How does $f:z\mapsto az+b$ send circles in $\mathbb{C}$ to circles in $\mathbb{C}$?

Question

I am looking at the proof in the beginning of chapter 2 of Anderson's book Hyperbolic Geometry, however I don't understand it. In particular, he seems to assume that $\overline{z_1} \cdot \overline{z_2}=\overline{z_1z_2},$ with $z_1,z_2\in\mathbb{C},$ but I think I have proven that that identity is false. Is there some special case where this is true?

Answer 1

To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.

Answer 2

Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.

$z'= f(z)=az+b$, $z_0'=az_0+b$;

$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$

i.e. a circle with center $z_0'$, and radius $|a|r$.