Question

Points $z_1$ and $z_2$ are adjacent vertices of a regular polygon of n sides. Find the vertex $z_3$ adjacent to $z_2(z_3\ne z_1).$

• A. $z_2+(z_2-z_1)[\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n}].$
• B. $z_2-\left(z_1\right)[\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n}].$
• C. $z_2+(z_2-z_1)[\sin \frac{2\pi }{n}\pm i\cos \frac{2\pi }{n}].$
• D. $z_2-\left(z_1\right)[\sin \frac{2\pi }{n}\pm i\cos \frac{2\pi }{n}].$

Answer 1

The correct option is A $z_2+(z_2-z_1)[\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n}].$

Let $C\left(z_0\right)$ be the centre of the polygon and $A_1\left(z_1\right),A_3\left(z_3\right)$ be two vertices on either side of $A_2\left(z_2\right)$ as shown in figures I and II.
$z_1,z_2$ being known to be adjacent and we have to find the vertex $z_3$ in terms of $z_1$ and $z_2.$
From fig. (1), rotation being anticlockwise, we have
$z_2-z_0=(z_1-z_0)e^{2\pi i/n}$
$z_3-z_0=(z_2-z_0)e^{2\pi i/n}.$ Subtracting we get
$z_3-z_2=(z_2-z_1)e^{2\pi i/n}$
$\therefore z_3=z_2+(z_2-z_1)e^{2\pi i/n}$ ...(1)
Similarly, proceeding as above for the second figure in which the rotation is clockwise, we have
$z_3=z_2+(z_2-z_1)e^{-2\pi i/n}$ ...(2)
$\therefore z_3=z_2+(z_2-z_1)e^{\pm 2\pi i/n}$
$=z_2+(z_2-z_1)[\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n}].$
Ans: A