Question

If $Z_1$ and $Z_2$ are two non-zero complex numbers such that $| Z_1 + z_2| = | z_1| + | z_2|,$ then arg $Z_1$ - arg $z_2$ is equal to :

• A. $- \frac{\pi}{2}$
• B. $0$
• C. $-\pi$
• D. $-\frac{\pi}{2}$

Answer 1

Answer: (B)

Let $z _{1}=\cos \theta_{1}+ i \sin \theta_{1}$
$ \begin{array}{l} z _{2}=\cos \theta_{2}+ i \sin \theta_{2} \\ \therefore z _{1}+ z _{2}=\left(\cos \theta_{1}+\cos \theta_{2}\right)+ i \left(\sin \theta_{1}+\sin \theta_{2}\right) \end{array} $
Now,
$ \begin{array}{l} \left| z _{1}+ z +2\right|=\left| z _{1}\right|+\left| z _{2}\right| \\ \Rightarrow \sqrt{\left(\cos \theta_{1}\right)^{2}+\left(\sin \theta_{1}+\sin \theta_{2}\right)^{2}}=1+1 \end{array} $
On squaring both side
$ \begin{array}{l} \Rightarrow 2\left(1+\cos \left(\theta_{1}-\theta_{2}\right)\right)=4 \\ \Rightarrow \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \Rightarrow \theta_{1}-\theta_{2}=0 (\because \cos 0=1) \\ \Rightarrow \operatorname{argz}_{1}-\operatorname{argz}_{2}=0 \end{array} $