Question

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. $5$
• D. $3$

Answer 1

The correct option is D $3$

first rationalize the number

$(\frac{1+i\sqrt{3}}{1-i\sqrt{3}})\times (\frac{1+i\sqrt{3}}{1+i\sqrt{3}})=(\frac{-2+i2\sqrt{3}}{4})=(\frac{1-i\sqrt{3}}{-2})................\left(1\right)$

$(\frac{1+i\sqrt{3}}{1-i\sqrt{3}})\times (\frac{1-i\sqrt{3}}{1-i\sqrt{3}})=(\frac{4}{-2-i2\sqrt{3}})=(\frac{-2}{1+i\sqrt{3}})................\left(2\right)$

using (1) and (2)

$(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}{)}^3=(\frac{1+i\sqrt{3}}{1-i\sqrt{3}})\times (\frac{1+i\sqrt{3}}{1-i\sqrt{3}})\times (\frac{1+i\sqrt{3}}{1-i\sqrt{3}})=(\frac{1+i\sqrt{3}}{1-i\sqrt{3}})\times (\frac{-2}{1+i\sqrt{3}})\times (\frac{1-i\sqrt{3}}{-2})=1$

Therefore correct Answer is $3$

so correct option is $D$