I am able to make mathematica plot the solution to a complex inequality as the interior or exterior of a circle :

Let $a$ be a positive real constant, and $f(z) = \frac{z+1}{z-1}$. I want to get the equation of the image of the disc $D(a) = \{ z \in \mathbb X:\, \lvert z \rvert \le a\}$ by $f$.

If $a<1$, its also a disc, located in the left half plane :

ComplexRegionPlot[Abs[(z + 1)/(z - 1)] <= 0.9, {z, 20}]

If $a > 1$, its the complementary of a disc, located in the right half plane:

ComplexRegionPlot[Abs[(z + 1)/(z - 1)] <= 1.1, {z, 30}]

And if $a=1$, it's the full left half plane, which in a certian sens is also a disc.

How can i get the center and radius of this disc depending on the value of $a$ ?

I am able to make Mathematica plot the solution to a complex inequality as the interior or exterior of a circle :

Let $a$ be a positive real constant, and $f(z) = \frac{z+1}{z-1}$. I want to get the equation of the image of the disc $D(a) = \{ z \in \mathbb X:\, \lvert z \rvert \le a\}$ by $f$.

If $a<1$, its also a disc, located in the left half plane :

ComplexRegionPlot[Abs[(z + 1)/(z - 1)] <= 0.9, {z, 20}]

If $a > 1$, it's the complementary of a disc, located in the right half plane:

ComplexRegionPlot[Abs[(z + 1)/(z - 1)] <= 1.1, {z, 30}]

And if $a=1$, it's the full left half plane, which in a certain sense is also a disc.

How can I get the center and radius of this disc depending on the value of $a$ ?