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(C) First,simplify the expression $\frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $\frac{(\sqrt{3}+i)^2}

Vedclass · Accepted Answer

$\sqrt{13}$

Vedclass · Answer

(C) First,simplify the expression $\frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $\frac{(\sqrt{3}+i)^2}{3+1} = \frac{3-1+2\sqrt{3}i}{4} = \frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$.Given $(e^{i\pi/3})^n = -1 = e^{i\pi}$,so $n\pi/3 = \pi + 2k\pi$. For the least positive integer $n$,$n/3 = 1 \implies n = 3$. Thus,$p = 3$.Next,simplify $\frac{1-\sqrt{3}i}{1+\sqrt{3}i}$. Multiplying by $1-\sqrt{3}i$,we get $\frac{(1-\sqrt{3}i)^2}{1+3} = \frac{1-3-2\sqrt{3}i}{4} = \frac{-2-2\sqrt{3}i}{4} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = e^{i4\pi/3}$.Given $(e^{i4\pi/3})^m = \operatorname{cis}(2\pi/3) = e^{i2\pi/3}$,so $4m\pi/3 = 2\pi/3 + 2k\pi$. Dividing by $2\pi/3$,we get $2m = 1 + 3k$. For $k=1$,$2m = 4 \implies m = 2$. Thus,$q = 2$.Finally,$\sqrt{p^2+q^2} = \sqrt{3^2+2^2} = \sqrt{9+4} = \sqrt{13}$.

If the least positive integer $n$ satisfying the equation $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{n}=-1$ is $p$ and the least positive integer $m$ satisfying the equation $\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^m=\operatorname{cis} \frac{2 \pi}{3}$ is $q$,then $\sqrt{p^2+q^2}=$

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