FINDING ROOTS OF COMPLEX NUMBERS 

Problem 1 :

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 = 1 cos 𝜋 + 2(1)𝜋4 + i sin𝜋 + 2(1)𝜋4 = cos 3𝜋4+ i sin 3𝜋4 If k = 2 cos 𝜋 + 2(2)𝜋4 + i sin 𝜋 + 2(2)𝜋4 = cos 5𝜋4+ i sin 5𝜋4 If k = 3 cos𝜋 + 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

(1)   1    (2)  -1   (3)  √3i    (4)   -√3i

Solution :

Given, 1111-ω2 - 1ω21ω2ω7 = 3kω7 = ω32 · ωω7= ω1111-ω2 - 1ω21ω2ω = 3kLet ω3 = 1ω = -1 + 3i2ω2 = -1 - 3i2On applying R1 R1 + R2 + R3 we get11 + ω + ω21 + ω + ω21-ω2 - 1ω21ω2ω = 3k3001-ω2 - 1ω21ω2ω = 3k3-ω3 - 𝜔 - ω4 = 3k-ω3 - 𝜔 - ω4 = k-1 - 𝜔 - ω = k-1 - 2𝜔 = k -2𝜔 - 1 = k- 2-1 + 3i2 - 1 = k 2 - 23i2 - 1 = k1 - 3i - 1 = kk = -3i

So, option (4) is correct.

Problem 4 :

The value of 1 + 3i1 - 3i10is
(1) cis 2𝜋3
(3) -cis 2𝜋3
(2) cis 4𝜋3
(4) -cis 4𝜋3

Solution :

Given, 1 + 3i1 - 3i101 + i3 = r (cos 𝜃 + i sin 𝜃)r = x2 + y2r = (1)2 + 32r = 1 + 3r = 4r = 2𝜃 = tan-1 yx= tan-1 31𝜃 = tan-1 3𝜃 = 𝜋31 + i3 = 2cos 𝜋3 + i sin 𝜋31 - i3 = 2cos 𝜋3 - i sin 𝜋3= 2cos -𝜋3 + i sin -𝜋31 + 3i1 - 3i10 = 2 cis 𝜋32 cis -𝜋310= cis 𝜋3 + 𝜋310 = cis 2𝜋310= cis 20𝜋3= cis 6𝜋 + 2𝜋31 + 3i1 - 3i10= cis 2𝜋3

So, option (1) is correct.

Problem 5 :

If 𝜔 = cis 2𝜋3, then the number of distinct roots of z + 1𝜔𝜔2𝜔z + ω2 1𝜔21z + ω = 0

(1)  1   (2)   2   (3)   3   (4)   4

Solution :

𝜔 = cis 2𝜋/3

𝜔 = cos 2𝜋/3 + i sin 2𝜋/3

𝜔 = -1/2 + i(√3/2)

Given, z + 1𝜔𝜔2𝜔z + 𝜔2 1𝜔21z + 𝜔On applying C1 C1 + C2 + C3 and 1 + 𝜔 + 𝜔2 = 0z + 1 + 𝜔 + 𝜔2𝜔𝜔2𝜔 + z + 𝜔2 + 1z + 𝜔2 1𝜔2 + 1 + z + 𝜔1z + 𝜔 = 0z + 1 + 𝜔 + 𝜔21𝜔𝜔21 z + 𝜔2 111z + 𝜔 = 0z1𝜔𝜔21 z + 𝜔2 111z + 𝜔= 0

R2  → R2 – R1

R3  → R3 – R1

z1𝜔𝜔20 z + 𝜔2 - 𝜔1 - 𝜔201- 𝜔z + 𝜔 - 𝜔2= 0 zz + 𝜔 2 - 𝜔 z + 𝜔 - 𝜔 2 - 1 - 𝜔 2 (1 - 𝜔 ) = 0z z2- 𝜔 - 𝜔22 - 1 - 𝜔 - 𝜔2 + 𝜔 3 = 0zz2 - 𝜔 2 + 𝜔 - 2 - 3= 0zz2 =0z3 = 0z = 0

One distinct roots.

So, option (1) is correct.

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