In the complex plane,let z_1=sqrt{3}+i and z_ | vedclass

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(B) Given that $z_1=\sqrt{3}+i$ and $z_2=\sqrt{3}-i$ are two adjacent vertices of an $n$-sided regular polygon centered at the origin.First,we find th

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$6$

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(B) Given that $z_1=\sqrt{3}+i$ and $z_2=\sqrt{3}-i$ are two adjacent vertices of an $n$-sided regular polygon centered at the origin.First,we find the arguments of $z_1$ and $z_2$:$\arg(z_1) = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$$\arg(z_2) = 	an^{-1}\left(-\frac{1}{\sqrt{3}}ight) = -\frac{\pi}{6}$The angle subtended by the side $z_1z_2$ at the origin $O$ is:$	heta = |\arg(z_1) - \arg(z_2)| = \left|\frac{\pi}{6} - \left(-\frac{\pi}{6}ight)ight| = \frac{\pi}{3}$For a regular $n$-sided polygon centered at the origin,the angle subtended by any side at the center is $\frac{2\pi}{n}$.Therefore,$\frac{2\pi}{n} = \frac{\pi}{3}$.Solving for $n$,we get $n = 6$.

In the complex plane,let $z_1=\sqrt{3}+i$ and $z_2=\sqrt{3}-i$ be two adjacent vertices of an $n$-sided regular polygon centered at the origin. Then,$n$ equals

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