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 prove that $c={\overline{z}}_{1}b+z_2\overline{b}$.

Answer 1

The given line is $\overline{b}z+b\overline{z}=c$ ...(1)
Let $A\left(z_1\right)$ be a reflection of $B\left(z_2\right)$ in the line (1)
Let $P\left(z\right)$ be any point on the line (1), we have

$AP=BP\Rightarrow {\left|AP\right|}^2={\left|BP\right|}^2\Rightarrow {|z-z_1|}^2={|z-z_2|}^2\Rightarrow (z-z_1)(\overline{z}-\overline{z_1})=(z-z_2)(\overline{z}-\overline{z_2})$
$\Rightarrow (\overline{z_2}-\overline{z_1})z+(z_2-z_1)\overline{z}+z_1\overline{z}-z_2\overline{z_2}=0$ ...(2)

Since (1) represent (2) the same line, we get
$\frac{\overline{b}}{\overline{z_2}-\overline{z_1}}=\frac{b}{z_2-z_1}=\frac{c}{z_1\overline{z_1}-z_2\overline{z_2}}=k$(let)
$\Rightarrow k(\overline{z_2}-\overline{z_1})=\overline{b},k(z_2-z_1)=b,k(z_1\overline{z_1}-z_2\overline{z_2})=c$

Now $\overline{z_1}b+z_2\overline{b}$

$=\overline{z_1}\left[k(z_2-z_1)\right]+z_2\left(k(\overline{z_2}-\overline{z_1})\right)=k(\overline{z_1}{z}_2-z_1\overline{z_1}+z_2\overline{z_2}-z_2\overline{z_1})=k(z_2\overline{z_2}-z_1\overline{z_1})=c$