Let $\mathbb{D}=\{z: |z| <1\}$ and $\mathbb{T}=\{z: |z|=1\}$. Suppose $D$ and $E$ are polynomials of degree atmost $n$ with complex coefficients such that
- $|E(z)| \leq |D(z)|$ for all $z \in \mathbb{D}$
- $D(z) \ne 0$ for all $z \in \mathbb{D}$.
Edit: 3. $E(z)=z^n\overline{E(1/\overline{z})}$ for all $z \in \mathbb{C}$.
Prove that there exists a polynomial $A$ that does not vanish on $\mathbb{D}$ and $|A(z)|^2=|D(z)|^2-|E(z)|^2$ for all $z \in \overline{\mathbb{D}}$.
My attempt: By continuity arguments, the trigonometric polynomial $|D(z)|^2-|E(z)|^2 \geq 0$ for all $z \in \overline{\mathbb{D}}$. In particular, $|D(z)|^2-|E(z)|^2 \geq 0$ for all $z \in \mathbb{T}$ . By Fejer-Riesz Theorem, there exists a polynomial $A$ that does not vanish on $\mathbb{D}$ and $|A(z)|^2=|D(z)|^2-|E(z)|^2$ for all $z \in \mathbb{T}$.
From here onwards, I think one can use maximum-modulus principle to obtain the desired inequality for $\mathbb{D}$. Any hints or suggestions.
This is not true; for example take $E(z)=2z-1, D(z)=z-2$ It is clear that if $|z|=1$ we have $|E(z)|=|D(z)|$ (multiply with $|\bar z|=1$ either term) but if $|z| <1$ we have $|E(z)|<|D(z)|$ (squaring for example), while $D$ doesn't vanish in the unit disc.
Then any $A$ st $|A(z)|^2=|D(z)|^2-|E(z)|^2$ would need to be identically zero on the unit circle, hence identically zero in the plane, so the equality cannot hold inside the unit disc
After the edit in the OP which changed the problem one can take $E(z)=(2z-1)(z-2)$ and $D(z)=(z-2)^2$ and now $E$ satisfies the extra condition and still the result fails