SATHEE: Complex Numbers 4 Question 13

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 $P (z_{1})$ in the given line. If $O (z)$ be any point on the given line then by definition $O R$ is right bisector of $Q P$ .

$∴$

$O P = O Q$ or $| z - z_{1} | = | z - z_{2} |$

$\begin{array}{llrl} \Rightarrow & {| z - z_{1} |}^{2} & = {| z - z_{2} |}^{2} \\ \Rightarrow & (z - z_{1}) (\bar{z} - {\bar{z}}_{1}) & = (z - z_{2}) (\bar{z} - {\bar{z}}_{2}) \\ \Rightarrow & z ({\bar{z}}_{1} - {\bar{z}}_{2}) + \bar{z} (z_{1} - z_{2}) & = z_{1} {\bar{z}}_{1} - z_{2} {\bar{z}}_{2} \end{array}$

Comparing with given line $z \bar{b} + \bar{z} b = c$

$\begin{matrix} \frac{{\bar{z}}_{1} - {\bar{z}}_{2}}{\bar{b}} = \frac{z_{1} - z_{2}}{b} = \frac{z_{1} {\bar{z}}_{1} - z_{2} {\bar{z}}_{2}}{c} = λ, \\ \frac{{\bar{z}}_{1} - {\bar{z}}_{2}}{λ} = \bar{b}, \frac{z_{1} - z_{2}}{λ} = b, \frac{z_{1} {\bar{z}}_{1} - z_{2} {\bar{z}}_{2}}{λ} = c \\ ∴ {\bar{z}}_{1} b + z_{2} \bar{b} = {\bar{z}}_{1} \frac{z_{1} - z_{2}}{λ} + z_{2} \frac{{\bar{z}}_{1} - {\bar{z}}_{2}}{λ} \\ = \frac{z {\bar{z}}_{1} - z_{2} {\bar{z}}_{2}}{λ} = c \end{matrix}$

Complex Numbers 4 Question 13

13.

Solution:

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Complex Numbers 4 Question 13

13.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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