Verified Solution • 2025–26 Session moderate

QJEEM18Maf19e: JEE Main 2018 Mathematics - Class 11 JEEM18Maf19e Solution: The least positive integer n for which is :

The least positive integer n for which (1+i31−i3)n=1,{\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1, is :

Multiple Choice Options

  • A 2
  • B 3
  • C 5
  • D 6

Step-by-Step Verified Solution

(1+i31−i3)n=1{\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1

⇒     (1+i321−i32)n=1\Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1

We know, ω\omega = −- 1−i32{{1 - i\sqrt 3 } \over 2}

and ω\omega 2 = −- (1+i3)2{{\left( {1 + i\sqrt 3 } \right)} \over 2}

⇒\Rightarrow     \,\,\,\, (−ω2−ω)n=1{\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1

⇒\Rightarrow ( ω\omega )n = 1 = ω\omega 3

⇒\Rightarrow     \,\,\,\, n = 3

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