On analytic functions $f,g$ with the sum of their norms is constant I am new here. I don't know if this problem is easy or not or is that suitable for this site or not. But I have difficulties dealing with it. 
Let $A\subseteq \Bbb{C}$ be open and connected. Also let $f,g :A \longrightarrow \Bbb{C}$ be analytic functions such that $|f| + |g|$ is constant. How can I show that $f$ and $g$ are constant (if its true) ?
 A: We have that $|f|+|g|=\sqrt{f\overline{f}}+\sqrt{g\overline{g}}$ is constant. 
Applying $\frac{\partial}{\partial\bar{z}}$,
$$\frac{\frac{\partial f}{\partial\bar{z}}\bar{f}+ f\frac{\partial \bar{f}}{\partial\bar{z}}}{\sqrt{f\bar{f}}}+\frac{\frac{\partial g}{\partial\bar{z}}\bar{g}+g\frac{\partial\bar{g}}{\partial\bar{z}}}{\sqrt{g\bar{g}}}=0$$
Since $f$ and $g$ are analytic $\frac{\partial f}{\partial\bar{z}}=\frac{\partial g}{\partial\bar{z}}=0$ (these are the Cauchy-Riemann equations), and we get
$$\frac{f\frac{\partial \bar{f}}{\partial\bar{z}}}{\sqrt{f\bar{f}}}+\frac{g\frac{\partial\bar{g}}{\partial\bar{z}}}{\sqrt{g\bar{g}}}=0$$
So 
$$g\bar{g}f^2\left(\frac{\partial\bar{f}}{\partial\bar{z}}\right)^2=-f\bar{f}g^2\left(\frac{\partial\bar{g}}{\partial\bar{z}}\right)^2.$$
or
$$\bar{g}f\left(\frac{\partial\bar{f}}{\partial\bar{z}}\right)^2=-\bar{f}g\left(\frac{\partial\bar{g}}{\partial\bar{z}}\right)^2.$$
So, the meromorphic function $f/g$ is equals to an anti-meromorphic function, i.e.  $$-\bar{f}\left(\frac{\partial\bar{g}}{\partial\bar{z}}\right)^2/\bar{g}\left(\frac{\partial\bar{f}}{\partial\bar{z}}\right)^2.$$
Hence $f/g$ is constant. You can see this by expanding in Laurent series. The series for $f/z$ is going to be $\sum_{n\in\mathbb{Z}} a_n(z-a)^n$ and the one of the other function $\sum_{n\in\mathbb{Z}}b_n(\bar{z}-a)^n$. They are only equal if both are constant.
Then $|f|$ and $|g|$ are constant follows because if $f=cg$, for some constant $c$, then $|f|+|g|=(1+|c|)|f|$. From this the result follows again applying Cauchy-Riemann. In fact, $|f|=\sqrt{f\bar{f}}$ is constant. Then taking $\frac{\partial}{\partial\bar{z}}$ and using that $\frac{\partial f}{\partial\bar{z}}=0$, we get $f\frac{\partial\bar{f}}{\partial\bar{z}}=0$. If $f=0$ it is constant. If $\frac{\partial\bar{f}}{\partial\bar{z}}=0$ then $f$ is constant too.
A: If $A=\mathbb C$ then the result holds: Since $|f+g|\leq |f|+|g|$, if $|f|+|g|$ is constant you have that $f+g$ is bounded, hence by Liouville's theorem we have that $f+g$ is constant, say $f+g=c$. Then
$$|f|+|g|+|c|=|f|+|c-f|+|-c|\geq |f|+|-f|=2|f|$$
hence $f$ is bounded as well, so by Liouville's theorem again we have that $f$ is constant, hence $g=c-f$ is constant as well.
