\int_{\gamma}|g_1|^2 dz =\frac{1}{2\pi}\int_{\gamma}cdz \Longrightarrow \\
&c_1 = \sum_{n=0}^{\infty}|a_n|^2r^{2n}+\sum_{n=0}^{\infty}|b_n|^2r^{2n} = \sum_{n=0}^{\infty}\Bigl(|a_n|^2+|b_n|^2\Bigr)r^{2n} \quad \forall r\in(0,R)
\end{eqnarray}
where $\; c_1\in \mathbb C\;$ is a constant, so all $a_n,b_n = 0$ for $n\geq 1$, and you have the result.
A: By the maximum modulus principle $|f(z)|$ has no local minimum and maximum except at its zeros. So if $f(a)=0$ then $a$ is a local minimum of $|f(z)|$ so it is a local maximum of $|g(z)|$ and $g(a) = 0$, i.e. $f(z) = g(z)= 0$.
$\implies$ If $f(z)$ is non-constant then $f(z),g(z)$ have no zeros, and we can look at the holomorphic functions $F(z) = \log f(z), G(z) = \log g(z)$ and $H (z) = e^{Re(F(z))}+e^{Re(G(z))}=C$. 
Differentiating : $$\partial_x H(z) = Re(F'(z)) e^{Re(F(z))}+Re(G'(z))e^{Re(G(z))}=0 $$ $$ \partial_y H(z) = Im(F'(z)) e^{Re(F(z))}+Im(G'(z))e^{Re(G(z))}=0$$
so that $$0= F'(z)e^{Re(F)}+G'(z)e^{Re(G(z))}=(F'(z)-G'(z))e^{Re(F(z))}+G'(z)(e^{Re(F(z))}+e^{Re(G(z))})$$ $$ = (F'(z)-G'(z))e^{Re(F(z))}+C G'(z)$$
Finally $(G'(z)-F'(z))e^{Re(F(z))}=C G'(z)$ means that $(G'(z)-F'(z))e^{Re(F(z))}$ and $e^{Re(F(z))}$ is holomorphic, i.e. $Re(F(z))$ and $F(z),f(z)$ are constant.
