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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.

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Was a definition of reflection given to you? – Git Gud Jan 26 '13 at 20:54
Is there maybe an $a$ missing with $2ic$? – k.stm Jan 26 '13 at 21:08
Probably not, if you reflect $z = 0$ at the vertical line $2·x + 0·y = 2$, this formula yields $\tfrac{4·i}{2·i} = 2$, the correct result, the formula of my calculations give you $4$, which is false. – k.stm Jan 26 '13 at 21:28
up vote 2 down vote accepted

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.

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There must be something wrong with this, but I don't know what it is. If I don't findit , I'll delete this post. Or maybe I won't, so someone else can find it. I'm really interested. – k.stm Jan 26 '13 at 21:27
Keep it please, hopefully someone can go over it. THanks a lot! – Q.matin Jan 26 '13 at 23:50
@K.Stm The problem is in both translations. First subtract $\frac{c}{a}$ and at last add $\frac{c}{a}$. – RicardoCruz Jan 27 '13 at 16:38
@RicardoCruz Thanks, I see my mistake now. I updated the answer. – k.stm Jan 27 '13 at 22:38

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