Determine all $z \in \mathbb{C} $, so that:$ | \frac{z-a}{1-\bar{a} z} | = 1,$ where $a \in \mathbb{C}$, $|a| < 1$.

Question

Determine all $z \in \mathbb{C} $, so that: $$ \left| \frac{z-a}{1-\bar{a} z} \right| = 1,$$ where $a \in \mathbb{C}$, $|a| < 1$.

Let $ w = \frac{z-a}{1-\bar{a} z} $. As $|w| = 1$, I though of multiplying $w* \bar{w} (=1)$ but I got stuck at the step $|z|^{2} = \frac {1- |a|^{2}}{1+|a|^{2}}$. Any hint helps!

Answer 1

$|z|^{2} = \frac {1- |a|^{2}}{1+|a|^{2}}$ is not correct. Correct is

$|z|^{2} = \frac {1- |a|^{2}}{1-|a|^{2}}$, hence $|z|=1.$

Answer 2

Hint: We have $$|z-a|=|1-\bar{a}z|$$ with $$z=x+iy,a=u+iv$$ we get $$|x-u+i(y-v)|=|1-xu-vy+i(vx+uy)|$$ Can you proceed? We get $$u^2 \left(-x^2\right)-u^2 y^2+u^2-4 u v x y-v^2 x^2-v^2 y^2+v^2+x^2+y^2-1=0$$

Answer 3

$$ \left| \frac{z-a}{1-\bar{a} z} \right|^2 = \frac{|z|^2+|a|^2-2\Re{\bar{a} z}}{1+|a|^2|z|^2-2\Re{\bar{a} z}}=1$$ if and only if $$|z|^2+|a|^2-2\Re{\bar{a} z}=1+|a|^2|z|^2-2\Re{\bar{a} z}$$ if and only if $$|z|^2(1-|a|^2)=(1-|a|^2)$$ if and only if $|z|=1.$