If n is an integer then show that ( 1 + i ) 2 n + ( 1 - i ) 2 n = 2 n + 1 cos n π 2 .

To prove: ( 1 + i ) 2 n + ( 1 - i ) 2 n = 2 n + 1 cos n π 2 From LHS, we have, = ( 1 + i ) 2 n + ( 1 - i ) 2 n = 2 1 2 + 1 2 i 2 n + 2 1 2 - 1 2 i 2 n = ( 2 ) 2 n cos π 4 + i sin π 4 2 n + ( 2 ) 2 n cos π 4 - i sin π 4 2 n = 2 n cos 2 n π 4 + i sin 2 n π 4 + cos 2 n π 4 - i sin 2 n π 4 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = 2 n 2 cos 2 n π 4 = 2 n + 1 cos n π 2 = RHS Hence, proved.

If α , β are the roots of the equation x 2 - 2 x + 4 = 0 , then for any n ∈ N , show that α n + β n = 2 n + 1 cos n π 3 .

Given quadratic equation: x 2 - 2 x + 4 = 0 and to prove: α n + β n = 2 n + 1 cos n π 3 ⇒ x = 2 ± 4 - 16 2 = 2 ± 2 i 3 2 = 1 ± i 3 ∵ x = - b ± b 2 - 4 a c 2 a , when a x 2 + b x + c = 0 Given that, α , β are the roots of the given quadratic equation i.e., α = 1 + i 3 = 2 1 2 + i 3 2 = 2 cos π 3 + i sin π 3 and β = 1 - i 3 = 2 1 2 - i 3 2 = 2 cos π 3 - i sin π 3 Now, from LHS, we have, α n + β n = 2 cos π 3 + i sin π 3 n + 2 cos π 3 - i sin π 3 n = 2 n cos n π 3 + i sin n π 3 + cos n π 3 - i sin n π 3 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = 2 n 2 cos n π 3 = 2 n + 1 cos n π 3 = RHS Hence, proved.

If n is an integer and z = cis θ , θ ≠ ( 2 n + 1 ) π 2 , then show that z 2 n - 1 z 2 n + 1 = i tan n θ .

Given that : z = cis θ = cos θ + i sin θ , θ ≠ ( 2 n + 1 ) π 2 , where n is an integer. To prove: z 2 n - 1 z 2 n + 1 = i tan n θ From, LHS, we have, = z 2 n - 1 z 2 n + 1 = ( cos θ + i sin θ ) 2 n - 1 ( cos θ + i sin θ ) 2 n + 1 = cos 2 n θ + i sin 2 n θ - 1 cos 2 n θ + i sin 2 n θ + 1 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = - ( 1 - cos 2 n θ ) + i sin 2 n θ ( 1 + cos 2 n θ ) + i sin 2 n θ = i 2 2 sin 2 n θ + 2 i sin n θ · cos n θ 2 cos 2 n θ + 2 i sin n θ · cos n θ ∵ - 1 = i 2 , 1 + cos 2 x = 2 cos 2 x , 1 - cos 2 x = 2 sin 2 x & sin 2 x = 2 sin x cos x = 2 i sin n θ { cos n θ + i sin n θ } 2 cos n θ { cos n θ + i sin n θ } = i tan n θ = RHS 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.

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.

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Algebra>Chapter 2 - De Moivre's Theorem>Exercise

Embibe Experts Solutions for Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: Exercise

Attempt the practice questions from Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 1: Exercise with Hints & Solutions

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 1

If $n$ is an integer then show that $(1 + i)^{2 n} + (1 - i)^{2 n} = 2^{n + 1} \cos \frac{n π}{2}$ .

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 2

If $α, β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$ , then for any $n \in N$ , show that $α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$ .

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 3

If $n$ is an integer and $z = cis θ, (θ \neq (2 n + 1) \frac{π}{2})$ , then show that $\frac{z^{2 n} - 1}{z^{2 n} + 1} = i \tan n θ$ .

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 4

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

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 5

Solve the following equation: $x^{4} - 1 = 0$ .

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 6

Solve $(x - 1)^{n} = x^{n}$ , where $n$ is positive integer.

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 7

Solve the following equation: $x^{9} - x^{5} + x^{4} - 1 = 0$ .

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - De Moivre's Theorem>Exercise>Q 11

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

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