Complex functions with similar magnitude Let $f$ and $g$ be complex-analytic functions on the unit disk $D_1$, and suppose that
$$
\sup_{|z|\leq 1} \big||f(z)|-|g(z)|\big| \leq \epsilon.
$$
I am curious whether there exists some $\theta\in \mathbb{R}$ such that on the half-disk
$$
\sup_{|z|\leq 1/2} |f(z)- e^{i\theta} g(z)| \leq C \epsilon.
$$
I can show that this holds when $g=1$ is the constant function.  In this case I can use the fact that $|f|+\epsilon -1$ is subharmonic and positive in order to apply Cacciopoli's inequality to get a bound on $\int_{D_{3/4}} \big|\nabla |f|\big|^2$.  Then I can use the fact that $\big|\nabla |f|\big| = |f'|$ to show that on $D_{1/2}$,
$$
|f'|\leq C\delta.
$$  This shows that $f$ is close to a constant, as desired.
This argument does not work when $g$ is not constant because $|f|-|g|+\epsilon$ is not subharmonic.  I am curious if there is another way to proceed.
One barrier that makes the argument difficult to find is the example $f(z)=z$ and $g(z)=\overline{z}$.  In this case $g$ is not analytic so the hypotheses do not hold, but it does show that the desired bound will not follow from only using bounds such as the maximum principle on $f$ and $g$ in isolation (perhaps one can use that $fg$ is also analytic, for example).
 A: Some partial results; wlog we can assume $f,g$ analytic on the closed disc (so they have analytic extension slightly beyond) as we can just use $f(rz), r \to 1, r<1$ etc;
We claim that if $M\epsilon \le A \le |f|, |g| \le B, M \ge 2$ the result holds for any radius $r$ st $1-r \ge C, 8\epsilon \le AC$ (so for $r=1/2, M=16$) with a constant $K$ that depends only on $A,B,C$ so for example it holds for $r=1/2, g=1, \epsilon \le 1/16, K=16$ as in the OP and we will give a short proof below.
On the other hand if $f,g$ can have small absolute values (eg zeroes) then one has counterexamples for the unit disc but not for the disc of radius $1/2$ as requested in the OP of the type $f=z^k, g=z^{k+1}$ where $\max_{|z|=1}|f-\alpha g| =2$ but $\max_{|z| \le 1}||f|-|g||=c_k/(k+1), c_k \to 1/e$ (and of course one can slightly perturb them with small analytic functions to distribute the zeroes), so that case definitely bears further investigation
Assume $|g| \ge A \ge 2\epsilon$ (and also $1-r \ge C, 8\epsilon \le AC$ as above) and choose $|\alpha|=1,|\beta|=1, \alpha f(0), \beta g(0) > 0$ and replacing $f,g$ by $\alpha f, \beta g$ we can assume $f(0), g(0) >0$ and prove $\sup_{z\in D_{r}} |f(z)- g(z)| \le K\epsilon $ (hence in the original problem $\sup_{z\in D_{r}} |f-\bar \alpha \beta g|\le K\epsilon$) with $K$ depending on $A,B,C,r$
We note that $1/2<|f/g| \le 1+\frac{\epsilon}{A} \le 1+1/2$. Then $\log (f/g)$ is analytic in the unit disc where we choose the branch for which $\log f/g(0)$ is the usual one for positive numbers and $\Re \log (f/g)=\log |f/g| \le 2\epsilon/A$, while $\arg f/g(0)=0$ so $|\log f/g(0)|=\log |f/g(0)| \le 2\epsilon/A $
By Borel Caratheodory one has $\sup_{z\in D_{r}} |\log (f/g)(z)| \le \frac{4\epsilon }{A(1-r)} \le 4\epsilon/ AC \le 1/2$. But if $|w| \le 1/2$ one has $|e^w-1| \le e^{|w|}-1\le 2|w|$ so on $D_r$ one has  $|f/g-1| \le 8\epsilon/ AC$ or $|f-g| \le K\epsilon$ with $K=8B/AC$
