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

Answer 1

The correct option is D $0$

A] Let $z_1=cos{\theta }_1+isin{\theta }_1$

$z_2=cos{\theta }_2+isin{\theta }_2$

$z_1+z_2=(cos{\theta }_1+cos{\theta }_2)+i(sin{\theta }_1+sin{\theta }_2)$

As $|z_1+z_2|=|z_1|+|z_2|$

$=\sqrt{(cos{\theta }_1+cos{\theta }_2{)}^2+(sin{\theta }_1+sin{\theta }_2{)}^2}=1+1$

$=2(1+cos({\theta }_1-{\theta }_2\left)\right)=4$ [squaring]

$=cos({\theta }_1-{\theta }_2)=1$

$={\theta }_1-{\theta }_2=0$ [as $cos{0}^{\circ }=1$]

$=Arg{z}_1-Arg{z}_2=0$