If z, iz and z + iz are the vertices of a tri | vedclass

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(B) Let $z = x + iy$. Then $iz = i(x + iy) = -y + ix$ and $z + iz = (x - y) + i(x + y)$.These vertices are $A(x, y)$,$B(-y, x)$,and $C(x - y, x + y)$.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $z = x + iy$. Then $iz = i(x + iy) = -y + ix$ and $z + iz = (x - y) + i(x + y)$.These vertices are $A(x, y)$,$B(-y, x)$,and $C(x - y, x + y)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.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 $= \frac{1}{2} |-xy - xy - (x - y)^2| = \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)| = \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} |x^2 + y^2| = \frac{1}{2} |z|^2$.Given that the area is $2$,we have $\frac{1}{2} |z|^2 = 2$.Therefore,$|z|^2 = 4$,which implies $|z| = 2$.

If $z, iz$ and $z + iz$ are the vertices of a triangle whose area is $2$ units,then the value of $|z|$ is

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