Let $z=x+iy$
$iz=i(x+iy)=xi+i^2y=-y+xi$
$-iz=-i(x+iy)=-xi-i^2y=y-xi$
Then the vertices of the triangle are given by
$A(z)=(x,y), \:\:B(iz)=(-y,x)\:\:and\:\:C(-z)=(y,-x)$
Area of $\Delta \:ABC=\large \frac{1}{2}$$\left|\begin {array}{ccc}x & y & 1\\-y & x &1\\y &-x&1\end {array}\right|$
$=\large\frac{1}{2}$$|1(xy-xy)-1(-x^2-y^2)+1(x^2+y^2)|$
$=x^2+y^2=|z|^2$