If a - i b a + i b = 1 + i 1 - i , then prove that a + b = 0 .

Given, a - i b a + i b = 1 + i 1 - i We know that, if a b = c d ⇒ a + b a - b = c + d c - d (Componendo and Dividendo rule) ∴ a - i b + a + i b a - i b - a - i b = 1 + i + 1 - i 1 + i - 1 + i ⇒ 2 a - 2 i b = 2 2 i ⇒ a - b = 1 ⇒ a = - b ⇒ a + b = 0 Hence, proved.

If ( a + i b ) ( x + i y ) = a 2 + b 2 i , a 2 + b 2 ≠ 0 , then show that x = b and y = a .

Given, ( a + i b ) ( x + i y ) = a 2 + b 2 i ⇒ a x + a i y + b i x + i 2 b y = a 2 + b 2 i ⇒ a x + a i y + b i x - b y = a 2 + b 2 i ∵ i 2 = - 1 ⇒ ( a x - b y ) + i ( b x + a y ) = 0 + a 2 + b 2 i Now equate real an imaginary parts on each side, we get ⇒ a x - b y = 0 and b x + a y = a 2 + b 2 ⇒ y = a b x and b x + a · a b x = a 2 + b 2 ⇒ y = a b x and a 2 + b 2 b x = a 2 + b 2 ⇒ y = a b x and x = b ⇒ x = b and y = a Hence, proved.

Find the complex numbers z satisfying the equations z - 4 z - 8 = 1 and z - 12 z - 8 i = 5 3 .

Given equations, z - 4 z - 8 = 1 → 1 and z - 12 z - 8 i = 5 3 → 2 Putting z = x + i y in 1 , we get ⇒ x + i y - 4 x + i y - 8 = 1 ⇒ x - 4 + i y x - 8 + i y = 1 ⇒ x - 4 + i y x - 8 + i y = 1 ∵ z 1 z 2 = z 1 z 2 , z 2 ≠ 0 ⇒ x - 4 2 + y 2 x - 8 2 + y 2 = 1 ∵ | x + i y | = x 2 + y 2 ⇒ x - 4 2 + y 2 = x - 8 2 + y 2 ⇒ x - 4 2 + y 2 = x - 8 2 + y 2 ⇒ x 2 - 8 x + 16 + y 2 = x 2 - 16 x + 64 + y 2 ⇒ - 8 x + 16 = - 16 x + 64 ⇒ 8 x = 48 ⇒ x = 6 → 3 From 2 , we have, z - 12 z - 8 i = 5 3 Putting z = x + i y , we get ⇒ x + i y - 12 x + i y - 8 i = 5 3 ⇒ x - 12 + i y x + i y - 8 = 5 3 ⇒ x - 12 + i y x + i ( y - 8 ) = 5 3 ∵ z 1 z 2 = z 1 z 2 , z 2 ≠ 0 ⇒ x - 12 2 + y 2 x 2 + y - 8 2 = 5 3 ∵ x + i y = x 2 + y 2 ⇒ ( x - 12 ) 2 + y 2 x 2 + ( y - 8 ) 2 = 25 9 (Squaring on both sides) ⇒ 9 ( 6 - 12 ) 2 + y 2 = 25 36 + ( y - 8 ) 2 From 3 ⇒ 9 36 + y 2 = 25 36 + ( y - 8 ) 2 ⇒ 324 + 9 y 2 = 900 + 25 ( y - 8 ) 2 ⇒ 324 + 9 y 2 = 900 + 25 y 2 - 16 y + 64 ⇒ 324 + 9 y 2 = 900 + 25 y 2 - 400 y + 1600 ⇒ 0 = 16 y 2 - 400 y + 2176 ⇒ y 2 - 25 y + 136 = 0 (Divide by 16 on both sides) ⇒ y 2 - 17 y - 8 y + 136 = 0 ⇒ y y - 17 - 8 y - 17 = 0 ⇒ y - 8 y - 17 = 0 ⇒ y = 8 or y = 17 Hence, the required complex numbers are 6 + 8 i or 6 + 17 i .

Find the square root of - 2 + 2 - 3 .

Let - 2 + 2 - 3 = - 2 + i 2 3 = ± x + i y , x , y ∈ R Squaring on both sides, we get ⇒ - 2 + i 2 3 = ( x + i y ) 2 ⇒ - 2 + i 2 3 = x 2 + ( i y ) 2 + 2 x i y ⇒ - 2 + i 2 3 = x 2 - y 2 + 2 x i y ∵ i 2 = - 1 Now equate real and imaginary parts on each side, we get ⇒ x 2 - y 2 = - 2 → 1 , 2 x y = 2 3 → 2 Now, x 2 + y 2 2 = x 2 - y 2 2 + 4 x 2 y 2 ⇒ x 2 + y 2 2 = x 2 - y 2 2 + 2 x y 2 ⇒ x 2 + y 2 2 = 4 + 12 ⇒ x 2 + y 2 = 16 ⇒ x 2 + y 2 = ± 4 As x and y are real numbers, so x 2 + y 2 cannot be negative. ∴ x 2 + y 2 = 4 → 3 Now, add 1 and 3 , we get ⇒ 2 x 2 = 2 ⇒ x 2 = 1 ⇒ x = ± 1 From 2 , we have y = 3 x = 3 ± 1 = ± 3 ∴ - 2 + i 2 3 = ± ( 1 + 3 i ) .

Find the square root of 1 + 2 6 i .

Let 1 + 2 6 i = ± x + i y , x , y ∈ R Squaring on both sides, we get ⇒ 1 + 2 6 i = ( x + i y ) 2 ⇒ 1 + 2 6 i = x 2 + ( i y ) 2 + 2 x i y ⇒ 1 + 2 6 i = x 2 - y 2 + 2 x i y ∵ i 2 = - 1 Now equate real and imaginary parts on each side, we get ⇒ x 2 - y 2 = 1 → 1 , 2 x y = 2 6 → 2 Now, x 2 + y 2 2 = x 2 - y 2 2 + 4 x 2 y 2 ⇒ x 2 + y 2 2 = x 2 - y 2 2 + 2 x y 2 ⇒ x 2 + y 2 2 = 1 + 24 ⇒ x 2 + y 2 = 25 ⇒ x 2 + y 2 = ± 5 As x and y are real numbers, so x 2 + y 2 cannot be negative. ∴ x 2 + y 2 = 5 → 3 Now, add 1 and 3 , we get ⇒ 2 x 2 = 6 ⇒ x 2 = 3 ⇒ x = ± 3 From 2 , we have y = 6 x = 6 ± 3 = ± 2 ∴ 1 + 2 6 i = ± ( 3 + 2 i ) .

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST

M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 8: CHAPTER TEST

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 8: CHAPTER TEST

Attempt the practice questions on Chapter 5: Complex Numbers, Exercise 8: CHAPTER TEST with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 8: CHAPTER TEST with Hints & Solutions

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 9

If $\frac{a - i b}{a + i b} = \frac{1 + i}{1 - i}$ , then prove that $a + b = 0$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 10

If $(a + i b) (x + i y) = (a^{2} + b^{2}) i, a^{2} + b^{2} \neq 0$ , then show that $x = b$ and $y = a$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 12

Find the complex numbers $z$ satisfying the equations $|\frac{z - 4}{z - 8}| = 1 and |\frac{z - 12}{z - 8 i}| = \frac{5}{3}$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 13

Find the square root of $- 5 - 12 i$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 13

Find the square root of $- 2 + 2 \sqrt{- 3}$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 13

Find the square root of $1 + 2 \sqrt{6} i$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 14

If $ω$ is a cube root of unity, then find the conjugate of $2 ω - 3 i$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 5 - Complex Numbers>CHAPTER TEST>Q 15

If $x = a + b, y = a ω + b ω^{2} and z = a ω^{2} + b ω$ , where $ω$ and $ω^{2}$ are cube roots of unity, then show that $x^{2} + y^{2} + z^{2} = 6 a b$ .

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