If x + i y = a + i b c + i d , then write the value of x 2 + y 2 2 .

Given, x + i y = a + i b c + i d Taking modulus on both the sides, | x + i y | = a + i b c + i d = a + i b c + i d x 2 + y 2 = a 2 + b 2 c 2 + d 2 ∵ | x + i y | = x 2 + y 2 ⇒ x 2 + y 2 = a + i b c + i d Squaring again, we get, x 2 + y 2 2 = a + i b c + i d

Find z , if | z | = 4 and arg ( z ) = 5 π 6 .

Given, | z | = 4 and arg ( z ) = 5 π 6 We know that, z = | z | cos arg ( z ) + i sin arg ( z ) ⇒ z = 4 cos 5 π 6 + i sin 5 π 6 = 4 - cos π 6 + i sin π 6 = 4 - 3 2 + 1 2 i ∴ z = - 2 3 + 2 i

Write the argument of ( 1 + 3 ) ( 1 + i ) ( cos θ + i sin θ ) .

Let the argument of ( 1 + i 3 ) be α . Then, tan α = 3 1 = tan π 3 ⇒ a = π 3 Let the argument of ( 1 + i ) be β . Then, tan β = 1 1 = tan π 4 ⇒ β = π 4 Let the argument of ( cos θ + i sin θ ) b e γ tan γ = sin θ cos θ = tan θ ⇒ γ = θ Hence, argument is π 4 + π 3 + θ = 7 π 12 + θ .

E M B I B E

MATHEMATICS for CLASS XI>Chapter 13 - Complex Numbers>VERY SHORT ANSWER QUESTIONS (VSAQs)

R D Sharma Solutions for Chapter: Complex Numbers, Exercise 5: VERY SHORT ANSWER QUESTIONS (VSAQs)

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Complex Numbers, Exercise 5: VERY SHORT ANSWER QUESTIONS (VSAQs)

Attempt the practice questions on Chapter 13: Complex Numbers, Exercise 5: VERY SHORT ANSWER QUESTIONS (VSAQs) with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.