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Question
Let z1 and z2 be two complex numbers such that |z1 + z2| = |z1| + |z2|. Then show that arg(z1) – arg(z2) = 0.
Solution
Let z1 = r1(cosθ1 + isin θ1) and z2 = r2(cosθ2 + isin θ2)
Where r1 = |z1|, arg(z1) = θ1, r2 = |z2|, arg(z2) = θ2.
We have |z1 + z2| = |z1| + |z2|
= `|r_1(cos theta_1 + cos theta_2) + r_2 (cos theta_2 + sin theta_2)|`
= r1 + r2
= `r_1^2 + r_2^2 + 2r_1r_2 cos(theta_1 - theta_2)`
= (r1 + r2)2
⇒ `cos(theta_1 - theta_2)` = 1
⇒ `theta_1 - theta_2` i.e. argz1 = argz2
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