Q47 of 66 Page 91

|z1 + z2| = |z1| + |z2| is possible if

Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)


Since |z1 + z2| = |z1| + |z2|


⇒ z1 + z2 = r1 cos θ1 + ir1 sin θ1+ r2 cos θ2 + ir2 sin θ2




But |z1 + z2| = |z1| + |z2|



Squaring both sides,



⇒ 2r1r2 – 2r1r2 cos (θ1 – θ2) = 0


⇒ 1 – cos (θ1 – θ2) = 0


⇒ cos (θ1 – θ2) = 1


⇒ (θ1 – θ2) = 0


⇒ θ1 = θ2


∴ arg (z1) = arg (z2)

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