The area of the triangle whose vertices are i | vedclass

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

Question

(D) The vertices of the triangle are $z_1 = i$,$z_2 = \omega$,and $z_3 = \omega^2$. We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\o

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4}$

Vedclass · Answer

(D) The vertices of the triangle are $z_1 = i$,$z_2 = \omega$,and $z_3 = \omega^2$. We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. The area of a triangle with vertices $z_1, z_2, z_3$ in the Argand plane is given by $\frac{1}{2} |	ext{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$. Substituting the values: $z_1 = 0 + i$,$z_2 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$,$z_3 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. $\bar{z_1}z_2 = (-i)(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} + i\frac{1}{2}$. $\bar{z_2}z_3 = (-\frac{1}{2} - i\frac{\sqrt{3}}{2})(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = \frac{1}{4} - \frac{3}{4} + i\frac{\sqrt{3}}{2} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. $\bar{z_3}z_1 = (-\frac{1}{2} + i\frac{\sqrt{3}}{2})(i) = -\frac{\sqrt{3}}{2} - i\frac{1}{2}$. Sum $= (\frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}) + i(\frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{1}{2}) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. The imaginary part is $\frac{\sqrt{3}}{2}$. Area $= \frac{1}{2} |\frac{\sqrt{3}}{2}| = \frac{\sqrt{3}}{4}$ sq.units.

The area of the triangle whose vertices are $i, \omega$ and $\omega^2$ is (Where $\omega$ is a complex cube root of unity other than $1$,$i$ is an imaginary number) . . . . . . sq.units

Similar Questions

If a complex number $z=x+iy$ represents a point $P$ on the Argand plane and $\operatorname{Arg}\left(\frac{z-(3-2i)}{z-(-2+3i)}\right)=\frac{\pi}{4}$,then the locus of $P$ is a

If $z_1 = 10 + 6i$,$z_2 = 4 + 6i$ and $z$ is any complex number such that the argument of $\frac{z - z_1}{z - z_2}$ is $\frac{\pi}{4}$,then

If $|z - 3 + 2i| \leq 4$,then the difference between the greatest value and the least value of $|z|$ is

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

The solutions of the equation in $z$,$|z|^2 - (z + \bar{z}) + i(z - \bar{z}) + 2 = 0$ are $(i = \sqrt{-1})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module