JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 12)
The least positive integer n for which $${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1,$$ is :
2
3
5
6
Explanation
$${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$$
$$ \Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$$
We know, $$\omega $$ = $$-$$ $${{1 - i\sqrt 3 } \over 2}$$
and $$\omega $$2 = $$-$$ $${{\left( {1 + i\sqrt 3 } \right)} \over 2}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$$
$$ \Rightarrow $$ ($$\omega $$)n = 1 = $$\omega $$3
$$ \Rightarrow $$ $$\,\,\,\,$$ n = 3
$$ \Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$$
We know, $$\omega $$ = $$-$$ $${{1 - i\sqrt 3 } \over 2}$$
and $$\omega $$2 = $$-$$ $${{\left( {1 + i\sqrt 3 } \right)} \over 2}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$$
$$ \Rightarrow $$ ($$\omega $$)n = 1 = $$\omega $$3
$$ \Rightarrow $$ $$\,\,\,\,$$ n = 3
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