Prove that if |z|=1 , then |(a_2z^2+a_1z+a_0)/(bar…

Prove that if $|z|=1$, then $\ \left|\dfrac{a_2z^2+a_1z+a_0}{\bar a_0z^2+\bar a_1z+\bar a_2}\right|=1$. Generalize to polynomials of degree $n$.

Step 1 · Set up for degree $2$

Let $P(z)=a_2z^2+a_1z+a_0$ and let $Q(z)=\bar a_0z^2+\bar a_1z+\bar a_2$ be its reversed-conjugate polynomial. Since $|z|=1$, $\bar z=\dfrac1z$.

Step 2 · Rewrite $Q(z)$ using $\bar z=1/z$

$$ Q(z)=\bar a_0z^2+\bar a_1z+\bar a_2=z^2\left(\bar a_0+\bar a_1\bar z+\bar a_2\bar z^2\right)=z^2\,\overline{a_0+a_1z+a_2z^2}=z^2\,\overline{P(z)} $$

Step 3 · Compute the modulus of the quotient

$$ \left|\frac{P(z)}{Q(z)}\right|=\left|\frac{P(z)}{z^2\overline{P(z)}}\right|=\frac{|P(z)|}{|z|^2\,|\overline{P(z)}|}=\frac{|P(z)|}{1\cdot|P(z)|}=1 $$

using $|z|=1$ and $|\overline{P(z)}|=|P(z)|$ (assuming $P(z)\neq0$, so the quotient is defined).

Step 4 · Generalize to degree $n$

For $P(z)=a_nz^n+\cdots+a_1z+a_0$ and its reversed-conjugate $Q(z)=\bar a_0z^n+\cdots+\bar a_{n-1}z+\bar a_n$, the identical argument gives, for $|z|=1$:

$$ Q(z)=z^n\left(\bar a_0+\bar a_1\bar z+\cdots+\bar a_n\bar z^n\right)=z^n\,\overline{P(z)} $$

$$ \boxed{\left|\frac{P(z)}{Q(z)}\right|=\left|\frac{P(z)}{z^n\overline{P(z)}}\right|=\frac{|P(z)|}{|z|^n|P(z)|}=1} $$

Yes — the identity holds for every degree $n$, whenever $P(z)\neq0$.

Verification

Checked numerically for random complex coefficients $a_0,a_1,a_2$ and several random $z$ on the unit circle: $\left|\dfrac{P(z)}{Q(z)}\right|$ equals $1$ to full floating-point precision in every trial.