The area of triangle on the complex plane formed by the complex numbers z,−iz  and z−iz  is:

Let z=x+iy  where x,y∈R So, −iz=−i(x+iy)=y−ixAnd z+iz=(x+iy)+i(x+iy)=x−y+i(x+y)So required area=12xy1y−x1x−yx+y1                       =12[x(−x−x−y)−y(y−x+y)+(xy+y2+x2−xy)]                       =12[−2x2−xy−2y2+xy+x2+y2]                       =−12[x2+y2]                       =12|z|2