SATHEE: Complex Numbers 2 Question 23

Complex Numbers 2 Question 23

24.

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $| z_{1} | = | z_{2} |$ . If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ may be

(1986, 2M)

(a) zero

(b) real and positive

(d) purely imaginary

Show Answer

Answer:

Correct Answer: 24. (a,d)

Solution:

Given, $| z_{1} | = | z_{2} |$

Now, $\frac{z_{1} + z_{2}}{z_{1} - z_{2}} \times \frac{{\bar{z}}_{1} - {\bar{z}}_{2}}{{\bar{z}}_{1} - \overset{―}{z_{2}}} = \frac{z_{1} {\bar{z}}_{1} - z_{1} {\bar{z}}_{2} + z_{2} {\bar{z}}_{1} - z_{2} {\bar{z}}_{2}}{{| z_{1} - z_{2} |}^{2}}$

$\begin{aligned} = \frac{{| z_{1} |}^{2} + (z_{2} {\bar{z}}_{1} - z_{1} {\bar{z}}_{2}) - {| z_{2} |}^{2}}{{| z_{1} - z_{2} |}^{2}} \\ = \frac{z_{2} {\bar{z}}_{1} - z_{1} {\bar{z}}_{2}}{{| z_{1} - z_{2} |}^{2}} [∵ {| z_{1} |}^{2} = {| z_{2} |}^{2}] \end{aligned}$

As, we know $z - \bar{z} = 2 i Im (z)$

$\begin{aligned} ∴ z_{2} {\bar{z}}_{1} - z_{1} {\bar{z}}_{2} = 2 i Im (z_{2} {\bar{z}}_{1}) \\ ∴ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{2 i Im (z_{2} {\bar{z}}_{1})}{{| z_{1} - z_{2} |}^{2}} \end{aligned}$

which is purely imaginary or zero.

Complex Numbers 2 Question 23

24.

Answer:

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Complex Numbers 2 Question 23

24.

Answer:

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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