The centre of a regular polygon of n sides is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $O$ be the origin $(z=0)$ and $A$ be the vertex with affix $z_1$. Since it is a regular polygon of $n$ sides,the angle subtended by any side a

Vedclass · Accepted Answer

$z_1 \left( \cos \frac{2\pi}{n} \pm i \sin \frac{2\pi}{n} \right)$

Vedclass · Answer

(A) Let $O$ be the origin $(z=0)$ and $A$ be the vertex with affix $z_1$. Since it is a regular polygon of $n$ sides,the angle subtended by any side at the centre is $\frac{2\pi}{n}$.$z_2$ can be obtained by rotating $z_1$ about the origin through an angle of $\pm \frac{2\pi}{n}$.Using the rotation formula,$z_2 = z_1 e^{\pm i \frac{2\pi}{n}}$.Applying Euler's formula,$z_2 = z_1 \left( \cos \frac{2\pi}{n} \pm i \sin \frac{2\pi}{n} \right)$.

The centre of a regular polygon of $n$ sides is located at the point $z = 0$ and one of its vertices $z_1$ is known. If $z_2$ is the vertex adjacent to $z_1$,then $z_2$ is equal to

Similar Questions

The points in the Argand plane given by $Z_1 = -3 + 5i$,$Z_2 = -1 + 6i$,$Z_3 = -2 + 8i$,and $Z_4 = -4 + 7i$ form a:

The set of all real values of $c$ for which the equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ represents a circle is

Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $w = \frac{5 + 3z}{5(1 - z)}$,then

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

Let $A = \{z \in \mathbb{C} : 1 \leq |z - (1 + i)| \leq 2\}$ and $B = \{z \in A : |z - (1 - i)| = 1\}$. Then,$B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module