Question

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

• A. $\frac{{\overline{z}}_{1}b+z_2\overline{b}}{2}$
• B. ${\overline{z}}_{1}b+z_2\overline{b}$
• C. $\frac{3({\overline{z}}_{1}b+z_2\overline{b})}{2}$
• D. $2({\overline{z}}_{1}b+z_2\overline{b})$

Answer 1

The correct option is B ${\overline{z}}_{1}b+z_2\overline{b}$

Given that $z_1$ is the reflection of $z_2$ through the line $b\overline{z}+\overline{b}z=c$

Therefore, for any arbitrary point $z$ on the line, we must have
$|z-z_1|=|z-z_2|$
$\Rightarrow |z-z_1{|}^2=|z-z_2{|}^2$
$\Rightarrow |z{|}^2+|z_1{|}^2-z{\overline{z}}_1-\overline{z}{z}_1=|z{|}^2+|z_2{|}^2-z{\overline{z}}_2-\overline{z}{z}_2$
$\Rightarrow ({\overline{z}}_2-{\overline{z}}_1)z+(z_2-z_1)\overline{z}=|z_2{|}^2-|z_1{|}^2$
Thus we have
$b=z_2-z_1\text{and}c=|z_2{|}^2-|z_1{|}^2$
Now, ${\overline{z}}_{1}b+z_2\overline{b}={\overline{z}}_1(z_2-z_1)+z_2({\overline{z}}_2-{\overline{z}}_1)=|z_2{|}^2-|z_1{|}^2=c$