Write a formula for $z^*$ with respect to line $l=\{ax+by=c\}$.

Question

Write a formula for $z^*$ with respect to line $l=\{ax+by=c\}$. ($z^*$ is the symmetric point of $z$ with respect to $l$) At first, I thought of solving it like I used to in High-School, but then recalled we were just learning about Mobius transformations and thought I better approach it with analogy to the real line, with is the simpler case. The thing is: I really don't manage to find a proper Mobius transformation. I try to take arguments and the origin into account but it just won't go very well. What do I do with the origin? Maybe it is all just a misconception of the whole subject? Have some of you got any idea as for how to correctly approach this in terms of the complex analysis?

Answer 1

First you should write the line $l$ in terms of $z$ instead of $x$ and $y$: $$ ax + by + c = a\frac{z + \overline z}{2} + b \frac{z - \overline z}{2i} \\ = \frac 12 \left( (a-ib)z + (a+ib) \bar z\right) + c = \text{Re} ( Az + c) = \text{Im}(i(Az+c)) $$ with $A := a - ib$.

So the (linear) Möbius transformation $w = i(Az +c)$ transforms the line $l$ to the real axis $ \text{Im}(w) = 0$. The symmetric point of $w$ with respect to the real axis is $w^* = \overline w$. Then transform $w^*$ back to $z^*$.