The area of the triangle whose vertices are c | vedclass

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(B) Let $z = x + iy$. Then the vertices are $z = (x, y)$,$iz = i(x + iy) = -y + ix = (-y, x)$,and $z + iz = (x - y) + i(x + y) = (x - y, x + y)$.The a

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$1/2|z|^2$

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(B) Let $z = x + iy$. Then the vertices are $z = (x, y)$,$iz = i(x + iy) = -y + ix = (-y, x)$,and $z + iz = (x - y) + i(x + y) = (x - y, x + y)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the coordinates:Area $= \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$$= \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$$= \frac{1}{2} |-xy - xy - (x - y)^2|$$= \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$$= \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$$= \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2)$.Since $|z|^2 = x^2 + y^2$,the area is $\frac{1}{2} |z|^2$.

The area of the triangle whose vertices are complex numbers $z, iz, z + iz$ in the Argand diagram is

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