Let a complex number be w=1-3i . Let another complex number z be such that |zw|=1 and arg(z)-arg(w)=π2. Then the area of the triangle with vertices origin, z and w is equal

Given w=1−3iw=1+3=4zw=1⇒z.w=1⇒z=1w=12argw=−tan−13            =−π3Given argz=π2+argw                   =π2−π3                   =π6z=143+iThe area of the triangle formed by the points O, z, w is 12x1y2−x2y1=1234+14=121=12 square units