Question

Let $\overline{b}z+b\overline{z}=c,b\ne 0,$ be a line in the complex plane, where $\overline{b}$ is the complex conjugate of $b$. If a point $z_1$ is the reflection of a point $z_2$ through the line, then show that $c={\overline{z}}_{1}b+z_2\overline{b}$.

Answer 1

Let $Q$ be $z_2$ and its reflection be the point $P\left(z_1\right)$ in the given line. If $O\left(z\right)$ be any point on the given
line then by definition $OR$ is right bisector of $QP$.
$\therefore OP=OQ$ or $|z-z_1|=|z-z_2|$
or ${{|z-z_1|}^2=|z-z_2|}^2$
or $(z-z_1)(\overline{z}-{\overline{z}}_1)=(z-z_2)(\overline{z}-{\overline{z}}_2)$
Comparing with given line $z\overline{b}+\overline{z}b=c$
or $\frac{{\overline{z}}_1-{\overline{z}}_2}{\overline{b}}=\frac{z_1-z_2}{b}=\frac{z_1{\overline{z}}_1-z_2{\overline{z}}_2}{c}=\lambda$, say
$\frac{{\overline{z}}_1-{\overline{z}}_2}{\lambda }=\overline{b},\frac{z_1-z_2}{\lambda }=b,\frac{z_1{\overline{z}}_1-z_2{\overline{z}}_2}{\lambda }=c$
Also ${\overline{z}}_{1}b+z_2\overline{b}={\overline{z}}_1\frac{z_1-z_2}{\lambda }+z_2\frac{{\overline{z}}_2-{\overline{z}}_2}{\lambda }$
.$\frac{{\overline{z}}_1{\overline{z}}_1-{\overline{z}}_2{\overline{z}}_2}{\lambda }=c$ by $\left(1\right)$