The area of the triangle whose vertices are r | vedclass

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(B) The vertices of the triangle are $z_1 = 0$,$z_2 = z$,and $z_3 = z e^{i\alpha }$. The area of a triangle with vertices $z_1, z_2, z_3$ in the compl

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$\frac{1}{2}|z|^2 \sin \alpha $

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(B) The vertices of the triangle are $z_1 = 0$,$z_2 = z$,and $z_3 = z e^{i\alpha }$. The area of a triangle with vertices $z_1, z_2, z_3$ in the complex plane is given by $\frac{1}{2} |	ext{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$. Since $z_1 = 0$,the area is $\frac{1}{2} |	ext{Im}(\bar{z} \cdot z e^{i\alpha })|$. Substituting $\bar{z}z = |z|^2$,we get: Area $= \frac{1}{2} |	ext{Im}(|z|^2 e^{i\alpha })|$. Since $e^{i\alpha } = \cos \alpha + i \sin \alpha$,we have: Area $= \frac{1}{2} |z|^2 |	ext{Im}(\cos \alpha + i \sin \alpha)|$. Area $= \frac{1}{2} |z|^2 \sin \alpha$ (since $0  0$).

The area of the triangle whose vertices are represented by the complex numbers $0, z,$ and $z{e^{i\alpha }}$ $(0 < \alpha < \pi )$ is equal to:

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