Prove that z 1 - z 2 1 - z 1 z 2 ¯ < 1 if z 1 < 1 , and z 2 > 1 or z 1 > 1 and z 2 < 1

Given, z 1 - z 2 1 - z 1 z 2 ¯ < 1 ⇒ z 1 - z 2 < 1 - z 1 z 2 ¯ On squaring both sides of the inequality, we have z 1 - z 2 2 < 1 - z 1 z 2 ¯ 2 We know that z z ¯ = z 2 ⇒ z 1 - z 2 z 1 - z 2 ¯ < 1 - z 1 z 2 ¯ 1 - z 1 z 2 ¯ ¯ ⇒ z 1 - z 2 z 1 ¯ - z 2 ¯ < 1 - z 1 z 2 ¯ 1 - z 1 ¯ z 2 ¯ ¯ ⇒ z 1 - z 2 z 1 ¯ - z 2 ¯ < 1 - z 1 z 2 ¯ 1 - z 1 ¯ z 2 ∵ z ¯ ¯ = z ⇒ z 1 z 1 ¯ - z 1 z 2 ¯ - z 2 z 1 ¯ + z 2 z 2 ¯ < 1 - z 1 ¯ z 2 - z 1 z 2 ¯ + z 1 z 1 ¯ z 2 z 2 ¯ ⇒ z 1 2 - z 1 z 2 ¯ - z 2 z 1 ¯ + z 2 2 < 1 - z 1 ¯ z 2 - z 1 z 2 ¯ + z 1 2 z 2 2 ∵ z z ¯ = z 2 ⇒ z 1 2 + z 2 2 < 1 + z 1 2 z 2 2 ⇒ z 1 2 - z 1 2 z 2 2 + z 2 2 - 1 < 0 ⇒ z 1 2 1 - z 2 2 - 1 1 - z 2 2 < 0 ⇒ z 1 2 - 1 1 - z 2 2 < 0 ⇒ z 1 2 - 1 < 0 and 1 - z 2 2 < 0 or z 1 2 - 1 > 0 and 1 - z 2 2 > 0 ⇒ z 1 2 < 1 and z 2 2 > 1 or ⇒ z 1 2 > 1 and z 2 2 < 1 ⇒ z 1 < 1 and z 2 > 1 or ⇒ z 1 > 1 and z 2 < 1

Solve ( 1 - i ) x + 3 i 2 + i + ( 3 + 2 i ) y + i 2 - i = - i where x , y , ∈ R

Given, ( 1 - i ) x + 3 i 2 + i + ( 3 + 2 i ) y + i 2 - i = - i Let z 1 = ( 1 - i ) x + 3 i 2 + i = ( 1 - i ) x + 3 i 2 + i × 2 - i 2 - i [Rationalizing the denominator] = 2 ( 1 - i ) x + 3 i - i ( 1 - i ) x + 3 i 2 + i 2 - i = 2 ( 1 - i ) x + 6 i - i ( 1 - i ) x - 3 i 2 2 + i 2 - i = 2 x - 2 x i + 6 i - i x + i 2 x - 3 i 2 2 2 - i 2 ∵ a 2 - b 2 = a - b a + b = 2 x - 2 x i + 6 i - i x - x + 3 4 - - 1 ∵ i 2 = - 1 = x + 3 + - 3 x + 6 i 5 . . . . . . i Let z 2 = ( 3 + 2 i ) y + i 2 - i = ( 3 + 2 i ) y + i 2 - i × 2 + i 2 + i = 2 3 + 2 i y + 2 i + i 3 + 2 i y + i 2 2 2 - i 2 = 6 y + 4 i y + 2 i + 3 i y + 2 i 2 y - 1 4 - - 1 = 6 y + 4 i y + 2 i + 3 i y - 2 y - 1 5 = 4 y - 1 + i 7 y + 2 5 . . . . . . . i i Now, z 1 + z 2 = - i (given) ⇒ x + 3 + - 3 x + 6 i 5 + 4 y - 1 + i ( 7 y + 2 ) 5 = - i ⇒ x + 3 + - 3 x + 6 i + 4 y - 1 + i ( 7 y + 2 ) 5 = - i ⇒ x + 3 - 3 i x + 6 i + 4 y - 1 + 7 i y + 2 i 5 = - i ⇒ x + 4 y + 2 - 3 i x + 8 i + 7 i y 5 = - i ⇒ x + 4 y + 2 + i - 3 x + 8 + 7 y 5 = - i ⇒ x + 4 y + 2 5 + i - 3 x + 7 y + 8 5 = 0 - i ⇔ x + 4 y + 2 = 0 . . . . i i i and - 3 x + 7 y + 8 = - 1 . . . . i v Applying 3 i i i + i v , we have 12 y + 7 y + 6 + 8 = - 1 ⇒ 19 y + 14 = - 1 ⇒ y = - 15 19 Substituting the value of y in equation i i i , we have x + 4 - 15 19 + 2 = 0 ⇒ x - 60 19 + 2 = 0 ⇒ x = 22 19 Hence, x = 22 19 , and y = - 15 19

If 1 + x + x 2 n = p 0 + p 1 x + p 2 x 2 + … + p 2 n x 2 n , then prove that p 0 + p 3 + p 6 + … = 3 n - 1

Given, 1 + x + x 2 n = p 0 + p 1 x + p 2 x 2 + … + p 2 n x 2 n . . . . . i To Prove : p 0 + p 3 + p 6 + … = 3 n - 1 Put x = 1 in equation i 1 + 1 + 1 2 n = p 0 + p 1 + p 2 + … + p 2 n ⇒ 3 n = p 0 + p 1 + p 2 + … + p 2 n . . . . i i Put x = ω in equation i , where ω is an imaginary cube root of unity 1 + ω + ω 2 n = p 0 + p 1 ω + p 2 ω 2 + p 3 ω 3 + p 4 ω 4 + … + p 2 n ω 2 n ⇒ 0 = p 0 + p 1 ω + p 2 ω 2 + p 3 + p 4 ω 3 · ω + p 5 ω 3 · ω 2 + … + p 2 n ω 2 n ∵ 1 + ω + ω 2 = 0 , ω 3 = 1 ⇒ 0 = p 0 + p 1 ω + p 2 ω 2 + p 3 + p 4 ω + p 5 ω 2 + … + p 2 n ω 2 n . . . i i i Put x = ω 2 in equation i , 1 + ω 2 + ω 4 n = p 0 + p 1 ω 2 + p 2 ω 4 + p 3 ω 6 + p 4 ω 8 + … + p 2 n ω 2 2 n ⇒ 1 + ω 2 + ω 3 · ω n = p 0 + p 1 ω 2 + p 2 ω 3 · ω + p 3 ω 3 2 + p 4 ω 3 2 · ω 2 + … + p 2 n ω 4 n ⇒ 0 = p 0 + p 1 ω 2 + p 2 ω + p 3 + p 4 ω 2 + … + p 2 n ω 4 n . . . . i v ∵ 1 + ω + ω 2 = 0 , ω 3 = 1 Adding i i , i i i , and i v , we have 3 n = 3 p 0 + p 1 1 + ω + ω 2 + p 2 1 + ω + ω 2 + 3 p 3 + p 4 1 + ω + ω 2 + p 5 1 + ω + ω 2 + 3 p 6 + . . . . We know that 1 + ω + ω 2 = 0 ⇒ 3 n = 3 p 0 + 3 p 3 + 3 p 6 + . . . . ⇒ 3 n = 3 p 0 + p 3 + p 6 + . . . ⇒ p 0 + p 3 + p 6 + … = 3 n - 1 Hence, Proved.

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 6 - Complex Numbers and Quadratic Equations>EXERCISES 6 (b)

Odisha Board Solutions for Chapter: Complex Numbers and Quadratic Equations, Exercise 2: EXERCISES 6 (b)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Complex Numbers and Quadratic Equations, Exercise 2: EXERCISES 6 (b)

Attempt the free practice questions on Chapter 6: Complex Numbers and Quadratic Equations, Exercise 2: EXERCISES 6 (b) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.

Questions from Odisha Board Solutions for Chapter: Complex Numbers and Quadratic Equations, Exercise 2: EXERCISES 6 (b) with Hints & Solutions

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 6 - Complex Numbers and Quadratic Equations>EXERCISES 6 (b)>Q 23

If $ω$ , and $ω^{2}$ are imaginary cube roots of unity then find the value of $\arg ω + \arg ω^{2}$ .

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 6 - Complex Numbers and Quadratic Equations>EXERCISES 6 (b)>Q 24

Prove that $|\frac{z_{1} - z_{2}}{1 - z_{1} \bar{z_{2}}}| < 1$ if $|z_{1}| < 1, and |z_{2}| > 1$ or $|z_{1}| > 1$ and $|z_{2}| < 1$

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 6 - Complex Numbers and Quadratic Equations>EXERCISES 6 (b)>Q 25

If $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} - z_{1} z_{2} - z_{2} z_{3} - z_{3} z_{1} = 0$ , show that
$|z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{1}| \cdot$