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Question: \(|z_{1} + z_{2}| = |z_{1}| + |z_{2}|\)is possible if...

∣z1+z2∣=∣z1∣+∣z2∣|z_{1} + z_{2}| = |z_{1}| + |z_{2}|is possible if

A

z2=zˉ1z_{2} = {\bar{z}}_{1}

B

z2=1z1z_{2} = \frac{1}{z_{1}}

C

arg⁡(z1)=arg⁡(z2)\arg(z_{1}) = \arg(z_{2})

D

∣z1∣=∣z2∣|z_{1}| = |z_{2}|

Answer

arg⁡(z1)=arg⁡(z2)\arg(z_{1}) = \arg(z_{2})

Explanation

Solution

Sol. Squaring both sides, we get

∣z1∣2+∣z2∣2+2∣z1∣∣z2∣cos⁡(θ1−θ2)=∣z1∣2+∣z2∣2+2∣z1∣∣z2∣|z_{1}|^{2} + |z_{2}|^{2} + 2|z_{1}||z_{2}|\cos(\theta_{1} - \theta_{2}) = |z_{1}|^{2} + |z_{2}|^{2} + 2|z_{1}||z_{2}|

⇒ 2∣z1∣∣z2∣cos⁡(θ1−θ2)=2∣z1∣∣z2∣2|z_{1}||z_{2}|\cos(\theta_{1} - \theta_{2}) = 2|z_{1}||z_{2}| ⇒ cos⁡(θ1−θ2)=1\mathbf{\cos}\mathbf{(}\mathbf{\theta}_{\mathbf{1}}\mathbf{-}\mathbf{\theta}_{\mathbf{2}}\mathbf{) = 1}

⇒ θ1−θ2=0o\theta_{1} - \theta_{2} = 0^{o}⇒ θ1=θ2\theta_{1} = \theta_{2}

Hence arg (z1)=(z_{1}) = arg (z2)(z_{2})

Trick: Let z1 and z2 are the two sides of a triangle. By applying triangle inequality (z1+z2)(z_{1} + z_{2}) is the third side. Equality holds only when θ1=θ2\theta_{1} = \theta_{2} i.e.,z1andz2z_{1}\text{and}z_{2} are parallel.