If z 2 + z + 1 = 0 , where z is a complex number, prove that z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 .

Given equation: z 2 + z + 1 = 0 , where z is a complex number, so clearly ω , ω 2 are the two roots which satisfies the given equation. Let z = ω To prove: z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 Now, from LHS, we have z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 4 + 1 ω 4 2 + ω 5 + 1 ω 5 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 3 · ω + 1 ω 3 · ω 2 + ω 3 · ω 2 + 1 ω 3 · ω 2 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 + ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 ∵ ω 3 = 1 , ω 6 = 1 = ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 + 8 + ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 = ω + ω 2 2 + ω 2 + ω 2 + 8 + ω + ω 2 2 + ω 2 + ω 2 = - 1 2 + - 1 2 + 8 + - 1 2 + - 1 2 ∵ 1 + ω + ω 2 = 0 = 12 = RHS Hence, proved.

If 1 , α , α 2 , α 3 , … , α n - 1 be the n th roots of unity, then prove that 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 0 ; if p ≠ k n n ; if p = k n , where p , k ∈ N .

We know that, if x is n th root of unity, then it satisfies the relation x n - 1 = 0 It is given that, 1 , α , α 2 , α 3 , … , α n - 1 are the n th roots of unity, therefore, α n = 1 and use α = e i 2 π n Case (i): when p ≠ k n ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = e i 2 π n n p - 1 e i 2 π n p - 1 (Since α = e i 2 π n and S n = a r n - 1 r - 1 , r > 1 ) ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = e i 2 π p - 1 e i 2 π n p - 1 ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 1 - 1 e i 2 π n p - 1 = 0 [Since, p , k ∈ N and e i 2 π p = cos 2 π p + i sin 2 π p = 1 + 0 = 1 , Since denominator cannot be zero because p ≠ k n ] Case (ii): when p = k n Therefore, 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 1 k n + α k n + α 2 k n + α 3 k n + … + α n - 1 k n = 1 + α n k + α n 2 k + α n 3 k + … + α n n - 1 k = 1 + 1 + 1 + 1 + … + 1 [since α n = 1 ] = n Hence, proved.

If n is a positive integer, show that P + i Q 1 n + P - i Q 1 n = 2 P 2 + Q 2 1 2 n · cos 1 n tan - 1 Q P .

To prove: P + i Q 1 n + P - i Q 1 n = 2 P 2 + Q 2 1 2 n · cos 1 n tan - 1 Q P Let P + i Q = r { cos θ + i sin θ } , here r cos θ = P , r sin θ = Q ⇒ r 2 = P 2 + Q 2 ∴ r = P 2 + Q 2 and similarly P - i Q = r { cos θ - i sin θ } Thus, cos θ = P P 2 + Q 2 and sin θ = Q P 2 + Q 2 , therefore tan θ = Q P ⇒ θ = tan - 1 Q P Therefore, ( P + i Q ) 1 n + ( P - i Q ) 1 n = { r ( cos θ + i sin θ ) } 1 n + { r ( cos θ - i sin θ ) } 1 n = r 1 n cos θ n + i sin θ n + cos θ n - i sin θ n ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = P 2 + Q 2 1 n 2 cos 1 n θ = 2 P 2 + Q 2 1 2 n cos 1 n tan - 1 Q P = RHS Hence, proved.

Show that 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = - 1 .

To prove: 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = - 1 We have LHS = 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = 1 + cos π 2 - π 8 + i sin π 2 - π 8 1 + cos π 2 - π 8 - i sin π 2 - π 8 8 3 = 1 + cos 3 π 8 + i sin 3 π 8 1 + cos 3 π 8 - i sin 3 π 8 8 3 = 2 cos 2 3 π 16 + 2 i sin 3 π 16 cos 3 π 16 2 cos 2 3 π 16 - 2 i sin 3 π 16 cos 3 π 16 8 3 ∵ 1 + cos x = 2 cos 2 x 2 , sin x = 2 sin x 2 cos x 2 = 2 cos 3 π 16 cos 3 π 16 + i sin 3 π 16 2 cos 3 π 16 cos 3 π 16 - i sin 3 π 16 8 3 = cos 3 π 16 + i sin 3 π 16 cos 3 π 16 + i sin 3 π 16 cos 3 π 16 - i sin 3 π 16 cos 3 π 16 + i sin 3 π 16 8 3 {Multiplying and dividing with the conjugate of the denominator} = cos 3 π 16 + i sin 3 π 16 2 cos 2 3 π 16 + sin 2 3 π 16 8 3 = cos 3 π 8 + i sin 3 π 8 8 3 ∵ sin 2 x + cos 2 x = 1 , x ∈ R = cos π + i sin π = - 1 {Using De Moivre's Theorem when N is a Fractional Number} = RHS Hence, proved.

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 2 - De Moivre's Theorem>Exercise 2(b)

Andhra Pradesh Board Solutions for Exercise 2: Exercise 2(b)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 2: Exercise 2(b)

Attempt the free practice questions from Exercise 2: Exercise 2(b) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 2: Exercise 2(b) with Hints & Solutions

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q II

If the cube roots of unity are $1, ω, ω^{2}$ , then find the roots of the equation $(x - 1)^{3} + 8 = 0$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q II

Find the product of all the values of $(1 + i)^{\frac{4}{5}}$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q II

If $z^{2} + z + 1 = 0$ , where $z$ is a complex number, prove that ${(z + \frac{1}{z})}^{2} + {(z^{2} + \frac{1}{z^{2}})}^{2} + {(z^{3} + \frac{1}{z^{3}})}^{2} + {(z^{4} + \frac{1}{z^{4}})}^{2} + {(z^{5} + \frac{1}{z^{5}})}^{2} + {(z^{6} + \frac{1}{z^{6}})}^{2} = 12$ .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q III

If $1, α, α^{2}, α^{3}, \dots, α^{n - 1}$ be the $n^{th}$ roots of unity, then prove that $1^{p} + α^{p} + {(α^{2})}^{p} + {(α^{3})}^{p} + \dots + {(α^{n - 1})}^{p} = \{\begin{cases} 0; if p \neq k n \\ n; if p = k n \end{cases}$ , where $p, k \in N$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q III

Prove the sum of $99^{th}$ powers of the roots of the equation $x^{7} - 1 = 0$ is zero and hence deduce the roots of $x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 = 0$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 2 - De Moivre's Theorem>Exercise 2(b)>Q III

If $n$ is a positive integer, show that ${(P + i Q)}^{\frac{1}{n}} + {(P - i Q)}^{\frac{1}{n}} = 2 {(P^{2} + Q^{2})}^{\frac{1}{2 n}} \cdot \cos [\frac{1}{n} \tan^{- 1} \frac{Q}{P}]$ .