Let z1=cosθ1+isinθ1,z2=cosθ2+isinθ2 ⇒z1+z2=(cosθ1+cosθ2)+i(sinθ1+sinθ2) As |z1+z2|=|z1|+z2| √(cosθ1+cosθ2)2+(sinθ1+sinθ2)2=1+1=2 By squaring on both the sides we get ⇒2(1+cos(θ1−θ2))=4 ⇒cos(θ1−θ2)=1 ⇒θ1−θ2=0[ as cos0°=1] ⇒Argz1−Argz2=0