Magnitude of function in $z$ domain I am newbie to $\mathcal Z$-transform, I searched to find the magnitude of a function in $z$-domain, but I couldn't find anything, for example when we have 
$$
H(z) = \frac{z-3}{z-0.5}
$$ 
How do you compute $\lvert H(z)\rvert^2$? I know that $\lvert H(z)\rvert^2 = H(z)\cdot H\left(z^{-1}\right)$, is it right? when I calculate it for my example:
$$
H(z)\cdot H\left(z^{-1}\right) = \frac{z-3}{z-0.5}\cdot\frac{z^{-1}-3}{z^{-1}-0.5} = \frac{10-3z-3z^{-1}}{1.25-0.5z-0.5z^{-1}} 
$$
but it is not a numerical value like Fourier domain.
The $z$-transform can be evaluated at any point on the complex plane that is also in the ROC of the $z$-transform. To find the magnitude of $H(z)$, you can find the magnitude of numerator and denumerator separately, and then divide the results.

Let's say $H(z)=\frac{A(z)}{B(z)}$
So

$$|H(z)|=\frac{|A(z)|}{|B(z)|}$$
and let's assume $A(z)=z-a$ and $B(z)=z-b$ .
$$|A(z)|=|z-a|$$
Let $z=re^{j\phi}$. So
$$\begin{align}
|A(z)|&amp;=|re^{j\phi}-a|\\&amp;=|(r\cos(\phi)-a)+j r\sin(\phi)|\\
&amp;=\sqrt{(r\cos(\phi)-a)^2+r^2\sin^2(\phi)}\\
&amp;=\sqrt{r^2(\sin^2(\phi)+\cos^2(\phi))+a^2-2ar\cos(\phi)}\\
&amp;=\sqrt{r^2+a^2-2ar\cos(\phi)}
\end{align}$$
Similarly $$|B(z)|=\sqrt{r^2+b^2-2br\cos(\phi)}$$
and

$$|H(z)|=\frac{\sqrt{r^2+a^2-2ar\cos(\phi)}}{\sqrt{r^2+b^2-2br\cos(\phi)}}$$