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In the set of rational polynomial functions $H(z)$ of a complex number $z$, there exist functions whose magnitude $|H(z)|^2$ is a constant $C$, but whose denominator and numerator are not constants. In engineering, systems that implement these functions on the Z-transform of a discrete time-series input are referred to as all-pass filters, because they do not change the frequency content of the input (i.e. $|Y(z)|^2 = |H(z)|^2|X(z)|^2 = C|X(z)|^2$).

It is often stated in textbooks (I have two here, Mitra and Papoulis) that give this identity without proof:

$|H(z)|^2 = H(z)H^*(z) = H(z)H(\frac{1}{z})$

This identity can be used, say, to prove that the canonical example of a first-order all-pass filter $H(z) = \frac{1 - cz^{-1}}{1+cz^{-1}}$ is an all-pass filter, but I would like to go the other way.

I know that if $z$ is purely complex, then $\frac{1}{z} = z^*$ -- sometimes you see the hand-waving proof where $e^{j\omega}$ is substituted in for $z$ which produces the expected result, but that's not enough. How can I prove the identity?

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If by $z^*$ you mean the conjugate of $z$, then $1/z=z^*$ isn't true for every purely complex $z$. Example $z=2i$ then $1/z=(1/2)(1/i)=-(1/2)i$ but $z^*=-2i$. What one can say is that if $|z|=1$ then $1/z=z^*$. –  coffeemath Mar 5 '13 at 1:12
That would make sense with respect to the $\exp(j\omega)$ argument, but what's going on with these textbooks then? –  Trevor Alexander Mar 5 '13 at 4:22
@coffeemath I thought about what you said some more; I think there is a required normalizing condition on the coefficients of the polynomials in $H(z)$. Edited to clarify, will do more later. Thanks. –  Trevor Alexander Mar 5 '13 at 8:45
There are no nonconstant analytic functions whose absolute value is constant on some domain. Maybe you want this only on the real axis or on the unit circle. –  Christian Blatter Mar 5 '13 at 9:24
How would I go about proving that statement on the unit circle? –  Trevor Alexander Mar 5 '13 at 18:18
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