PrepRung
Sign inSign up free

JEE Main 2017 · Mathematics · Complex Numbers · easy

If ω≠1 is a root of z^2+z+1=0, then ω^2017+(1)/(ω^2017) equals

Single correct option

If ω≠1\omega\ne1 is a root of z2+z+1=0z^2+z+1=0, then ω2017+1ω2017\omega^{2017}+\dfrac{1}{\omega^{2017}} equals
  1. A. 11
  2. B. 22
  3. C. −1-1
  4. D. −2-2
Show answer and solution

Answer: (C) −1-1

The roots of z2+z+1=0z^2+z+1=0 are the non-real cube roots of unity, so ω3=1\omega^3=1. Since 2017=3⋅672+12017=3\cdot672+1, ω2017=ω\omega^{2017}=\omega and 1/ω2017=1/ω=ω21/\omega^{2017}=1/\omega=\omega^2. Hence the required value is ω+ω2=−1\omega+\omega^2=-1 from the sum of the roots of the equation.
Report a mistake

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake