The area of the triangle formed by the comple | vedclass

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(A) Let $z = x + iy$. Then $iz = -y + ix$ and $z + iz = (x - y) + i(x + y)$.The vertices of the triangle are $(x, y)$,$(-y, x)$,and $(x - y, x + y)$.T

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

Vedclass · Answer

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

The area of the triangle formed by the complex numbers $z$,$iz$,and $z+iz$ as vertices in the Argand diagram is:

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