Solution:
SOLUTION
If z=x+iy, then the point represented by z is (x,y)
i2 = −1
$$|z|=\sqrt{x^2+y^2}$$
Area of the triangle with vertices (x1,y1),(x2,y2)and(x3,y3) is given by $$\large \frac{1}{2}\left|\begin {array}{ccc}x_1 & y_1 & 1\\x_2 & y_2 &1\\x_3 &y_3&1\end {array}\right|$$
Let z = x + iy
iz=i(x+iy)=xi+i2y=−y+xi
−iz=−i(x+iy)=−xi−i2y=y−xi
Then the vertices of the triangle are given by
A(z)=(x,y),B(iz)= (−y,x)andC(−z)=(y,−x)
Area of ΔABC = $$\large \frac{1}{2}\left|\begin {array}{ccc}x & y & 1\\-y & x &1\\y &-x&1\end {array}\right|$$
=1/2|1(xy−xy)−1(−x2−y2)+1(x2+y2)|
=x2+y2=|z|2