If z 1 = - 1 and z 2 = i , then find Arg z 1 z 2

Given, z 1 = - 1 and z 2 = i We know that, argument of purely real number with negative real part is π . Argument of purely imaginary part with positive imaginary part is π 2 . Now, Arg z 1 z 2 = Arg z 1 - Arg z 2 = π - π 2 = π 2 ∴ Arg z 1 z 2 = 90 °

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 .

Show that the points in the Argand diagram represented by the complex numbers 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i are the vertices of an equilateral triangle.

Given, 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i Let A ( 2 , 2 ) , B ( - 2 , - 2 ) , C ( - 2 3 , 2 3 ) be points represents given complex numbers in the Argand plane. A B = ( 2 + 2 ) 2 + ( 2 + 2 ) 2 ⇒ A B = 4 2 B C = ( - 2 + 2 3 ) 2 + ( - 2 - 2 3 ) 2 ⇒ B C = 4 + 12 - 8 3 + 4 + 12 + 8 3 ⇒ B C = 4 2 A C = ( 2 + 2 3 ) 2 + ( 2 - 2 3 ) 2 ⇒ A C = 4 + 12 + 8 3 + 4 + 12 - 8 3 ⇒ A C = 4 2 ∴ A B = A C = B C So, △ A B C is equilateral.

Express the complex number - 1 - i 3 in modulus amplitude form.

Let, - 1 - i 3 = r cos θ + r i sin θ Comparing real and imaginary parts, we get r cos θ = - 1 ...(i) and r sin θ = - 3 ...(ii) Here, r is the modulus of the given complex number ⇒ r = | - 1 - i 3 | = 1 + ( - 3 ) 2 = 1 + 3 = 4 = 2 Now, ( i i ) ( i ) = tan θ = 3 ⇒ θ = 4 π 3 , since - 1 - i 3 = - 1 , - 3 is in III quadrant. ∴ - 1 - i 3 = 2 cos 4 π 3 + i sin 4 π 3

If z = 2 - i 7 then, show that 3 z 3 - 4 z 2 + z + 88 = 0

Given, z = 2 - i 7 . . . i To Prove: 3 z 3 - 4 z 2 + z + 88 = 0 Equation i can be written as z - 2 = - i 7 ⇒ ( z - 2 ) 2 = ( - i 7 ) 2 ⇒ z 2 - 4 z + 4 = - 7 ∵ i 2 = - 1 ⇒ z 2 - 4 z + 11 = 0 . . . . i i Multiplying with 3 z on both sides, ⇒ 3 z 3 - 12 z 2 + 33 z = 0 ⇒ 3 z 3 - 4 z 2 + z + 88 + - 8 z 2 + 32 z - 88 = 0 ⇒ 3 z 3 - 4 z 2 + z + 88 - 8 z 2 - 4 z + 11 = 0 ∴ 3 z 3 - 4 z 2 + z + 88 = 0 [From i i ] Hence, proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Algebra>Chapter 1 - Complex Numbers>Telangana Board-2020

Embibe Experts Solutions for Exercise 1: Telangana Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: Telangana Board-2018

Attempt the free practice questions from Exercise 1: Telangana Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 1: Telangana Board-2018 with Hints & Solutions

EASY

12th Telangana Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Telangana Board-2018>Q 1

If $z = 2 - 3 i$ , show that, $z^{2} - 4 z + 13 = 0$ .

EASY

12th Telangana Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Telangana Board-2018>Q 2

If $z_{1} = - 1$ and $z_{2} = i$ , then find $Arg (\frac{z_{1}}{z_{2}})$

EASY

12th Telangana Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Telangana Board-2018>Q 3

If $x = cis θ$ , then find the value of $x^{6} + \frac{1}{x^{6}}$ .

MEDIUM

12th Telangana Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Complex Numbers>Telangana Board-2018>Q 4

Show that the points in the Argand diagram represented by the complex numbers $2 + 2 i, - 2 - 2 i, - 2 \sqrt{3} + 2 \sqrt{3} i$ are the vertices of an equilateral triangle.