Question

The centre of regular polygon of $n$ sides is located at $Z=0$ and one of its vertices is $Z_1$ lf $Z_2$ is the vertex adjacent to $Z_1$ then $Z_2=$

• A. $Z_1(\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n})$
• B. $Z_1(\cos \frac{\pi }{n}\pm i\sin \frac{\pi }{n})$
• C. $Z_1(\cos \frac{\pi }{2n}\pm i\sin \frac{\pi }{2n})$
• D. $Z_1(\cos \frac{\pi }{3n}\pm i\sin \frac{\pi }{3n})$

Answer 1

The correct option is A $Z_1(\cos \frac{2\pi }{n}\pm i\sin \frac{2\pi }{n})$

Polygon of n

sides

$\Rightarrow$ subsequent vertices subtend angle of $\frac{2\pi }{n}$ at
origin.

$[\because 2\pi$ is total angle at centre and there are n sides
$\Rightarrow$ each side subtends $\frac{2\pi }{n}]$

So,

$z_2$ is adjacent to $z_1$

$\Rightarrow$ $z_2$ can either be clockwise (cw) or anticlockwise
(acw) adjacent.

So,

$z_2=z_1{e}^{\frac{i2\pi }{2}}$ or $z_2=z_1{e}^{\frac{-i2\pi }{2}}$

$\Rightarrow$ $z_2=z_1(cos\frac{2\pi }{n}+isin\frac{2\pi }{n})$ or

$z_2=z_1(cos\frac{2\pi }{n}-isin\frac{2\pi }{n})$

$=z_1(cos\frac{2\pi }{n}\pm +isin\frac{2\pi }{n})$