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

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$3$

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(D) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.We recognize this as $\omega$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is a complex cube root of unity.Thus,$z = \omega$.The equation becomes $\omega^n = 1$.The smallest positive integer $n$ for which $\omega^n = 1$ is $n = 3$.

The least positive integer $n$ for which $\left( \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} \right)^n = 1$ is?

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