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Question
|z1 + z2| = |z1| + |z2| is possible if ______.
Options
`z_2 = barz_1`
`z_2 = 1/z_1`
arg(z1) = arg(z2)
|z1| = |z2|
Solution
|z1 + z2| = |z1| + |z2| is possible if arg (z1) = arg (z2).
Explanation:
Let z1 = r1(cosθ1 + isin θ1) and z2 = r2(cosθ2 + isin θ2)
Since |z1 + z2| = |z1| + |z2|
|z1 + z2| = r1cosθ1 + ir1sinθ1 + r2cosθ2 + ir2sinθ2
|z1 + z2| = `sqrt(r_1^2 cos^2 theta_ + r_2^2 cos^2 theta_2 + 2r_1r_2 cos theta_1 cos theta_2 + r_1^2 sin^2 theta_1 + r_2^2 sin^2 theta_2 + 2r_1r_2 sin theta_1 sin theta_2)`
= `sqrt(r_1^2 + r_2^2 + 2r_1r_2 cos(theta_1 - theta_2))`
But |z1 + z2| = |z1| + |z2|
So `sqrt(r_1^2 + r_2^2 + 2r_1r_2 cos(theta_1 - theta_2))` = r1 + r2
Squaring both sides, we get
`r_1^2 + r_2^2 + 2r_1r_2 cos(theta_1 - theta_2) = r_1^2 + r_2^2 + 2r_1r_2`
⇒ `2r_1r_2 - 2r_1r_2 cos(theta_1 - theta_2)` = 0
⇒ 1 – cos(θ1 – θ2) = 0
⇒ cos(θ1 – θ2) = 1
⇒ θ1 – θ2 = 0
⇒ θ1 = θ2
So, arg(z1) = arg(z2)
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