If {left( {frac{{1 + isqrt 3 }}{{1 - isqrt 3 | vedclass

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Question

(C) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:$z = \f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.This is the value of $\omega$,the complex cube root of unity.Thus,$z^n = \omega^n$.For $\omega^n$ to be an integer,$n$ must be a multiple of $3$ because $\omega^3 = 1$.The smallest positive integer $n$ is $3$.

If ${\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n}$ is an integer,then the smallest positive integer $n$ is

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