A value of n such that left(frac{sqrt{3}}{2}+ | vedclass

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Question

(A) Given,$\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$. We can write the complex number in polar form as $\cos\left(\frac{\pi}{6}\right) + i\sin\

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Given,$\left(\frac{\sqrt{3}}{2}+\frac{i}{2}ight)^n=1$. We can write the complex number in polar form as $\cos\left(\frac{\pi}{6}ight) + i\sin\left(\frac{\pi}{6}ight) = e^{i\pi/6}$. So,the equation becomes $(e^{i\pi/6})^n = 1$,which is $e^{in\pi/6} = 1$. For $e^{i	heta} = 1$,$	heta$ must be an integer multiple of $2\pi$. Thus,$\frac{n\pi}{6} = 2k\pi$ for some integer $k$. $n = 12k$. For $k=1$,$n=12$. Therefore,$n=12$ is a value that satisfies the equation.

$A$ value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is

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