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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?

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  • 2
    $\begingroup$ You mean a more elementary approach. This way is definitely not easier. $\endgroup$ – Cocopuffs Aug 25 '13 at 1:12
  • $\begingroup$ Kinda weird, seeing the algebra-precalculus and complex-analysis tags on the same question. $\endgroup$ – Gerry Myerson Aug 25 '13 at 1:14
  • $\begingroup$ I don't think your equation about the conjugate of R is correct. The conjugate is both on R and on 1/z ,but not on the coefficients of P and Q. $\endgroup$ – Betty Mock Aug 25 '13 at 2:18
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If we denote $$\sum_{k=0}^n \overline{a_k}z^{n-k} = z^n\overline{P(1/\overline{z})}$$ by $P^\ast(z)$, and similar for $Q$, the identity

$$R(z) \cdot \overline{R(1/\overline{z})} = 1, \quad \lvert z\rvert = 1$$

becomes

$$z^mP(z)P^\ast(z) = z^nQ(z)Q^\ast(z)$$

for $\lvert z\rvert = 1$, in other words, the polynomial

$$z^mP(z)P^\ast(z) - z^nQ(z)Q^\ast(z)$$

has infinitely many zeros, hence is the zero polynomial.

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