You can notice that all vertices are defined as $z$ multiplied by some constant:
-z & = z \cdot (-1)\\
iz & = z \cdot i\\
z-iz &= z \cdot (1-i)
So you can analyze the triangle with vertices $-1, i, (1-i)$ (complex), that is $(-1,0), (0,1), (1,-1)$ (cartesian) first. You will find it is isosceles with a base of $\sqrt 2$ and a height $\frac 32 \sqrt 2$, so its area is $\frac 32$.
When you multiply all vertices by arbitrary $z$, they all get scaled with respect to $(0,0)$ by factor $|z|$ and rotated by angle $\arg z$. Rotation does not affect the area of a figure, but scaling applies to the area with a square, hence the answer $$\frac 32 \cdot |z|^2$$