MCQ
The least positive integer $n$ for which $\left( \frac{1 + i\sqrt 3 }{1 - i\sqrt 3 }\right)^n = 1,$ is?
- A$2$
- B$6$
- C$5$
- ✓$3$
$\therefore l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1+i \sqrt{3}}\right)$
$=\left(\frac{-2+i 2 \sqrt{3}}{4}\right)=\left(\frac{1-i \sqrt{3}}{-2}\right)$
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 $3$ .
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