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JEE Main 201610 AprilOnline
Exam
JEE Main
Year
2016
Date
10 April
Mode
Online
Subject
Mathematics
Chapter
Complex Number

JEE Main 2016 · 10 April · Online · Mathematics · Complex Number

Let z = 1 + ai , be a complex number, a > 0 , such that z ^ 3 is a...

Let z=1+aiz = 1 + ai, be a complex number, a>0a > 0, such that z3{z}^{3} is a real number. Then, the sum 1+z+z2+….+z111 + z + {z}^{2} + \ldots . + {z}^{11} is equal to :

Options

  1. A13653i1365\sqrt{3}i
  2. B−13653i- 1365\sqrt{3}i
  3. C−12503i- 1250\sqrt{3}i
  4. D12503i1250\sqrt{3}i

Correct answer

B. −13653i- 1365\sqrt{3}i

Step-by-step solution

z=1+aiz = 1 + ai
Cubing both sides,
z3=(1−a2)+2ai+(1−a2)ai−2a2{z}^{3} = \left(1 - {a}^{2}\right) + 2ai + \left(1 - {a}^{2}\right)ai - 2{a}^{2}
z3=1−3a2+(3a−a3)i{z}^{3} = 1 - 3{a}^{2} + ( 3a - {a}^{3} ) i
∵z3∵ {z}^{3} is a real number.
a(3−a2)=0⇒a=3(a>0)a\left(3 - {a}^{2}\right) = 0 ⇒ a = \sqrt{3} \left(a > 0\right)

1+z+z2…..z11=z12−1z−1=(1+3i)12−11+3i−1=(1+3i)12−13i1 + z + {z}^{2} \ldots . . {z}^{11} = \dfrac{{z}^{12} - 1}{z - 1} = \dfrac{{\left(1 + \sqrt{3}i\right)}^{12} - 1}{1 + \sqrt{3}i - 1} = \dfrac{{\left(1 + \sqrt{3}i\right)}^{12} - 1}{\sqrt{3}i}

(1+3i)12=212(12+32i)12=212(cosπ3+isinπ3)12=212(cos⁡4π+isin4π)=212{\left(1 + \sqrt{3}i\right)}^{12} = {2}^{12}{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)}^{12} = {2}^{12}{\left(cos \dfrac{\pi}{3} + isin \dfrac{\pi}{3}\right)}^{12} = {2}^{12}\left(\cos 4\pi + isin 4\pi\right) = {2}^{12}

=212−13i=−13653i= \dfrac{{2}^{12} - 1}{\sqrt{3}i} = - 1365\sqrt{3}i
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