Let $D$ be a bounded region. Suppose $f$ and $g$ are both analytic on $D\cup\partial D$. Let $D$ be a bounded region. Suppose $f$ and $g$ are both analytic on $D\cup\partial D$. Show that $|f(z)|+|g(z)|$ takes its maximum on the boundary.
Consider $f(z)e^{i\alpha}+g(z)e^{i\beta}$ for appropriate $\alpha$ and $\beta$.
 A: Let be $z_0$ a maximum of $|f| + |g|$ in $D\cup\partial D$. Choose $\alpha$, $\beta$ s.t.
$|f(z_0)| = f(z_0)e^{i\alpha}$, $|g(z_0)| = g(z_0)e^{i\beta}$.
Then, $h(z) = f(z)e^{i\alpha} + g(z)e^{i\beta}$ is analytic in the same domain and for all $z\in D\cup\partial D$ and we have:
$$
|h(z)|\le |f(z)| + |g(z)|\le |f(z_0)| + |g(z_0)| = ||f(z_0)| + |g(z_0)|| = |f(z_0)e^{i\alpha} + g(z_0)e^{i\beta}| = |h(z_0)|.
$$
Now, apply the maximum modulus principle to $h$.
A: Suppose that  $\lvert {f(z)} \rvert + \lvert {g(z)} \rvert$ is maximized over $\overline{D}$ at  $z_{0} \in D$.
Define $h : \overline{D} \rightarrow \mathbb{C} : z \mapsto \lvert {f(z_{0})} \rvert + \lvert {g(z_{0})} \rvert - g(z) - f(z).$
Clearly, $h \in \text{Hol}(\overline{D})$ and $h(z_{0}) = 0$. 
$\underline{\text{Case 1:}}$ If $h \not\equiv$ constant on $\overline{D}$, then there exists $\delta > 0$ so that $B(z_{0},\delta) \subset D$ and $\vert h(z) \rvert > 0$ for all $z \in B^{'}(z_{0},\delta)$. (This is because $z_{0}$ minimizes $h(z)$ on $B(z_{0},\delta)$.) 
In which case, $h(B(z_{0},\delta))$ is not an open set, which contradicts the open-mapping theorem, since $h$ is holomorphic, non-constant on $B(z_{0},\delta)$.
$\underline{\text{Case 2:}}$ Suppose that $h \equiv $ constant on $\overline{D}$. Then, 
$f(z) = \alpha - g(z)$ where $\alpha = f(z_{0}) + g(z_{0})$. 
So, we have reduced the original problem to showing that for $f \in \text{Hol}(\overline{D})$ and $\alpha \in \mathbb{C}$ that
$\lvert f(z) \rvert + \lvert \alpha - f(z) \rvert$ achieves its maximum on $\partial D$.
If $\alpha = 0$, then by the Maximum Modulus principle, $2\cdot \lvert f(z) \rvert$ achieves its maximum on $\partial D$ provided that $f$ is not constant on $D$.
Suppose that $\alpha \neq 0$. Put $v := \frac{\alpha}{\lvert \alpha \rvert}$.
Pick $w_{0} \in f(D)$. Because $f(D)$ is open (provided $f$ is non-constant), there exists $w \in f(D)$ with $w = w_{0} + iv\epsilon$ for some $\epsilon > 0$.
Because $\{v,iv\}$ span $\mathbb{C}$, we may write 
$w_{0} = cv + idv$ where $c,d \in \mathbb{R}$.
Now, it is straight-forward to check that
$$\lvert w_{0} \rvert + \lvert \alpha - w_{0} \rvert < \lvert w \rvert + \lvert \alpha - w \rvert.$$
Since $w_{0} \in f(D)$ was arbitrary, it follows that $\lvert f(z) \rvert + \lvert \alpha - f(z) \rvert$ is maximized on $\partial D$.
The intuitive idea behind second half of this argument is that taking $w_{0} \in f(D)$, we may translate $w_{0}$ by a vector perpendicular to $\vec{\alpha}$ and remain in $f(D)$. Then, this new vector, $w$, is easily seen to be of a greater distance from $0$ and $\alpha$ than $w_{0}$.

