If α and β are the roots of x2 + x + 1 = 0, then α2020 + β2020 is
(1) -2 (2) -1 (3) 1 (4) 2
Solution :
Given, x2 + x + 1 = 0
Let α = ω, β = ω2
α2020 + β2020 = ω2020 + (ω2)2020
= ω3(673) + 1 + (ω3(673) + 1)2
= ω + ω2
α2020 + β2020 = -1
So, option (2) is correct.
Problem 2 :
The product of all four values of cos 𝜋3+ i sin 𝜋334is
(1) -2 (2) -1 (3) 1 (4) 2
Solution :
(cos 𝜃 + i sin 𝜃)n= cos n 𝜃 + i sin n 𝜃cos 𝜋3+ i sin 𝜋334= cos 34×𝜋3+ i sin 34×𝜋3= cos 𝜋4+ i sin 𝜋4= cos 𝜋 + 2k𝜋4+ i sin 𝜋 + 2k𝜋4(k =0, 1, 2, 3) If k = 0cos 𝜋 + 2(0)𝜋4+ i sin 𝜋 + 2(0)𝜋4= cos 𝜋4+ i sin 𝜋4 If k = 1cos 𝜋 + 2(1)𝜋4+ i sin𝜋 + 2(1)𝜋4= cos 3𝜋4+ i sin 3𝜋4 If k = 2cos 𝜋 + 2(2)𝜋4+ i sin 𝜋 + 2(2)𝜋4= cos 5𝜋4+ i sin 5𝜋4 If k = 3cos𝜋 + 2(3)𝜋4+ i sin𝜋 + 2(3)𝜋4= cos 7𝜋4+ i sin 7𝜋4Using De'moivre's theoremcos 𝜋4+ i sin 𝜋4· cos 3𝜋4+ i sin 3𝜋4·cos 5𝜋4+ i sin 5𝜋4·cos 7𝜋4+ i sin 7𝜋4= cos𝜋4+3𝜋4+5𝜋4+7𝜋4+ i sin 𝜋4+3𝜋4+5𝜋4+7𝜋4= cos 164𝜋 + i sin 164 𝜋= cos 4𝜋 + i sin 4𝜋= 1 + i0= 1cos 𝜋3+ i sin 𝜋334= 1
So, option (2) is correct.
Problem 3 :
If ω ≠ 1 is a cubic root of unity and 1111-ω2- 1ω21ω2ω7= 3k, then k is equal to