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

Given 

$A=-z$, $B=iz$,$C=z-iz$

let 

$z=x+iy$ then the values of $\left|A-B\right|,\left|B-C\right| and \left|C-A\right|$ forms an isosceles triangles with $AC=BC$

It is a property of an isosceles triangle that of we join ‘C’ and midpoint of ‘A’ and ‘B’ 

(let midpoint be P); it will be perpendicular to AB

$Area=\frac{1}{2}\times AB\times PC$

P=midpoint of A & B $=\frac{A+B}{2}=\frac{-z+iz}{2}$

now

$$\begin{gathered} PC=\left|-z+iz-\frac{(z-iz)}{2}\right|\Rightarrow \left|\frac{-2z+2iz-z+iz}{2}\right| \\ \Rightarrow \left|\frac{3iz-3z}{2}\right| \\ AB=\left|-z-iz\right| \end{gathered}$$

Therefore Area 

$$\begin{gathered} =\frac{1}{2}\times \left|\frac{3iz-3z}{2}\right|\times \left|-z-iz\right| \\ =\frac{1}{2}\times \left|\frac{-z+iz}{2}\right|\times \left|-z-iz\right| \\ =\frac{3}{2}\times \left|\frac{{(-z)}^2-{\left(iz\right)}^2}{2}\right| \\ =\frac{3}{2}{\left|z\right|}^2 \end{gathered}$$