Complex Numbers 4 Question 13
13.
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 the point $z _2$ through the line, then show that $c=\bar{z} _1 b+z _2 \bar{b}$.
(1997C, 5M)
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Solution:
- Let $Q$ be $z _2$ and its reflection be the point $P\left(z _1\right)$ in the given line. If $O(z)$ be any point on the given line then by definition $O R$ is right bisector of $Q P$.
$\therefore$
$O P=O Q$ or $\left|z-z _1\right|=\left|z-z _2\right|$
$ \begin{array}{llrl} & \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 & z\left(\bar{z} _1-\bar{z} _2\right)+\bar{z}\left(z _1-z _2\right) & =z _1 \bar{z} _1-z _2 \bar{z} _2 \end{array} $
Comparing with given line $z \bar{b}+\bar{z} b=c$
$ \begin{gathered} \frac{\bar{z} _1-\bar{z} _2}{\bar{b}}=\frac{z _1-z _2}{b}=\frac{z _1 \bar{z} _1-z _2 \bar{z} _2}{c}=\lambda, \\ \frac{\bar{z} _1-\bar{z} _2}{\lambda}=\bar{b}, \frac{z _1-z _2}{\lambda}=b, \frac{z _1 \bar{z} _1-z _2 \bar{z} _2}{\lambda}=c \\ \therefore \bar{z} _1 b+z _2 \bar{b}=\bar{z} _1 \frac{z _1-z _2}{\lambda}+z _2 \frac{\bar{z} _1-\bar{z} _2}{\lambda} \\ =\frac{z \bar{z} _1-z _2 \bar{z} _2}{\lambda}=c \end{gathered} $
