Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 1.
Find the value of (1 – i)⁸(Mar. ’07)
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 1

Question 2.
If x = cis θ, then find the value of [x⁶ + \(\frac{1}{x^{6}}\)]
Solution:
∵ x = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶
= cos 6θ + i sin 6θ
⇒ \(\frac{1}{x^{6}}\) = cos 6θ – i sin 6θ
∴ x⁶ + \(\frac{1}{x^{6}}\) = 2 cos 6θ

Question 3.
If A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of XYZ. (AP Mar. ‘16, ’15)
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
x = cis A, y = cis B, Z = cis C
⇒ xyz = cis(A + B + C)
= cos(A + B + C) + i sin(A + B + C)
= cos(180°) + i sin (180°) .
= -1 + i(0) = -1
∴ xyz = -1

Question 4.
If 1, ω, ω² are the cube roots of unity, then prove that \(\frac{1}{2+\omega}\) – \(\frac{1}{1+2 \omega}\) = \(\frac{1}{1+\omega}\)
Solution:
ω is a cube root of unity
1 + ω + ω² = 0 and ω³ = 1
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 2

Question 5.
(2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – w¹¹) = 49. (TS Mar. ’17)
Solution:
∵ 1, ω, ω² are the cube roots of unity,
ω³ = 1 and 1 + ω + ω² = 0
2 – ω¹⁰ = 2 – ω⁹ . ω
= 2 – (ω³)³ . ω²
= 2 – (1)³ ω² = 2 – ω²
(2 – ω)(2 – ω²) = 4 – 2ω – 2ω² + ω³
= 4 – 2(ω + ω²) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1 = 7
∴ (2 – ω)(2 – ω²)(2 – ω¹⁰)(2 – ω¹¹)
= (2 – ω) (2 – ω²) (2 – ω) (2 – ω²)
= ((2 – ω) (2 – ω²))²
= 7² = 49

Question 6.
If α, β are the roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2^{n + 1} cos \(\left(\frac{n \pi}{3}\right)\) (Mar. ’14)
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 4

Question 7.
Show that one value of
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 6
is -1.
Solution:

Question 8.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{\mathrm{n} \pi}{4}\right)\). (A.P) (Mar. ’15)
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 10

Question 9.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} cosⁿ (θ/2) cos \(\left(\frac{n \theta}{2}\right)\) (May. ’11) (TS & AP Mar. ’17)
Solution:
L.H.S.
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 11

Question 10.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ. Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α + sin² β + ² γ. (AP. Mar. ’16; TS Mar. ’15, ‘13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i. 0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 12

-i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0.
(cos α + i sin α)² + (cos β + i sin β)² + (cos γ + i sinγ)² = 0
(cos 2α + i sin 2α) + (cos β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ ²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = \(\frac{3}{2}\)
∴ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin²α + sin²β + sin²β

Question 11.
If 1, ω, ω² are the cube roots of unity prove that
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b) (aω + bω²) (aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1

i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= 2⁶(ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128

Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= ( -ω – ω)⁷ + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷ (ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a+b) (a]ω + bω²)(aω² + bω) (AP) (Mar. ’17)
Solution:
= (a + b) [a²ω³ + abω⁴)4 + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
⇒ x + 2 = ω – ω²
⇒ (x + 2)² = ω² + ω⁴ – 2ω³
⇒ x² + 4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3
⇒ x² + 4x + 7 = 0

Question 12.
Simplify \(\frac{(\cos \alpha+i \sin \alpha)^{4}}{(\sin \beta+i \cos \beta)^{8}}\)
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 14

= (cos α + i sin α)⁴ (cos β – i sin β)^-8 [i⁸ = (i²)⁴ = (-1)⁴ = 1]
= (cos 4α + i sin 4α) (cos 8β + i sin 8β)
= cos (4α + 8β) + i sin (4α + 8β)

Question 13.
If m, n are integers and x = cos α + i sin α, y = cos β + i sin β, then prove that
x^m yⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ) and
x^m yⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2i sin (mα + nβ).
Solution:
∵ x = cos α + i sin α, y = cos β + i sin β
⇒ x^m = (cos α + i sin α)^m = cos mα + i sin mα
yⁿ = (cos β + i sin β)ⁿ = cos nβ + i sin nβ
∴ x^m yⁿ = (cos mα + i sin mα)(cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ) ——— (1)
\(\frac{1}{x^{m} \cdot y^{n}}\) = cos (mα + nβ) – i sin (mα + nβ) —— (2)
By adding (1) and (2).
x^myⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ)
By subtracting (2) from (1)
x^myⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2 sin (mα + nβ)

Question 14.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{n \pi}{4}\right)\) (A.P.) (Mar. ‘15)
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 16

Question 15.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} + cosⁿ (θ/2) cos \(\left(\frac{\mathrm{n} \theta}{2}\right)\). (May ’11)
Solution:
L.H.S.
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 18

Question 16.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α +
sin² β + sin² γ. (A.P. Mar. ‘16, T.S. Mar. ‘15, ’13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i.0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 19
Similarly \(\frac{1}{y}\) = cos β – i sin β
\(\frac{1}{z}\) = cos γ – i sin γ
∴ \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\) = (cos α + cos β + cos γ) – i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0
(cos α + i sin β)² + (cos β + i sin β)² + (cos γ + i sin γ)² = o
(cos 2α + i sin 2α) + (cos 2β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2 cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\)
∴ cos² α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin² α + sin²β + sin²β

Question 17.
Find all the values of (\(\sqrt{3}\) + i)^1/4.
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 20

Question 18.
Find all the roots of the equation
x¹¹ – x⁷ + x⁴ – 1 = 0.
Solution:
x¹¹ – x⁷ + x⁴ – 1 = 0
⇒ x⁷(x⁴ – 1) + 1 (x⁴ – 1) = 0
⇒ (x⁴ – 1) (x⁷ + 1) = 0
Case(i) : x⁴ – 1 = 0
x⁴ = 1 = (cos o + i sin 0)
⇒ x⁴ = (cos 2kπ + i sin 2kπ)
∴ x = (cos 2kπ + i sin 2kπ)^1/4
⇒ x = cis \(\left(\frac{2 k \pi}{4}\right)\) = cis \(\frac{\mathrm{k} \pi}{2}\), k = 0, 1, 2, 3.

Case (ii): x⁷ + 1 = 0
⇒ x⁷ = -1 = cos π + i sin π
⇒ x⁷ = cos (2kπ + n) + i sin (2kπ + π)
∴ x = [cos (2k + 1)π + i sin (2k + 1)π]^1/7
⇒ x = cis(2k + 1)\(\frac{\pi}{7}\), k = 0, 1, 2, 3, 4, 5, 6
The values of x are
Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 22

Question 19.
If 1, ω, ω² are the cube roots of unity prove that (TS. Mar. ‘16)
i) (1 – ω + (ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b)(aω + bω²)(aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ +(-ω² – ω²)⁶
= (-2ω)⁶ + (-2w²)⁶
= 2⁶ (ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128 .
Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= (- ω – ω) + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷(ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a + b) (aω + bω²) (aω² + bω)
= (a + b) [a²ω³ + abω⁴ + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
= x + 2 = ω – ω²
(x + 2)² = ω² – 2ω³
⇒ x² +4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3.
⇒ x² + 4x + 7 = 0

Question 20.
If α, β are the roots of the equation x² + x + 1 = 0 then prove that α⁴ + β⁴ + α^-1β^-1 = 0
Solution:
Since α, β are the complex cube roots of unity.
We take α = ω, β = ω²
∴ α⁴ + β⁴ + α^-1 + β^-1
= ω⁴ + (ω²)⁴ + \(\frac{1}{\omega} \cdot \frac{1}{\omega^{2}}\)
= ω + ω² + \(\frac{1}{\omega^{3}}\)
= (-1) + \(\frac{1}{1}\)
= -1 + 1 = 0
∴ α⁴ + β⁴ + α^-1 + β^-1 = 0