Question

Let $\bar{b} z+b \bar{z}=c, b \neq 0$, be a line in the complex plane, where $\bar{b}$ is the complex conjugate of $b$. If a point $z_{1}$ is the reflection of a point $z_{2}$ through the line, then $\bar{z}_{1} b+$ $z_{2} \bar{b}=$

• A. $4 c$
• B. $2 c$
• C. $c$
• D. None of these

Answer 1

Answer: (C)

The given line is
$\bar{b} z+b \bar{z}=c\,\,\,$(1)
Let $A\left(z_{1}\right)$ be a reflection of $B\left(z_{2}\right)$ in the line $(1)$.
Let $P(z)$ be any point on the line $(1)$.
We have, $A P=B P$
$\Rightarrow |A P|^{2}=|B P|^{2}$
$\Rightarrow \left|z-z_{1}\right|^{2}=\left|z-z_{2}\right|^{2}$
$\Rightarrow \left(z-z_{1}\right)\left(\bar{z}-\bar{z}_{1}\right)=\left(z-z_{2}\right)\left(\bar{z}-\bar{z}_{2}\right)$
$\Rightarrow \left(\bar{z}_{2}-\bar{z}_{1}\right) z+\left(z_{2}-z_{1}\right) \bar{z}+z_{1} \bar{z}-z_{2} \bar{z}_{2}=0\,\,\,$(2)
Since (1) and (2) represent the same line, we get
$\frac{\bar{b}}{\bar{z}_{2}-\bar{z}_{1}}=\frac{b}{z_{2}-z_{1}}=\frac{c}{z_{1} \bar{z}_{1}-z_{2} \bar{z}_{2}}=k($ say $)$
$\Rightarrow k\left(\bar{z}_{2}-\bar{z}_{1}\right)=\bar{b}, k\left(z_{2}-z_{1}\right)=b, k\left(z_{1} \bar{z}_{1}-z_{2} \bar{z}_{2}\right)=c$
Now, $\bar{z}_{1} b+ z _{2} \bar{b}$
$=\bar{z}_{1}\left\{k\left(z_{2}-z_{1}\right)\right\}+z_{2}\left\{k\left(\bar{z}_{2}-\bar{z}_{1}\right)\right\}$
$=k\left\{\bar{z}_{1} z_{2}-z_{1} \bar{z}_{1}+\bar{z}_{2} z_{2}-z_{2} \bar{z}_{1}\right\}$
$=k\left(z_{2} \bar{z}_{2}-z_{1} \bar{z}_{1}\right)=c$