A property of all-pass filters Let $G(z)$ be a real coefficient stable all-pass transfer-function with
degree greater than zero. Then it can be shown that $|G(z)| &lt; 1$ for
$|z| &gt; 1$: Given that the poles occur in complex conjugate pairs, one
way of establishing this property is as a natural generalization of the
case when

$$G_1(z) = \frac{1 − \alpha^*z}{z − \alpha}$$

where the superscript $∗$ denotes the complex conjugate operation.
By considering $1 − |G_1(z)|^2$ or otherwise show that $G_1(z)$ has this
property, i.e., $|G_1(z)| &lt; 1$ for $|z| &gt; 1$ .

I try to demonstrate  $|G(z)|^2 = G(z)G(z^{-1}) &gt; 1$, but i don't know how to deal with the complex conjugate  $\alpha$ and $\alpha^*$. or my solution is totally wrong?
First, observe that it's sufficient to show the property for a first-order all-pass filter, because any higher order all-pass filter can be written as a cascade of first-order all-pass sections. And if

$$|G_i(z)|&lt;1\quad\text{for}\quad |z|&gt;1$$

then also

$$\left|\prod_iG_i(z)\right|&lt;1\quad\text{for}\quad |z|&gt;1$$

must be satisfied.

Now derive an expression for the squared magnitude of the first-order all-pass transfer function $G_1(z)$:

$$|G_1(z)|^2=\frac{1-2\text{Re}\{\alpha^*z\}+|\alpha|^2|z|^2}{|z|^2-2\text{Re}\{\alpha^*z\}+|\alpha|^2}\tag{1}$$

Using $(1)$, the inequality $|G_1(z)|^2&lt;1$ can be written as

$$\begin{align}1-2\text{Re}\{\alpha^*z\}+|\alpha|^2|z|^2&amp;&lt;|z|^2-2\text{Re}\{\alpha^*z\}+|\alpha|^2\\
1+|\alpha|^2|z|^2&amp;&lt;|z|^2+|\alpha|^2\\0&amp;&lt;(|z|^2-1)(1-|\alpha|^2)\tag{2}\end{align}$$

Note that if the all-pass filter is stable, we must have $|\alpha|&lt;1$, because $\alpha$ is the pole location, and a causal and stable filter must have all its poles inside the unit circle of the complex plane. Consequently, the inequality $(2)$ is satisfied for $|z|&gt;1$ because both factors on the right-hand side are positive.