Q.
If z1,z2,z3 are three complex numbers, then z1Im(zˉ2z3)+z2Im(zˉ3z1)+z3Im(zˉ1z2) is equal to
1469
183
Complex Numbers and Quadratic Equations
Report Error
Solution:
Let z1=x1+iy1 z2=x2+iy2 z3=x3+iy3 zˉ2z3=(x2−iy2)(x3+iy3) =(x2x3+y2y3)+i(x2y3−x3y2) ⇒Im(zˉ2z3)=x2y3−x3y2 ∴z1Im(zˉ2z3)=(x1+iy1)(x2y3−x3y2)
Similarly z2Im(zˉ3z1)=(x2+iy2)(x3y1−x1y3) z3Im(zˉ1z2)=(x3+iy3)(x1y2−x2y1)
Adding we get the result =0