Complex polynomial identity with norm condition

Complex polynomial identity with norm condition

In this question, the following was shown:
If $R(z)=\dfrac{P(z)}{Q(z)}$, where $P,Q$ are polynomials in a complex
variable $z$, satisfies the condition that $|R(z)|=1$ whenever $|z|=1$,
then the identity $R(z)\overline{R(1/\overline{z})}=1$ holds for all
complex $z$.
The proof given there uses the identity theorem. Since I'm not familiar
with the theorem yet, I would like to find an easier approach.
So I try direct substitution. Writing
$P(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0$ and
$Q(z)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_0$, we find that
$$R(z)=\frac{P(z)}{Q(z)}=\frac{a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0}{b_mz^m+b_{m-1}z^{m-1}+\ldots+b_0}$$
and $$\overline{R(1/\overline{z})} =
\frac{\overline{P(1/\overline{z})}}{\overline{Q(1/\overline{z})}} =
\frac{\overline{a_n}+\overline{a_{n-1}}z+\ldots+\overline{a_0}z^n}{\overline{b_m}+\overline{b_{m-1}}z+\ldots+\overline{b_0}z^m}\cdot
z^{m-n}$$
So the identity $R(z)\overline{R(1/\overline{z})}=1$ that we wants to
prove can be turned into
$$z^{m-n}(a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0)(\overline{a_0}z^n + \ldots +
\overline{a_{n-1}}z + \overline{a_n}) =
(b_mz^m+b_{m-1}z^{m-1}+\ldots+b_0)(\overline{b_0}z^m+\ldots+\overline{b_{m-1}}z+\overline{b_m}).$$
where we have the condition that for any $|z|=1$,
$$|a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0| =
|b_mz^m+b_{m-1}z^{m-1}+\ldots+b_0|.$$
How can we proceed from here?