Would you be kind enough to verify/reject my answer to the following question?
If $D$ is the open unit disk in $\mathbb{C}$, can there be a holomorphic $f:D \rightarrow \mathbb{C}$ such that $$\lvert f(z) \rvert^2=1-\lvert z \rvert^2\quad ?$$
I thought that no: The absolute value is maximal at $z=0$ which is in the interior of $D$. So according to the maximum principle, a holomorphic $f$ is constant. But $f$ cannot be constant since $\lvert f(0)\rvert^2 =1$ and $\left\lvert f\left(\frac{1}{2}\right)\right\rvert^2 =\frac{3}{4}$. Contradiction.