Polynomial identity : $|P(z)|^2=|Q(z)|^2-|R(z)|^2$ for $z \in \mathbb{D}$

Question

Let us denote by $\mathbb{D}=\{z : |z| <1\}$ and $\mathbb{T}=\{z: |z|=1\}$. Suppose that $P, Q$ and $R$ are polynomials that satisfy the following:

1. $|Q(z)| \geq |R(z)| $ for all $z \in \overline{\mathbb{D}}$

2. $|P(z)|^2=|Q(z)|^2-|R(z)|^2$ for $z \in \mathbb{T}$

3. $P$ and $Q$ are non-vanishing on $\overline{\mathbb{D}}$

4. $|P(z)|^2 \leq |Q(z)|^2-|R(z)|^2$ for $z \in \mathbb{D}$.

My question : $|P(z)|^2=|Q(z)|^2-|R(z)|^2$ for $z \in \mathbb{D}$?

I am trying to prove it. I think some version of maximum-modulus or identity theorem can be useful here. But what functions to choose to apply maximum modulus. Any hint how to proceed.

My Attempt: Note that for any $z \in \mathbb{T}$, we have $P(z)\overline{P(\frac{1}{\overline{z}})}=|P(z)|^2$. Thus, condition (2) implies that $P(z)\overline{P(\frac{1}{\overline{z}})}= Q(z)\overline{Q(\frac{1}{\overline{z}})} -R(z)\overline{R(\frac{1}{\overline{z}})} $ for all $z \in \mathbb{T}$. Hence, the map

$F: \mathbb{D}\setminus \{0\} \to \mathbb{C}, F(z)= P(z)\overline{P(\frac{1}{\overline{z}})} - Q(z)\overline{Q(\frac{1}{\overline{z}})} - R(z)\overline{R(\frac{1}{\overline{z}})}$ is zero on $\mathbb{T}$.

Answer 1

This isn't true in general.

First, thanks to Conrad for pointing out a problem with my earlier answer that claimed to prove this. As he noted, if we let $Q$ be an arbitrary polynomial that doesn't vanish on the closed disk, and pick constants $a,b$ such that $|a|^2 + |b|^2 = 1$, then $P = aQ$ and $R = bQ$ satisfy the hypotheses and conclusion. This gives some examples of polynomials for which this works.

Suppose $P, Q, R$ are polynomials that satisfy the hypotheses and conclusion. Then $P, Q, z^n R$ also satisfy the hypotheses if $n$ is a positive integer:

1. $|Q(z)| \geq |R(z)| \geq |z^n R(z)|$ for all $z \in \overline{\mathbb{D}}$ as $|z^n| \leq 1$ there.
2. $|P(z)|^2 = |Q(z)|^2 - |R(z)|^2 = |Q(z)|^2 - |z^n R(z)|^2$ when $|z| = 1$.
3. $P$ and $Q$ do not vanish on the closed disk.
4. $|P(z)|^2 \leq |Q(z)|^2 - |R(z)|^2 < |Q(z)|^2 - |z^n R(z)|^2$ on $\mathbb{D}$.

However, as the strict inequality in the last point shows, the conclusion does not hold.

If the conclusion holds, then we can consider $f = P/Q$ and $g = R/Q$ on the closed disk. The equation $$ \partial \bar\partial \log(|f|^2 + |g|^2) = \frac{|fg' - f'g|^2}{(|f|^2 + |g|^2)^2} $$ and that $|f|^2 + |g|^2 = 1$ on the disk imply that $f'/f = g'/g$, so $g = a f$ for some constant $a$. Then $|f|^2$ is constant on the disk, so $P = bQ$ for some constant $b$, and we are in the situation Conrad described.