Solve the equation | z | = z + 1 + 2 i for z .

Given | z | = z + 1 + 2 i Put z = x + i y , we get | x + i y ∣ = x + i y + 1 + 2 i ⇒ x 2 + y 2 = ( x + 1 ) + i ( y + 2 ) ∵ | z | = x 2 + y 2 Comparing real and imaginary parts, we get x 2 + y 2 = x + 1 and 0 = y + 2 ⇒ y = - 2 Putting this value of y in x 2 + y 2 = x + 1 , we get x 2 + ( - 2 ) 2 = ( x + 1 ) 2 ⇒ x 2 + 4 = x 2 + 2 x + 1 ⇒ x = 3 2 ∴ z = x + i y = 3 2 - 2 i is the required solution.

Find the complex number satisfying the equation z + 2 | ( z + 1 ) | + i = 0 .

We have z + 2 | ( z + 1 ) | + i = 0 Putting z = x + i y we get, ⇒ ( x + y i ) + 2 | ( x + y i + 1 ) | + i = 0 ⇒ x + ( y + 1 ) i + 2 | ( x + 1 ) + y i ) | = 0 ⇒ x + ( y + 1 ) i + 2 ( x + 1 ) 2 + y 2 = 0 ⇒ x + ( y + 1 ) i + 2 x 2 + 2 x + 1 + y 2 = 0 + 0 i Separating real and imaginary parts we obtain, ⇒ x = - 2 x 2 + 2 x + 1 + y 2 and y = - 1 ⇒ x 2 = 2 x 2 + 2 x + 1 + y 2 ⇒ x 2 = 2 x 2 + 4 x + 2 + 2 y 2 ⇒ x 2 + 4 x + 2 + 2 y 2 = 0 ⇒ x 2 + 4 x + 2 + 2 ( - 1 ) 2 = 0 [ ∵ y = - 1 ] ⇒ x 2 + 4 x + 4 = 0 ⇒ ( x + 2 ) 2 = 0 ⇒ x + 2 = 0 ⇒ x = - 2 ∴ z = x + y i = - 2 - i .

What is the conjugate of 2 - i ( 1 - 2 i ) 2 ?

Given z = 2 - i ( 1 - 2 i ) 2 = 2 - i 1 + 4 i 2 - 4 i = 2 - i 1 - 4 - 4 i = 2 - i - 3 - 4 i By rationalisation, we get = 2 - i - 3 - 4 i × - 3 + 4 i - 3 + 4 i = - 6 + 8 i + 3 i - 4 i 2 ( - 3 ) 2 - ( 4 i ) 2 = - 6 + 11 i + 4 9 - 16 i 2 = - 2 + 11 i 9 + 16 = - 2 + 11 i 25 = - 2 25 + 11 25 i ∴ z ¯ = - 2 25 - 11 25 i

If a 2 + 1 2 2 a - i = x + i y , what is the value of x 2 + y 2 ?

We have a 2 + 1 2 2 a - i = x + i y . . . . . ( i ) Take modulus on both side ⇒ a 2 + 1 2 2 a - i = | x + i y | ⇒ a 2 + 1 2 ( 2 a ) 2 + ( - 1 ) 2 = x 2 + y 2 Squaring on both sides, ∴ x 2 + y 2 = a 2 + 1 4 4 a 2 + 1 Hence, the value of x 2 + y 2 = a 2 + 1 4 4 a 2 + 1 .

Where does z lie, if z - 5 i z + 5 i = 1

We have z - 5 i z + 5 i = 1 Let us assume z = x + y i ⇒ x + y i - 5 i x + y i + 5 i = 1 ⇒ x + ( y - 5 ) i x + ( y + 5 ) i = 1 ⇒ | x + ( y - 5 ) i | = | x + ( y + 5 ) i | ⇒ x 2 + y - 5 2 = x 2 + y + 5 2 Squaring on both sides ⇒ x 2 + ( y - 5 ) 2 = x 2 + ( y + 5 ) 2 ⇒ ( y - 5 ) 2 = ( y + 5 ) 2 ⇒ y 2 + 25 - 10 y = y 2 + 25 + 10 y ⇒ 20 y = 0 ⇒ y = 0 ∴ z lies on x - axis i.e., real axis.

E M B I B E

NCERT Exemplar Mathematics - Class 11>Chapter 5 - Complex Numbers and Quadratic Equations>EXERCISE

NCERT Solutions for Chapter: Complex Numbers and Quadratic Equations, Exercise 1: EXERCISE

Author:NCERT

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Complex Numbers and Quadratic Equations, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 5: Complex Numbers and Quadratic Equations, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. NCERT Exemplar Mathematics - Class 11 solutions are prepared by Experienced Embibe Experts.