Question about Maximum Modulus Principle applied to $|f|+|g|$ > Let the functions $f$ and $g$ be holomorphic in $U$ and continuous in $\overline{U}$. Show that $|f(z)| + |g(z)|$ attains its maximum on $\\{|z| = 1\\}$. Hint: consider the function $h = e^{iα}f + e ^{iβ}g$ with suitably chosen constants $α$ and $β$. Even with the hint I am lost. Of course $|h(z)|$ must attain its maximum when $|z|=1$, and $|h| \le |f|+|g|$, but that doesn't mean that $|f|+|g|$ couldn't attain its maximum inside the unit circle. ny help?

As Martin R. pointed out, the result follows from a similar problem. It boils down to the following statement.

**Proposition**. Assume $f,g$ are holomorphic in a domain $U \subset \mathbb C$, and $|f(z)|+|g(z)|$ achieves maximum at an interior point. Then $f$ and $g$ are constant.

Proof. Suppose $|f(z)|+|g(z)|$ assumes maximum at an interior point $z_0$. Let $\alpha =-\arg f(z_0)$ and $\beta=-\arg g(z_0)$. Define $$h(z)=e^{i\alpha}f(z)+e^{i\beta}g(z),$$ which is holomorphic in $U$. Then $$h(z_0)=|f(z_0)|+|g(z_0)|=|h(z_0)|.$$ Since $$|h(z)|\leq |f(z)|+|g(z)|\leq |f(z_0)|+|g(z_0)|,$$ and $h(z)$ achieves maximum modulus at an interior point, $h(z)$ is a constant, and necessarily $h(z)=h(z_0).$ It follows that $$|h(z_0)|=|h(z)|\leq |f(z)|+|g(z)|\leq |f(z_0)|+|g(z_0)|=|h(z_0)|$$ $$\Rightarrow |f(z)|+|g(z)|~{\rm is~a~constant.}$$ By the result proven here, both $f$ and $g$ are constant. QED