Show that $f(z)=\frac{2z+3}{z-4}$ maps the circle $|z-2i|=2$ to $|8z+(6+11i)|=11$

Question

Show that $f(z)=\frac{2z+3}{z-4}$ maps the circle $|z-2i|=2$ to $|8z+(6+11i)|=11$

What I have done is decompose $f$ into dilations $D_a:z=az$, translations $T_a:z=z+a$, and inversions $S:z=1/z$, such that $$f=T_2 D_{11} S T_{-4}$$

So we go from $|z-2i|=2$, to $|z-(4+2i)|=2$, to $|\frac{1}{z}-(4+2i)|=2$.

This is where I start to get messy, and I can't get anywhere.

Is my method correct? And what should$|\frac{1}{z}-(4+2i)|=2$ become?

Answer 1

Set $$w=\frac{2z+3}{z-4}$$ and rearrange to get $$z=\frac{4w+3}{w-2}$$

Substitute this into the given locus so you get $$\left|\frac{4w+3}{w-2}-2i\right|=2$$ $$\implies|4w+3-2iw+4i|=2|w-2|$$

Putting $w=u+iv$, and after a bit of algebra, you end up with $$16u^2+16v^2+24u+44v+9=0$$

And it is easily seen that this is the same circle as given in the question.

Answer 2

Note that $$ \left|\frac{z-z_1}{z-z_2}\right|=k $$ is the circle $|z-a|=R$ where the center and radius are $$a=\frac{z_1-k^2z_2}{1-k^2}, R=\frac{k(|z_1-z_2|}{|1-k^2|}.$$ From David Quinn's answer, one has $$\left|\frac{4w+3}{w-2}-2i\right|=2$$ or $$\left|\frac{(4-2i)w+3+4i}{w-2}\right|=2$$ or $$\left|\frac{w+\frac{3+4i}{4-2i}}{w-2}\right|=\frac{2}{|4-2i|}=\frac{1}{\sqrt5}.$$ So $$ w_1=-\frac{3+4i}{4-2i}=-\frac{2+11i}{10}, w_2=2, k=\frac1{\sqrt5}, $$ and the center and the radius are $$ a=\frac{w_1-k^2w_2}{1-k^2}=-\frac{6+11i}{8}, R=\frac{k|w_1-w_2|}{|1-k^2|}=\frac{11}{8} $$ repectively. Namely $$ |8w+(6+11i)|=11. $$