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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Complex Number

Let z = ( 3 2 + (i)/(2)) ^ 5 + ( 3 2 - (i)/(2)) ^ 5 . If R(z) and...

Let z=(32+i2)5+(32−i2)5.z = {\left(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\right)}^{5} + {\left(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right)}^{5} . If R(z)R\left(z\right) and I(z)I\left(z\right) respectively denote the real and imaginary parts of z,z, then

Options

  1. AI(z)=0I\left(z\right) = 0
  2. BR(z)<0R\left(z\right) < 0 and I(z)>0I\left(z\right) > 0
  3. CR(z)>0R\left(z\right) > 0 and I(z)>0I\left(z\right) > 0
  4. DR(z)=−3R\left(z\right) = - 3

Correct answer

A. I(z)=0I\left(z\right) = 0

Step-by-step solution

Let α=32+i2,\alpha = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2}, then z=α5+(α¯)5z = {\alpha}^{5} + {\left(\overset{¯}{\alpha}\right)}^{5}

We know z¯n=z¯n\overset{¯}{z}n = \overset{¯}{z}n,

then, z=2Re(α5)z = 2Re\left({\alpha}^{5}\right)

Hence, I(z)=0I\left(z\right) = 0.

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