The mobious transformation of ${z\over 2z-8}$ from circle $|z-2|=2$

Question

through mobious transformation $f(z)$=${z\over 2z-8}$ a have to know what is the image from circle $K=\{z:|z-2|=2\}$.

I know that transformation is injective and onto. So I try this:

$w\in f(K) \Leftrightarrow {w\over 2w-8} \in K $

$\hspace{2cm}\Leftrightarrow |{w\over 2w-8}-2|=2$

$\hspace{2cm}\Leftrightarrow 2|{2w-8}|=|16-3w| $

$\hspace{2cm}\Leftrightarrow 7|w|^2-16w-16\overline w=0$

I don't how to continue, because my partners say me that transformation give me a semiplane in the second and third quadrant. I don't see it. My solution is right? I receive suggest or comments. Than you

Answer 1

Your solution unfortunately doesn't work; $w \in f(K)$ means $f^{-1}(w) \in K$, not $f(w)$.

Answer 2

HINT

The circle has parametric equation $z=2e^{i\varphi}+2\,$ for $\varphi \in [0,2\pi).$ Then for $z\in K$ is $$f(z)=\frac{2e^{i\varphi}+2}{4e^{i\varphi}-4}.$$