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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Complex Number

If z = 3 2 + (i)/(2)(i = - 1 ), then (1 + iz + z ^ 5 + i z ^ 8 ) ^...

If z=32+i2(i=−1),z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\left(i = \sqrt{- 1}\right), then (1+iz+z5+iz8)9{\left(1 + iz + {z}^{5} + i{z}^{8}\right)}^{9} is equal to:

Options

  1. A−1- 1
  2. B11
  3. C00
  4. D{\left(- 1 + 2i\right)}^{9}

Correct answer

A. −1- 1

Step-by-step solution

z=32+i2∴iz=−12+32i=ωz = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2} ∴ iz = - \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i = \omega
∴(1+iz+z5+iz8)9=(1+ω+ω5i5+i×ω8i8)9∴ {\left(1 + iz + {z}^{5} + i{z}^{8}\right)}^{9} = {\left(1 + \omega + \dfrac{{\omega}^{5}}{{i}^{5}} + i \times \dfrac{{\omega}^{8}}{{i}^{8}}\right)}^{9}
=(1+ω−iω2+iω2)9= {\left(1 + \omega - i{\omega}^{2} + i{\omega}^{2}\right)}^{9}
=(−ω2)9=−ω18= {\left(- {\omega}^{2}\right)}^{9} = - {\omega}^{18}
=−1= - 1
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