Find the number $|z_1|$ where $\left|\frac{z_1-2z_2}{2-z_1\bar{z_2}}\right|=1$

Question

Let $z_1$ and $z_2$ be two complex numbers such that $$\left|\frac{z_1-2z_2}{2-z_1\bar{z_2}}\right|=1$$ where $|z_2|\neq1$. Find $|z_1|$.

I could not understand in this if we are not given $z_2$ then how we find $z_1$

Answer 1

Hint - write it as:

$$ \require{cancel} \begin{align} |z_1-2z_2|^2=|2-z_1 \bar z_2|^2 & \quad\iff\quad (z_1-2z_2)(\bar z_1-2 \bar z_2) = (2-z_1 \bar z_2)(2-\bar z_1 z_2) \\ & \quad\iff\quad |z_1|^2 - \cancel{2z_1\bar z_2}-\bcancel{2z_2 \bar z_1} + 4 |z_2|^2 = 4 - \bcancel{2\bar z_1 z_2} - \cancel{2 z_1 \bar z_2} +|z_1|^2|z_2|^2 \\ & \quad\iff\quad \left(|z_1|^2-4\right)\left(1-|z_2|^2\right)=0 \end{align} $$

Answer 2

Let $w=\dfrac{z_1}{2}$ thus $$f(w)=\frac{w-2z_2}{1-w\bar{z_2}}$$ is a bilinear transformation that carry circle to circle which $|f|=1$, so every points $|w|=1$ goes to $|f|=1$ so $|z_1|=2$.

Answer 3

WLOG $z_1-2z_2=e^{it}(2-z_1\overline{z_2})$

$\iff\dfrac{z_1}2=\dfrac{e^{it}+z_2}{1+e^{it}\overline{z_2}}$

Now $\left|\dfrac{e^{it}+z_2}{1+e^{it}\overline{z_2}}\right|^2=\dfrac{(\cos t+a)^2+(\sin t+b)^2}{(1+a\cos t+b\sin t)^2+(b\cos t-a\sin t)^2}=1$