If x = cis θ , then find the value of x 6 + 1 x 6 .

Given that, x = cis θ = cos θ + i sin θ = e i θ Now consider, x 6 + 1 x 6 = e 6 i θ + 1 e 6 i θ = e 6 i θ + e - 6 i θ ∵ 1 x = x - 1 = cos 6 θ + i sin 6 θ + cos - 6 θ + i sin - 6 θ = cos 6 θ + i sin 6 θ + cos 6 θ - i sin 6 θ ∵ cos - θ = cos θ a n d sin - θ = - sin θ = 2 cos 6 θ Therefore, x 6 + 1 x 6 = 2 cos 6 θ .

If the real part of z + 1 z + i is 1 , then find the locus of z where z = x + i y .

Given, z + 1 z + i Let, z = x + i y ⇒ z + 1 z + i = ( x + 1 ) + i y ( x ) + ( y + 1 ) i = [ ( x + 1 ) + i y ] [ x - ( y + 1 ) i ] [ x + ( y + 1 ) i ] [ x - ( y + 1 ) i ] = x ( x + 1 ) + y ( y + 1 ) + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 = x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 Given, real part = 1 ⇒ x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 = 1 ⇒ x 2 + y 2 + x + y = x 2 + ( y + 1 ) 2 ⇒ x 2 + y 2 + x + y = x 2 + y 2 + 2 y + 1 ⇒ x - y = 1 ∴ Locus of z is x - y = 1 .

Write z = - 3 + i in modulus-amplitude form.

Given that z = - 3 + i We know that the modulus-amplitude form of z = a + i b is z = r cos θ + i sin θ , where r = a 2 + b 2 and θ is the principal argument of complex number. Modulus of the given complex number is, r = - 3 2 + 1 2 = 3 + 1 = 2 The complex number z = - 3 + i is in the second quadrant. When a complex number z = x + i y is belongs to the second quadrant, the principal argument is θ = π - tan - 1 y x . For given complex number z = - 3 + i , θ = π - tan - 1 1 - 3 = π - tan - 1 1 3 ⇒ θ = π - π 6 = 5 π 6 Hence, the required modulus-amplitude form is z = 2 cos 5 π 6 + i sin 5 π 6 .

Show that the points in the Argand plane represented by the complex numbers - 2 + 7 i . - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i are the vertices of a rhombus.

Given, - 2 + 7 i , - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i Let the four points represented by the complex numbers be A - 2 , 7 , B - 3 2 , 1 2 , C 4 , - 3 , D 7 2 , 7 2 The distance between the given points are, A B = - 2 + 3 2 2 + 7 - 1 2 2 = 1 4 + 169 4 = 170 2 units B C = 4 + 3 2 2 + - 3 - 1 2 2 = 121 4 + 49 4 = 170 2 units C D = 4 - 7 2 2 + - 3 - 7 2 2 = 1 4 + 169 4 = 170 2 units A D = - 2 - 7 2 2 + 7 - 7 2 2 = 121 4 + 49 4 = 170 2 units So, all sides are equal, A B = B C = C D = A D . Length of diagonal A C = - 2 - 4 2 + 7 + 3 2 = 36 + 100 = 136 units Length of diagonal B D = - 3 2 - 7 2 2 + 1 2 - 7 2 2 = 100 4 + 36 4 = 136 4 units We know that the diagonals of a rhombus are not equal. ∴ A B C D is a rhombus.

Write z = - 3 + i in modulus-amplitude form.

Given that z = - 3 + i We know that the modulus-amplitude form of z = a + i b is z = r cos θ + i sin θ , where r = a 2 + b 2 and θ is the principal argument of complex number. Modulus of the given complex number is, r = - 3 2 + 1 2 = 3 + 1 = 2 The complex number z = - 3 + i is in the second quadrant. When a complex number z = x + i y is belongs to the second quadrant, the principal argument is θ = π - tan - 1 y x . For given complex number z = - 3 + i , θ = π - tan - 1 1 - 3 = π - tan - 1 1 3 ⇒ θ = π - π 6 = 5 π 6 Hence, the required modulus-amplitude form is z = 2 cos 5 π 6 + i sin 5 π 6 .

Show that the points in the Argand plane represented by the complex numbers - 2 + 7 i . - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i are the vertices of a rhombus.

Given, - 2 + 7 i , - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i Let the four points represented by the complex numbers be A - 2 , 7 , B - 3 2 , 1 2 , C 4 , - 3 , D 7 2 , 7 2 The distance between the given points are, A B = - 2 + 3 2 2 + 7 - 1 2 2 = 1 4 + 169 4 = 170 2 units B C = 4 + 3 2 2 + - 3 - 1 2 2 = 121 4 + 49 4 = 170 2 units C D = 4 - 7 2 2 + - 3 - 7 2 2 = 1 4 + 169 4 = 170 2 units A D = - 2 - 7 2 2 + 7 - 7 2 2 = 121 4 + 49 4 = 170 2 units So, all sides are equal, A B = B C = C D = A D . Length of diagonal A C = - 2 - 4 2 + 7 + 3 2 = 36 + 100 = 136 units Length of diagonal B D = - 3 2 - 7 2 2 + 1 2 - 7 2 2 = 100 4 + 36 4 = 136 4 units We know that the diagonals of a rhombus are not equal. ∴ A B C D is a rhombus.

If A , B , and C are angles of a triangle such that x = cis A , y = cis B , z = cis C , then find the value of x y z

Given that, A , B , C are angles of a triangle and x = cis A , y = cis B , z = cis C So, from angle sum property, we have A + B + C = 180 0 = π → 1 Now, x y z = cis A · cis B · cis C ⇒ x y z = cis A + B + C [ ∵ cis x · cis y · cis z = cis x + y + z ] ⇒ x y z = cis π {Using 1 } ⇒ x y z = cosπ + i sinπ = - 1 + 0 = - 1 Hence, the value of x y z = - 1 .

If x = cis θ , then find the value of x 6 + 1 x 6 .

Given that, x = cis θ = cos θ + i sin θ = e i θ Now consider, x 6 + 1 x 6 = e 6 i θ + 1 e 6 i θ = e 6 i θ + e - 6 i θ ∵ 1 x = x - 1 = cos 6 θ + i sin 6 θ + cos - 6 θ + i sin - 6 θ = cos 6 θ + i sin 6 θ + cos 6 θ - i sin 6 θ ∵ cos - θ = cos θ a n d sin - θ = - sin θ = 2 cos 6 θ Therefore, x 6 + 1 x 6 = 2 cos 6 θ .

If the real part of z + 1 z + i is 1 , then find the locus of z where z = x + i y .

Given, z + 1 z + i Let, z = x + i y ⇒ z + 1 z + i = ( x + 1 ) + i y ( x ) + ( y + 1 ) i = [ ( x + 1 ) + i y ] [ x - ( y + 1 ) i ] [ x + ( y + 1 ) i ] [ x - ( y + 1 ) i ] = x ( x + 1 ) + y ( y + 1 ) + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 = x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 Given, real part = 1 ⇒ x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 = 1 ⇒ x 2 + y 2 + x + y = x 2 + ( y + 1 ) 2 ⇒ x 2 + y 2 + x + y = x 2 + y 2 + 2 y + 1 ⇒ x - y = 1 ∴ Locus of z is x - y = 1 .

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 2 i b + 3 a i b 2 + i b 3 ∵ i 2 = - 1 ⇒ x + i y = a 3 - 3 a b 2 + i b 3 - 3 a 2 b Comparing real parts and imaginary parts, we have 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.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2019>Q 5

EASY

12th Andhra Pradesh Board

IMPORTANT

Earn 100

Find the multiplicative inverse of : $7 + 24 i$

Important Questions on Complex Numbers

EASY

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2019>Q 6

Simplify

i^{2} + i^{4} + i^{6} + . . . . . . . . . . + (2 n + 1)

terms.

EASY

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2019>Q 7

If

x = cis θ

, then find the value of

x^{6} + \frac{1}{x^{6}}

.

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2019>Q 8

If the real part of

(\frac{z + 1}{z + i})

is

1

, then find the locus of

z

where

z = x + i y

.

EASY

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2020>Q 9

Find the complex conjugate of

(3 + 4 i) (2 - 3 i)

.

EASY

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2020>Q 10

Write

z = - \sqrt{3} + i

in modulus-amplitude form.

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Andhra Pradesh Board-2020>Q 11

Show that the points in the Argand plane represented by the complex numbers $- 2 + 7 i . \frac{- 3}{2} + \frac{1}{2} i, 4 - 3 i, \frac{7}{2} (1 + i)$ are the vertices of a rhombus.