The area of Delta whose vertices are z, omega | vedclass

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

Question

(B) Let the vertices be $A = z$,$B = \omega z$,and $C = z + \omega z$.The side lengths are:$|AB| = |z - \omega z| = |z(1 - \omega)| = |z| |1 - \omega|

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4} |z|^2$

Vedclass · Answer

(B) Let the vertices be $A = z$,$B = \omega z$,and $C = z + \omega z$.The side lengths are:$|AB| = |z - \omega z| = |z(1 - \omega)| = |z| |1 - \omega| = |z| \sqrt{1^2 + 1^2 - 2 \cos(120^{\circ})} = |z| \sqrt{3}$.$|BC| = |(z + \omega z) - \omega z| = |z|$.$|AC| = |(z + \omega z) - z| = |\omega z| = |z|$.Since $|BC| = |AC| = |z|$ and $|AB| = |z|\sqrt{3}$,the triangle is an isosceles triangle with sides $|z|, |z|, |z|\sqrt{3}$.The angle between sides $BC$ and $AC$ is $120^{\circ}$.Area $= \frac{1}{2} 	imes |BC| 	imes |AC| 	imes \sin(120^{\circ})$Area $= \frac{1}{2} 	imes |z| 	imes |z| 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} |z|^2$.

The area of $\Delta$ whose vertices are $z, \omega z, z + \omega z$ is (where $\omega$ is a complex cube root of unity):

Similar Questions

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

If $|z - 2 - 3i| + |z + 2 - 6i| = 4$,where $i = \sqrt{-1}$,then the locus of $P(z)$ is

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

The number of points $z$ on the Argand plane which satisfy the conditions $\operatorname{Re}\left(\frac{z-2}{z-4i}\right)=0$ and $\operatorname{Im}\left(\frac{z-2}{z-4i}\right)=1$ simultaneously is

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