If x + i y 1 3 = a - i b , then show that x a - y b = 4 a 2 - b 2 .

Given, ( x + i y ) 1 3 = a - i b ⇒ x + i y = ( a - i b ) 3 ⇒ x + i y = a 3 - 3 a 2 i b + 3 a i b 2 - i b 3 ⇒ x + i y = a 3 - 3 a b 2 + i b 3 - 3 a 2 b ⇒ x = a 3 - 3 a b 2 and y = b 3 - 3 a 2 b To prove, x a - y b = 4 a 2 - b 2 LHS = x a - y b = a 3 - 3 a b 2 a - b 3 - 3 a 2 b b = a a 2 - 3 b 2 a - b b 2 - 3 a 2 b = a 2 - 3 b 2 - b 2 + 3 a 2 = 4 a 2 - b 2 = RHS Hence, proved.

Write a + i b a - i b 2 - a - i b a + i b 2 in the form x + i y .

Given, a + i b a - i b 2 - a - i b a + i b 2 = ( a + i b ) ( a + i b ) ( a - i b ) ( a + i b ) 2 - ( a - i b ) ( a - i b ) ( a + i b ) ( a - i b ) 2 = a 2 - b 2 + 2 a b i a 2 + b 2 2 - a 2 - b 2 - 2 a b i a 2 + b 2 2 We know that, m + n 2 - m - n 2 = 4 m n ∴ a + i b a - i b 2 - a - i b a + i b 2 = 8 a b a 2 - b 2 i a 2 + b 2 2 , which is in the form x + i y .

If x and y are real numbers such that ( 1 + i ) x - 2 i 3 + i + ( 2 - 3 i ) y + i 3 - i = i , then determine the values of x and y .

Given, ( 1 + i ) x - 2 i 3 + i + ( 2 - 3 i ) y + i 3 - i = i ⇒ [ ( 1 + i ) x - 2 i ] ( 3 - i ) ( 3 + i ) ( 3 - i ) + [ ( 2 - 3 i ) y + i ] ( 3 + i ) ( 3 - i ) ( 3 + i ) = i ⇒ x ( 3 - i + 3 i + 1 ) + y ( 6 + 2 i - 9 i + 3 ) - 6 i - 2 + 3 i - 1 9 + 1 = i ⇒ i ( 2 x - 7 y - 3 ) 10 + 4 x + 9 y - 3 10 = i ⇒ i ( 2 x - 7 y - 3 ) + 4 x + 9 y - 3 = 0 + 10 i ⇒ 4 x + 9 y - 3 = 0 and 2 x - 7 y - 3 = 10 ⇒ 4 x + 9 y - 3 = 0 ......(i) and 2 x - 7 y - 13 = 0 ......(ii) Solving (i) and (ii), 2 × (ii) - (i) ⇒ 4 x - 14 y - 26 - 4 x + 9 y - 3 = 0 ⇒ - 23 y - 23 = 0 ⇒ y = - 1 From (i), 4 x + 9 - 1 - 3 = 0 ⇒ x = 3 Therefore, x = 3 , y = - 1 .

If 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y , find x and y .

Given, 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y ⇒ ( 1 + i ) ( 1 + i ) ( 1 - i ) ( 1 + i ) 3 - ( 1 - i ) ( 1 - i ) ( 1 + i ) ( 1 - i ) 3 = x + i y ⇒ 1 + i 2 + 2 i 1 2 - i 2 3 - 1 + i 2 - 2 i 1 2 - i 2 3 = x + i y ⇒ 2 i 2 3 - - 2 i 2 3 = x + i y ⇒ - i - i = x + i y ∵ i 2 = - 1 , i 3 = - i & i 4 = 1 ⇒ - 2 i = x + i y ∴ x = 0 , y = - 2

Find the real values of θ in order that 3 + 2 i sin θ 1 - 2 i sin θ is a real number.

Let, z = 3 + 2 i sin θ 1 - 2 i sin θ ⇒ z = ( 3 + 2 isin θ ) ( 1 + 2 isin θ ) ( 1 - 2 isin θ ) ( 1 + 2 isin θ ) ⇒ z = 3 - 4 sin 2 θ + 8 i sin θ 1 + 4 sin 2 θ z is purely real ⇒ imaginary part = 0 ⇒ 8 sin θ 1 + 4 sin 2 θ = 0 ⇒ sin θ = 0 ∴ θ = n π , n ∈ I

Find the real values of θ in order that 3 + 2 i sin θ 1 - 2 i sin θ is a purely imaginary number.

Let, z = 3 + 2 i sin θ 1 - 2 i sin θ ⇒ z = ( 3 + 2 isin θ ) ( 1 + 2 isin θ ) ( 1 - 2 isin θ ) ( 1 + 2 isin θ ) ⇒ z = 3 - 4 sin 2 θ + 8 i sin θ 1 + 4 sin 2 θ z is purely imaginary ⇒ real part = 0 ⇒ 3 - 4 sin 2 θ 1 + 4 sin 2 θ = 0 ⇒ 3 - 4 sin 2 θ = 0 ⇒ sin 2 θ = 3 4 ⇒ sin 2 θ = sin 2 π 3 ∴ θ = n π ± π 3 , n ∈ I

Find the real values of x and y if x - 1 3 + i + y - 1 3 - i = i .

Given, x - 1 3 + i + y - 1 3 - i = i ⇒ ( x - 1 ) ( 3 - i ) ( 3 + i ) ( 3 - i ) + ( y - 1 ) ( 3 + i ) ( 3 - i ) ( 3 + i ) = i ⇒ ( x - 1 ) ( 3 - i ) 9 + 1 + ( y - 1 ) ( 3 + i ) 9 + 1 = i ⇒ 3 ( x - 1 ) + 3 ( y - 1 ) 10 + i ( 1 - x + y - 1 ) 10 = i ⇒ 3 x + 3 y - 6 10 + i - x + y 10 = i ⇒ 3 x + 3 y - 6 + i - x + y = 0 + 10 i Comparing real and imaginary parts, ⇒ 3 x + 3 y - 6 = 0 and y - x = 10 ⇒ x + y - 2 = 0 ....(i) and y - x = 10 ......(ii) Solving (i) and (ii), we get x = - 4 , y = 6 .

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 1 - Complex Numbers>Exercise 1(b)

Telangana Board Solutions for Exercise 2: Exercise 1(b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 2: Exercise 1(b)

Attempt the free practice questions from Exercise 2: Exercise 1(b) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 2: Exercise 1(b) with Hints & Solutions

MEDIUM

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 1 - Complex Numbers>Exercise 1(b)>Q II

If ${(x + i y)}^{\frac{1}{3}} = a - i b$ , then show that $\frac{x}{a} - \frac{y}{b} = 4 (a^{2} - b^{2})$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 1 - Complex Numbers>Exercise 1(b)>Q II

Write ${(\frac{a + i b}{a - i b})}^{2} - {(\frac{a - i b}{a + i b})}^{2}$ in the form $x + i y$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 1 - Complex Numbers>Exercise 1(b)>Q II

If $x$ and $= 4 (a^{2} - b^{2})$ are real numbers such that $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then determine the values of $x$ and $y$ .