Question

If $z_1$ $z_2$ 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. $-\pi /2$
• C. $0$
• D. $\pi /2$
• E. $\pi$

Answer 1

Answer: (C)

$0$

$\left|z_1\right|+\left|z_2\right|=\left|z_2{+z}_1\right|$ ...(1)
let $z_1=a(\cos A+i\sin A)\&z_2=b(\cos B+i\sin B)$ where $A=arg{z}_1\&B=arg{z}_2$
Substitute $z_1\&z_2$ in eq (1)
$\Rightarrow a+b=|a\cos A+b\cos B+i(a\sin A+b\sin B)|\Rightarrow a^2+b^2+2ab={(a\cos A+b\cos B)}^2+{(a\sin A+b\sin B)}^2\Rightarrow 2ab=2ab(\cos A\cos B+\sin A\sin B)\Rightarrow \cos (A-B)=1\Rightarrow A-B=0$
$\therefore arg{z}_1-arg{z}_2=0$
Hence, option 'C' is correct.