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$n$ का वह न्यूनतम धनपूर्णांक मान जिसके लिए $\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)^{ n }=1$ है
$2$
$6$
$5$
$3$
Solution
Let $l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)$
$\therefore l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1+i \sqrt{3}}\right)$
$=\left(\frac{-2+i 2 \sqrt{3}}{4}\right)=\left(\frac{1-i \sqrt{3}}{-2}\right)$
Also, $l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1-i \sqrt{3}}{1-i \sqrt{3}}\right)$
$=\left(\frac{4}{-2-i 2 \sqrt{3}}\right)=\left(\frac{-2}{1+i \sqrt{3}}\right)$
Now,
${\left( {\frac{{1 + i\sqrt 3 }}{{1 – i\sqrt 3 }}} \right)^3} = \left( {\frac{{1 + i\sqrt 3 }}{{1 – i\sqrt 3 }}} \right)\, \times $ $\,\left( {\frac{{1 + i\sqrt 3 }}{{1 – i\sqrt 3 }}} \right)\, \times \,\left( {\frac{{1 + i\sqrt 3 }}{{1 – i\sqrt 3 }}} \right)$
$ = \left( {\frac{{1 + i\sqrt 3 }}{{1 – i\sqrt 3 }}} \right) \times \left( {\frac{{ – 2}}{{1 + i\sqrt 3 }}} \right) \times \left( {\frac{{1 – i\sqrt 3 }}{{ – 2}}} \right) = 1$
$\therefore$ least positive integer $n$ is $3$ .
