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
$\begin{array}{l}\Rightarrow 2\left(1+\cos \left({\theta }_1-{\theta }_2\right)\right)=4\\ \Rightarrow \cos \left({\theta }_1-{\theta }_2\right)=1\\ \Rightarrow {\theta }_1-{\theta }_2=0(\because \cos 0=1)\\ \Rightarrow {\mathrm{argz}}_1-{\mathrm{argz}}_2=0\end{array}$