The area of the triangle with vertices A(z),B | vedclass

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(B) Let the vertices of the triangle be $A(z)$,$B(iz)$,and $C(z+iz)$.This triangle is formed by the origin $O(0)$,the point $P(z)$,and the point $Q(iz

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the vertices of the triangle be $A(z)$,$B(iz)$,and $C(z+iz)$.This triangle is formed by the origin $O(0)$,the point $P(z)$,and the point $Q(iz)$.Since $iz$ is obtained by rotating $z$ by $90^{\circ}$ about the origin,the vectors $\vec{OP}$ and $\vec{OQ}$ are perpendicular and have the same magnitude $|z|$.The point $C(z+iz)$ represents the fourth vertex of the square formed by $O, P, C, Q$.The triangle $ABC$ is the same as the triangle $OPQ$ if we consider the vertices as vectors from the origin,but specifically,the triangle with vertices $O(0)$,$P(z)$,and $Q(iz)$ is a right-angled triangle with area $\frac{1}{2} 	imes |z| 	imes |iz| = \frac{1}{2}|z|^2$.The triangle with vertices $A(z)$,$B(iz)$,and $C(z+iz)$ is congruent to the triangle $OPQ$ because it is a translation of the triangle $OPQ$ by the vector $z+iz$ is not correct; rather,the triangle with vertices $0, z, iz$ has area $\frac{1}{2}|z|^2$. The triangle with vertices $z, iz, z+iz$ is simply a shifted version of the triangle with vertices $0, z, iz$.Thus,the area is $\frac{1}{2}|z||iz| = \frac{1}{2}|z|^2$.

The area of the triangle with vertices $A(z)$,$B(iz)$,and $C(z+iz)$ is

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