Find the conjugate of the following: 1 ( 4 + 3 i )

Given, 1 ( 4 + 3 i ) Let z = 1 ( 4 + 3 i ) Multiply and divide by 4 - 3 i ⇒ z = 1 4 + 3 i × 4 - 3 i 4 - 3 i = 4 - 3 i 4 2 - 3 i 2 = 4 - 3 i 16 - 9 i 2 = 4 - 3 i 16 - 9 - 1 ∵ i 2 = - 1 = 4 - 3 i 25 = 4 25 - i 3 25 We know that the conjugate of z = a + i b , where a and b are real numbers is given by z ¯ = a - i b Thus, for the given complex number, we have z ¯ = 4 25 + i 3 25

Find the conjugate of the following: ( 1 + i ) 2 ( 3 - i )

Given, ( 1 + i ) 2 ( 3 - i ) Let z = ( 1 + i ) 2 ( 3 - i ) = 1 + i 2 + 2 i 3 - i = 1 - 1 + 2 i 3 - i = 2 i 3 - i Multiply and divide by 3 + i z = 2 i 3 - i × 3 + i 3 + i = 6 i + 2 i 2 3 2 - i 2 = - 2 + 6 i 9 - - 1 ∵ i 2 = - 1 = - 2 + 6 i 10 = - 1 5 + i 3 5 We know that the conjugate of complex number z = a + i b , where a and b are real numbers is given by z ¯ = a - i b Hence, the conjugate of the given complex number is - 1 5 - i 3 5 .

Find the conjugate of the following: ( 1 + i ) ( 2 + i ) ( 3 + i )

Given, ( 1 + i ) ( 2 + i ) ( 3 + i ) Let z = ( 1 + i ) ( 2 + i ) ( 3 + i ) = 2 + i + 2 i + i 2 3 + i = 1 + 3 i 3 + i ∵ i 2 = - 1 Rationalizing the denominator, we have z = 1 + 3 i 3 + i × 3 - i 3 - i = 3 - i + 9 i - 3 i 2 3 2 - i 2 = 3 + 8 i - 3 - 1 9 - - 1 = 3 + 8 i + 3 10 = 6 + 8 i 10 = 6 10 + i 8 10 = 3 5 + i 4 5 We know that the conjugate of z = a + i b , where a and b are real numbers is given by z ¯ = a - i b . Hence, the conjugate of the given complex number is 3 5 - 4 5 i .

Separate 3 + - 1 2 - - 1 into real and imaginary parts, and hence find its modulus.

Given, 3 + - 1 2 - - 1 We know that i = - 1 ∴ 3 + - 1 2 - - 1 = 3 + i 2 - i Multiply the numerator and denominator by 2 + i , we have = 3 + i 2 - i × 2 + i 2 + i = 6 + 3 i + 2 i + i 2 2 2 - i 2 ∵ a + b a - b = a 2 - b 2 = 6 + 5 i - 1 4 - - 1 ∵ i 2 = - 1 = 5 + 5 i 5 = 1 + i We know that the modulus of complex number z = a + i b , where a and b are real numbers is defined by z = a 2 + b 2 ∴ The modulus of 1 + i is given by 1 2 + 1 2 = 2

If z is a complex number such that | z | = 1 , prove that z - 1 z + 1 , where z ≠ - 1 is purely imaginary except when z = 1 .

Given, for a complex number z , z = 1 We need to prove that z - 1 z + 1 , where z ≠ - 1 is purely imaginary except when z = 1 . Consider z - 1 z + 1 ¯ = z - 1 ¯ z + 1 ¯ = z ¯ - 1 z ¯ + 1 Multiplying the numerator and denominator by z , we have = z z ¯ - z z z ¯ + z = z 2 - z z 2 + z = 1 - z 1 + z ∵ z = 1 = - z - 1 z + 1 Since z - 1 z + 1 ¯ = - z - 1 z + 1 Hence, the given complex number is purely imaginary except when z = 1 . At z = 1 z - 1 z + 1 = 1 - 1 1 + 1 = 0 , which is a purely real number. Hence Proved.

If ( x + i y ) 1 / 3 = ( a + i b ) , prove that x a + y b = 4 a 2 - b 2 .

Given, ( x + i y ) 1 / 3 = ( a + i b ) To Prove: x a + y b = 4 a 2 - b 2 Consider ( x + i y ) 1 / 3 = ( a + i b ) ⇒ x + i y = a + i b 3 = a 3 + i b 3 + 3 i a b a + i b = a 3 + i 3 b 3 + 3 i a 2 b + 3 i 2 a b 2 Using i 2 = - 1 and i 3 = i 2 · i = - i = a 3 - i b 3 + 3 i a 2 b - 3 a b 2 x + i y = a 3 - 3 a b 2 + i 3 a 2 b - b 3 We know that two complex numbers are equal if and only if their real and imaginary parts are respectively equal to each other. ⇒ x = a 3 - 3 a b 2 and y = 3 a 2 b - b 3 ⇒ x = a a 2 - 3 b 2 and y = b 3 a 2 - b 2 ⇒ x a = a 2 - 3 b 2 and y b = 3 a 2 - b 2 Then, x a + y b = a 2 - 3 b 2 + 3 a 2 - b 2 = 4 a 2 - 4 b 2 Hence, x a + y b = 4 a 2 - b 2

If x + i y = ( a + i b ) ( a - i b ) , prove that x 2 + y 2 = 1 .

Given, x + i y = ( a + i b ) ( a - i b ) To Prove: x 2 + y 2 = 1 Consider x + i y = ( a + i b ) ( a - i b ) Multiply and divide the numerator and denominator in the R.H.S. by a + i b = a + i b a - i b × a + i b a + i b = a + i b 2 a 2 - i b 2 ∵ x + y x - y = x 2 - y 2 = a 2 + i b 2 + 2 i a b a 2 - i 2 b 2 = a 2 + i 2 b 2 + i 2 a b a 2 - - 1 b 2 ∵ i 2 = - 1 = a 2 - b 2 + i 2 a b a 2 + b 2 x + i y = a 2 - b 2 a 2 + b 2 + i 2 a b a 2 + b 2 We know that two complex numbers are equal if and only if their real and imaginary parts are respectively equal to each other. ⇒ x = a 2 - b 2 a 2 + b 2 and y = 2 a b a 2 + b 2 Now, x 2 + y 2 = a 2 - b 2 a 2 + b 2 2 + 2 a b a 2 + b 2 2 = a 2 - b 2 2 a 2 + b 2 2 + 4 a 2 b 2 a 2 + b 2 2 = a 4 + b 4 - 2 a 2 b 2 + 4 a 2 b 2 a 2 + b 2 2 ∵ x + y 2 = x 2 + 2 x y + y 2 = a 4 + b 4 + 2 a 2 b 2 a 2 + b 2 2 = a 2 + b 2 2 a 2 + b 2 2 = 1 Hence Proved.

If z = ( x + i y ) and w = 1 - i z z - i such that w = 1 , then show that z is purely real.

Given, z = x + i y and w = 1 - i z z - i such that w = 1 Consider w = 1 - i z z - i ⇒ w = 1 - i z z - i ⇒ | w | = | ( 1 - i z ) | | ( z - i ) | ⇒ | w | 2 = | ( 1 - i z ) | 2 | ( z - i ) | 2 ⇒ | w | 2 = | ( 1 - i ( x + i y ) ) | 2 | ( x + i y ) - i | 2 ∵ z = x + i y ⇒ 1 2 = | 1 - i x - i 2 y | 2 | x + ( y - 1 ) i | 2 ∵ w = 1 ⇒ 1 = 1 - i x + y 2 x + y - 1 i 2 ∵ i 2 = - 1 ⇒ 1 = | ( 1 + y ) - i x | 2 | x + ( y - 1 ) i | 2 We know that the modulus of a complex number z = a + i b , where a and b are real numbers is defined by z = a 2 + b 2 ⇒ x 2 + ( y - 1 ) 2 = x 2 + ( 1 + y ) 2 ⇒ y - 1 2 - 1 + y 2 = 0 ⇒ y - 1 - 1 - y y - 1 + y + 1 = 0 ∵ a + b a - b = a 2 - b 2 ⇒ - 2 2 y = 0 ⇒ y = 0 ⇒ z = x + i y = x + i 0 We know that the imaginary part of a purely real number is zero. Hence, z is purely real.

E M B I B E

WBCHSE Mathematics for Class 11>Algebra>Chapter 4 - Complex Numbers and Quadratic Equations>EXERCISE 2B

R S Aggarwal Solutions for Exercise 2: EXERCISE 2B

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE 2B

Attempt the practice questions from Exercise 2: EXERCISE 2B with hints and solutions to strengthen your understanding. WBCHSE Mathematics for Class 11 solutions are prepared by Experienced Embibe Experts.