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Q.If z1 and z2 be two non-zero complex numbers such that |z1+z2| = |z1|+|z2|, then prove that arg z1 - arg z2 = 0.

West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 5mImportance★★★★★est
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Expand ∣z1+z2∣2|z_1+z_2|^2 in polar form; the given condition forces cos⁡(θ1−θ2)=1\cos(\theta_1-\theta_2)=1, i.e. equal arguments.

Let z1=r1(cos⁡θ1+isin⁡θ1), z2=r2(cos⁡θ2+isin⁡θ2)z_1=r_1(\cos\theta_1+i\sin\theta_1),\ z_2=r_2(\cos\theta_2+i\sin\theta_2) with r1,r2>0r_1,r_2>0.

∣z1+z2∣2=(z1+z2)(z1+z2)‾=∣z1∣2+∣z2∣2+z1zˉ2+zˉ1z2=r12+r22+2r1r2cos⁡(θ1−θ2)|z_1+z_2|^2=(z_1+z_2)\overline{(z_1+z_2)}=|z_1|^2+|z_2|^2+z_1\bar z_2+\bar z_1z_2=r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)

(using z1zˉ2+zˉ1z2=2Re(z1zˉ2)=2r1r2cos⁡(θ1−θ2)z_1\bar z_2+\bar z_1z_2=2\text{Re}(z_1\bar z_2)=2r_1r_2\cos(\theta_1-\theta_2)).

Given ∣z1+z2∣=∣z1∣+∣z2∣=r1+r2|z_1+z_2|=|z_1|+|z_2|=r_1+r_2, so ∣z1+z2∣2=(r1+r2)2=r12+2r1r2+r22|z_1+z_2|^2=(r_1+r_2)^2=r_1^2+2r_1r_2+r_2^2.

Equating the two expressions: …

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