The area of the triangle whose vertices are i | vedclass

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(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

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

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(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

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