Question

Statement I Let $z_1$ and $z_2$ be two complex numbers such that $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$, then $\arg \left(z_1\right)-\arg \left(z_2\right)=0$
Statement II $\arg \left(\frac{z_1}{z_2}\right)=\arg \left(z_1\right)-\arg \left(z_2\right)$.

• A. Statement I is correct
• B. Statement II is coprect
• C. Both are correct
• D. Neither I nor II is correct

Answer 1

Answer: (C)

Let $z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right)$ and $z_2=r_2\left(\cos \theta_2+i \sin \theta_2\right)$
where $ r_1=\left|z_1\right|, \arg \left(z_1\right)=\theta_1, r_2=\left|z_2\right|, \arg \left(z_2\right)=\theta_2$
we have, $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$ $=\left|r_1\left(\cos \theta_1+\cos \theta_2\right)+r_2\left(\cos \theta_2+\sin \theta_2\right)\right|=r_1+r_2$
$r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1 \theta_2\right)=\left(r_1+r_2\right)^2$
$\Rightarrow \cos \left(\theta_1-\theta_2\right)=1 $
$\Rightarrow \theta_1-\theta_2=0 \Rightarrow \theta_1=\theta_2 \text { i.e; } \arg z_1=\arg z_2$