If z = x + iy,then the area of the triangle w | vedclass

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(B) Let $z = x + iy$. The vertices of the triangle are $z = (x, y)$,$iz = (-y, x)$,and $z + iz = (x - y, x + y)$.The area $A$ of a triangle with verti

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

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

If $z = x + iy$,then the area of the triangle whose vertices are the points $z$,$iz$,and $z + iz$ is

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