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On a website

It is written as

The concept of reflection points of a straight line or reflection on a number line is important and a bit tricky too. It is entirely different from the point reflection. Two points say P and Q are said to be the reflection points for a given straight line if the given line is the perpendicular bisector of the segment PQ. Note that the two points denoted by the complex numbers z1 and z2 will be the reflection points for the straight line enter image description here

I could not understand how they have written this condition .

2 Answers 2

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Let $z_1$ and $z_2$ are reflection points of a straight line. The intersection segment $z_1z_2$ with line be $z_0\in$Line so $\color{red}{z_0=\dfrac{z_1+z_2}{2}}$ and then \begin{eqnarray} \bar{\alpha}z_0+\alpha\bar{z_0}+r=0 &\Leftrightarrow& \\ \bar{\alpha}\dfrac{z_1+z_2}{2}+\alpha\overline{\dfrac{z_1+z_2}{2}}+r=0 &\Leftrightarrow& \\ \bar{\alpha}z_1+\bar{\alpha}z_2+\alpha\bar{z_1}+\alpha\bar{z_2}+2r=0 &\Leftrightarrow& \\ 2{\bf Re}\,(\bar{\alpha}z_1+\alpha\bar{z_2})+2r=0 &\Leftrightarrow& \\ \bar{\alpha}z_1+\alpha\bar{z_2}+r=0 \end{eqnarray} where $r\in\mathbb{R}$.

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    How do you got${\bf Re}\,(\bar{\alpha}z_1+\alpha\bar{z_2})+r=0 \\$ – 2017-01-24
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    In last statement ${\bf Im}r=0$. – 2017-01-24
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    Can you please explain it more – 2017-01-24
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    ${\bf Re}\,(\bar{\alpha}z_1+\alpha\bar{z_2})+r=0 $ shows ${\bf Re}\,(\bar{\alpha}z_1+\alpha\bar{z_2})=-r$ so $r\in\mathbb{R}$. – 2017-01-24
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    I think there is two calculation mistake in third step . ... 1) you have replaced z$_2$ with z$_2$ ... 2) there should be 2r – 2017-01-24
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Basically it boils down to showing that:

If $\bar \alpha z + \alpha \bar z =r $ is a line in the complex plane such that if $z_1$ is the reflection of $z_2$ in the given line, then $r=\bar z_1 \alpha + z_2\bar \alpha $.

Proof: Let $P (z) $ be any point on $\bar \alpha z+\bar z \alpha=r $ and let $A $ and $B $ represent $z_1$ and $z_2$. Then we have $AP=BP \Rightarrow |z-z_1|^2=|z-z_2|^2 \Rightarrow (\bar z_2 -\bar z_1)z +(z_2-z_1)\bar z=z_2\bar z_2- z _1\bar z_1$. This is equivalent to the line $\bar \alpha z +\bar z \alpha=r$

Therfore $$\frac {\bar z_2- \bar z_1}{\bar \alpha}=\frac {z_2-z_1}{\alpha}=\frac {z_2\bar z_2 -z_1 \bar z_1}{c} = k $$ Now finding $\bar z_1\alpha +z_2\bar \alpha $ gives us $r $ proving the result to be true. Hope it helps.