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)

Andhra Pradesh Board Solutions for Exercise 2: Exercise 1(b)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh 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 Andhra Pradesh Board Solutions for Exercise 2: Exercise 1(b) with Hints & Solutions

MEDIUM

12th Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 $y$ .