Complex Numbers: Mixed MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

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MixedHardQuestion 230024

Question

If α and β are roots of z² − z + 1 = 0, then α¹⁰¹ + β¹⁰¹ is:

A
−1
Correct
B
1
C
2
D
0

Correct answer

−1

Explanation

Roots of z²−z+1=0 are z = (1±i√3)/2 = e^(±iπ/3), which are primitive 6th roots of unity ω and ω̄ where ω = e^(iπ/3). α¹⁰¹ = e^(101iπ/3). 101 = 6×16 + 5, so e^(101iπ/3) = e^(5iπ/3) = cos(300°) + i sin(300°) = 1/2 − i√3/2. Similarly β¹⁰¹ = 1/2 + i√3/2. Sum = 1. But checking: roots are e^(±iπ/3), period 6. 101 mod 6 = 5. α⁵ + β⁵ = 2cos(5π/3) = 2(1/2) = 1. So answer is 1.

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