AP DSC School Assistant Math Model Test 1

Concept:The expression $\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)^n = 1$ implies that the complex number inside the parentheses is a $2\pi$ multiple root of unity. We simplify the complex number using rationalization or polar form, then find the smallest positive integer $n$.Explanation:Let $z = \frac{1+i\sqrt{3}}{1-i\sqrt{3}}$. To simplify $z$, multiply numerator and denominator by the conjugate of the denominator, $1+i\sqrt{3}$:$$z = \frac{(1+i\sqrt{3})^2}{(1-i\sqrt{3})(1+i\sqrt{3})} = \frac{1 + 2i\sqrt{3} - 3}{1+3} = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}.$$Notice that $-\frac{1}{2} + i\frac{\sqrt{3}}{2} = \cos 120^\circ + i\sin 120^\circ = e^{i 2\pi/3}$. So $z = e^{i 2\pi/3}$.Now $z^n = e^{i 2n\pi/3} = 1$ when $\frac{2n\pi}{3}$ is a multiple of $2\pi$, i.e., $\frac{2n}{3}$ is an integer. The smallest positive integer $n$ satisfying $\frac{2n}{3} \in \mathbb{Z}$ is $n=3$ (since $2\times3/3 = 2$). Checking: $z^3 = (e^{i 2\pi/3})^3 = e^{i 2\pi} = 1$.Thus the least positive integer $n$ is $3$.Answer:$3$ (Option D)