# Solve the following

Question:

$\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$ is possible if

(a) $z_{2}=\bar{z}_{1}$

(b) $z_{2}=\frac{1}{z_{1}}$

(c) $\arg \left(z_{1}\right)=\arg \left(z_{2}\right)$

(d) $\left|z_{1}\right|=\left|z_{2}\right|$

Solution:

Since given $\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$

i.e $\left|z_{1}+z_{2}\right|^{2}=\left(\left|z_{1}\right|+\left|z_{2}\right|\right)^{2}$

i.e $\left|z_{1}+z_{2}\right|^{2}=\left(\left|z_{1}\right|+\left|z_{2}\right|\right)^{2}$

Since, $\left|z_{1}\right|=r_{1},\left|z_{2}\right|=r_{2} \quad$ where $z_{1}=r_{1} e_{1}^{i \theta} z_{2}=r_{2} e_{2}^{i \theta}$

$\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+2\left|z_{1}\right|\left|z_{2}\right| \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}$

i. e $r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}$

i. e $2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=2 r_{2}$

i. e $\cos \left(\theta_{1}-\theta_{2}\right)=1$

i. e $\theta_{1}-\theta_{2}=0$

i. e $\operatorname{ch} g z_{1}=\operatorname{ch} g z_{2}$

Hence, the correct answer is option $\mathrm{C}$.