Question

The area of the triangle on the complex plane formed by the complex numbers $z,-iz$ and $z+iz$ is

• A. ${\left|z^{}\right|}^2$
• B. ${\left|\overline{z^{}}\right|}^2$
• C. $\frac{{\left|z^{}\right|}^2}{2}$
• D. None of these

Answer 1

Answer: (C)

$\frac{{\left|z^{}\right|}^2}{2}$

Let $z=x+iy$.

Therefore, $-iz=y-ix$ and $z+iz=(x-y)+i(x+y)$.

So, we need to find the area of a triangle whose vertices are $A(x,y),B(y,-x)$ and $C(x-y,x+y)$.

By Shoelace formula,

Area = $\frac{1}{2}|{(x}_A-x_C\left)\right(y_B-y_A)-(x_A-x_B)-(y_C-y_A)|$

= $\frac{1}{2}|{-x}^2+xy+y^2+xy-y^2-x^2-xy+x^2-xy-y^2|$

= $\frac{1}{2}|-x^2-y^2|$

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

Therefore, the correct answer is C.