Q. The centre of a regular polygon of $n$ sides is located at the point $\mathcal{z} = 0$ and one of its vertex $\mathcal{z}_{1}\$is known. If $\mathcal{z}_{2}$ be the vertex adjacent to $\mathcal{z}_{1},\$then $\mathcal{z}_{2}$ is equal to
A) $\mathcal{z}_{1}\left( \cos{\frac{2\pi}{n} \pm i\sin{\frac{2\pi}{n}\ }} \right)$
B) $\mathcal{z}_{1}\left( \cos{\frac{\pi}{n} \pm i\sin{\frac{\pi}{n}\ }} \right)$
C) $\mathcal{z}_{1}\left( \cos{\frac{\pi}{2n} \pm i\sin{\frac{\pi}{2n}\ }} \right)$
D) None of these
Explanation
(a)
Let $A$ be the vertex with affix $\mathcal{z}_{1}$. There are two possibilities of $\mathcal{z}_{2}\ ie,\ \mathcal{z}_{2}$ can be obtained by rotating $\mathcal{z}_{1}$ through $\frac{2\pi}{n}$ either in clockwise or in anti-clockwise direction.
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$\therefore\ \frac{\mathcal{z}_{2}}{\mathcal{z}_{1}} = \left| \frac{\mathcal{z}_{2}}{\mathcal{z}_{1}} \right|e^{\pm \frac{i2\pi}{n}}$
$\Rightarrow \ \mathcal{z}_{2} = \mathcal{z}_{1}\left( \cos{\frac{2\pi}{n} \pm i\sin\frac{2\pi}{n}} \right)\ \ \ \ (\because\left| \mathcal{z}_{2} \right| = \left| \mathcal{z}_{1} \right|)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.