SATHEE: Complex Numbers 2 Question 24

Complex Numbers 2 Question 24

25.

If $z_{1} = a + i b$ and $z_{2} = c + i d$ are complex numbers such that $| z_{1} | = | z_{2} | = 1$ and $Re (z_{1} {\bar{z}}_{2}) = 0$ , then the pair of complex numbers $w_{1} = a + i c$ and $w_{2} = b + i d$ satisfies

(a) $| w_{1} | = 1$

(b) $| w_{2} | = 1$

(d) None of these

$(1985, 2 M)$

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Answer:

Correct Answer: 25. (a,b,c)

Solution:

Since, $z_{1} = a + i b$ and $z_{2} = c + i d$

$\Rightarrow {| z_{1} |}^{2} = a^{2} + b^{2} = 1$ and ${| z_{2} |}^{2} = c^{2} + d^{2} = 1$

$[∵ | z_{1} | = | z_{2} | = 1]$

Also, $Re (z_{1} {\bar{z}}_{2}) = 0 \Rightarrow a c + b d = 0$

$\Rightarrow \frac{a}{b} = - \frac{d}{c} = λ$

From Eqs. (i) and (ii), $b^{2} λ^{2} + b^{2} = c^{2} + λ^{2} c^{2}$

$\Rightarrow b^{2} = c^{2}$ and $a^{2} = d^{2}$

Also, given $w_{1} = a + i c$ and $w_{2} = b + i d$

Now, $| w_{1} | = \sqrt{a^{2} + c^{2}} = \sqrt{a^{2} + b^{2}} = 1$

$| w_{2} | = \sqrt{b^{2} + d^{2}} = \sqrt{a^{2} + b^{2}} = 1$

and $Re (w_{1} \overset{―}{w_{2}}) = a b + c d = (b λ) b + c (- λ c)$ [from Eq. (i)]

$= λ (b^{2} - c^{2}) = 0$

Complex Numbers 2 Question 24

25.

Answer:

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Complex Numbers 2 Question 24

25.

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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