Reflection of a line on complex plane > Show that the reflection of $z$ on the line $ax+by=c$, where $a,b,c \in \mathbb{R}$ is given by the following: $$\frac{2ic+(b-ai)\bar z}{b+ai}$$ I know that the conjugate of $z$, which is $\bar z$, will be the reflection of the point $z$ in the real axis, but I don't know how to proceed further in proving this.

This solutions depends on whether $a ≠ 0$ or $b ≠ 0$.

Translate the problem to the origin. $z ↦ z-\tfrac{c}{a}$ or $z ↦ z - i\tfrac{c}{b}$

Spin the line onto the real line. $z ↦ \tfrac{z}{b-ai}$

Reflect the plane at the real line. $z ↦ \overline z$

Spin back to the former line . $z ↦ (b-ai) z$

Translate to your former point. $z ↦ z + \tfrac{c}{a}$ or $z ↦ z + i\tfrac{c}{b}$

All in all I get: $(b-ai)\overline{\left(\tfrac{z-\tfrac{c}{a}}{b-ai}\right)} + \tfrac{c}{a} = \tfrac{2ic + (b-ai)\overline z}{b+ai}$ (And the same with the other variant.)

**Note** : I think something's wrong with this as it should be $2ic$ instead of $2iac$. I can't figure it out, but I really want to know what my error is.

_This is now corrected._