SATHEE: Complex Numbers 5 Question 2

Complex Numbers 5 Question 2

2.

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2} (i = \sqrt{- 1})$ , then ${(1 + i z + z^{5} + i z^{8})}^{9}$ is equal to

(2019 Main, 8 April II)

(a) 1

(b) $(- 1 + 2 i)^{9}$

(d) 0

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Answer:

Correct Answer: 2. (c)

Solution:

Key Idea: Use, $e^{i θ} = \cos θ + i \sin θ$

Given, $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i = \cos \frac{π}{6} + i \sin \frac{π}{6} = e^{i \frac{π}{6}}$

so, ${(1 + i z + z^{5} + i z^{8})}^{9}$

$= 1 + i e^{i \frac{π}{6}} + e^{i \frac{5 π}{6}} + i e^{i \frac{8 π}{6}}$

$= 1 + e^{i \frac{π}{2}} \cdot e^{i \frac{π}{6}} + e^{i \frac{5 π}{6}} + e^{i \frac{π}{2}} \cdot e^{i \frac{4 π}{3}}^{9} ∵ i = e^{i \frac{π}{2}}$

$= 1 + e^{i \frac{2 π}{3}} + e^{i \frac{5 π}{6}} + e^{i \frac{11 π}{6}}^{9}$

$= 1 + \cos \frac{2 π}{3} + i \sin \frac{2 π}{3} + \cos \frac{5 π}{6} + i \sin \frac{5 π}{6}$ $+ \cos \frac{11 π}{6} + i \sin \frac{11 π}{6}^{9}$

$= 1 - \frac{1}{2} + \frac{i \sqrt{3}}{2} - \frac{\sqrt{3}}{2} + \frac{1}{2} i + \frac{\sqrt{3}}{2} - {\frac{i}{2}}^{9}$

$= \frac{1}{2} + {\frac{\sqrt{3} i}{2}}^{9} = \cos \frac{π}{3} + i \sin {\frac{π}{3}}^{9}$

$= \cos 3 π + i \sin 3 π [∵$ for any natural number ’ $n$ ’ $(\cos θ + i \sin θ)^{n} = \cos (n θ) + i \sin (n θ)]$

$= - 1$

Complex Numbers 5 Question 2

2.

Answer:

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Complex Numbers 5 Question 2

2.

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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