If 1 , ω , ω 2 are three cube roots of unity, then prove that p + q ω + r ω 2 r + p ω + q ω 2 = ω 2 .

We have to prove that, p + q ω + r ω 2 r + p ω + q ω 2 = ω 2 From LHS, we have = p + q ω + r ω 2 r + p ω + q ω 2 = ω ω p + q ω + r ω 2 r + p ω + q ω 2 (Multiply by ω to numerator and denominator) = 1 ω p ω + q ω 2 + r ω 3 r + p ω + q ω 2 = 1 ω p ω + q ω 2 + r r + p ω + q ω 2 ∵ ω 3 = 1 = 1 ω = ω 3 ω ∵ ω 3 = 1 = ω 2 = RHS Hence, proved.

If x = a + b ω + c ω 2 , y = a ω + b ω 2 + c and z = a ω 2 + b + c ω , then prove that x 2 y z + y 2 z x + z 2 x y = 3 .

Given, x = a + b ω + c ω 2 , y = a ω + b ω 2 + c and z = a ω 2 + b + c ω Now, x + y + z = a + b ω + c ω 2 + a ω + b ω 2 + c + a ω 2 + b + c ω ⇒ x + y + z = a + a ω + a ω 2 + b + b ω + b ω 2 + c + c ω + c ω 2 ⇒ x + y + z = a 1 + ω + ω 2 + b 1 + ω + ω 2 + c 1 + ω + ω 2 ⇒ x + y + z = a . 0 + b . 0 + c . 0 ∵ 1 + ω + ω 2 = 0 ⇒ x + y + z = 0 We know that, if x + y + z = 0 ⇒ x 3 + y 3 + z 3 = 3 x y z → 1 Now, x 2 y z + y 2 z x + z 2 x y = x 3 + y 3 + z 3 x y z (Apply LCM) = 3 x y z x y z (From 1 ) = 3 . Hence, proved.

Find the value of ( - 1 + 3 i ) 100 + ( - 1 - 3 i ) 100 .

We know that cube roots of unity are 1 , ω , ω 2 . ∴ ω = - 1 + 3 i 2 , ω 2 = - 1 - 3 i 2 → 1 Given, ( - 1 + 3 i ) 100 + ( - 1 - 3 i ) 100 = ( 2 ω ) 100 + ( 2 ω 2 ) 100 (From 1 ) = 2 100 ω 100 + 2 100 ω 200 = 2 100 ω 100 + ω 200 = 2 100 ω 3 33 . ω + ω 3 66 . ω 2 = 2 100 ω + ω 2 ∵ ω 3 = 1 = 2 100 - 1 ∵ 1 + ω + ω 2 = 0 = - 2 100 Hence, the value of ( - 1 + 3 i ) 100 + ( - 1 - 3 i ) 100 is - 2 100 .

Prove that - 1 + i 3 2 n + - 1 - i 3 2 n is equal to 2 if n is a multiple of 3 .

We know that cube roots of unity are 1 , ω , ω 2 . ∴ ω = - 1 + 3 i 2 , ω 2 = - 1 - 3 i 2 → 1 Given, - 1 + i 3 2 n + - 1 - i 3 2 n = ω n + ω 2 n (From 1 ) = ω n + ω 2 n Since n is a multiple of 3 For n = 3 ∴ ω n + ω 2 n = ω 3 + ω 6 = 1 + 1 = 2 For n = 6 ∴ ω n + ω 2 n = ω 6 + ω 9 = 1 + 1 = 2 And so on. Hence, - 1 + i 3 2 n + - 1 - i 3 2 n = 2 , if n is a multiple of 3 .

Prove that - 1 + i 3 2 n + - 1 - i 3 2 n is equal to - 1 if n is not a multiple of 3 .

We know that cube roots of unity are 1 , ω , ω 2 . ∴ ω = - 1 + 3 i 2 , ω 2 = - 1 - 3 i 2 → 1 Given, - 1 + i 3 2 n + - 1 - i 3 2 n = ω n + ω 2 n (From 1 ) = ω n + ω 2 n Since n is not a multiple of 3 . For n = 1 , ∴ ω n + ω 2 n = ω 1 + ω 2 = - 1 ∵ 1 + ω + ω 2 = 0 For n = 2 , ∴ ω n + ω 2 n = ω 2 + ω 4 = ω 2 + ω 3 . ω = ω 2 + ω = - 1 ∵ ω 3 = 1 And so on. Hence, - 1 + i 3 2 n + - 1 - i 3 2 n = - 1 , if n is not a multiple of 3 .

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6

M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 6: EXERCISE 5.6

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 6: EXERCISE 5.6

Attempt the practice questions on Chapter 5: Complex Numbers, Exercise 6: EXERCISE 5.6 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 6: EXERCISE 5.6 with Hints & Solutions

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6>Q 4

If $1, ω, ω^{2}$ are three cube roots of unity, then prove that $\frac{p + q ω + r ω^{2}}{r + p ω + q ω^{2}} = ω^{2}$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6>Q 5

If $α, β$ are complex cube roots of unity other than $1$ , show that $α^{4} + β^{4} + α^{- 1} β^{- 1} = 0$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6>Q 6

If $p, q$ are complex cube roots of unity, then prove that $p^{2} + q^{2} - p q = - 2$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6>Q 7

If $x = a + b, y = a α + b β, z = a β + b α$ , where $α and β$ are complex cube roots of unity, then show that $x^{3} + y^{3} + z^{3} = 3 (a^{3} + b^{3})$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>EXERCISE 5.6>Q 8

If $x = a + b ω + c ω^{2}, y = a ω + b ω^{2} + c and z = a ω^{2} + b + c ω$ , then prove that $\frac{x^{2}}{y z} + \frac{y^{2}}{z x} + \frac{z^{2}}{x y} = 3$ .