If z1 and z2 are complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then z1+z2z1−z2may be

Let z1=a+ib and z2=c−id, where a > 0 and d > 0. Then,z1=z2⇒a2+b2=c2+d2                  (1)Now, z1+z2z1−z2=(a+ib)+(c−id)(a+ib)−(c−id)=[(a+c)+i(b−d)][(a−c)−i(b+d)][(a−c)+i(b+d)][(a−c)−i(b+d)]=a2+b2−c2+d2−2(ad+bc)ia2+c2−2ac+b2+d2+2bd=−(ad+bc)ia2+b2−ac+bd                               [Using (1)]Hence, z1+z2z1−z2 is purely imaginary. However, if ad + bc = 0, then z1+z2z1−z2will be equal to zero. According to the conditions of the equation, we can have ad + bc = 0.