Question

If for complex numbers $z_{1}$ and $z_{2},$ arg$(z_{1})$ - arg$(z_{2}) = 0,$
then $|z_{1}- z_{2}|$ is equal to

• A. $|Z_{1}|+|Z_{2}|$
• B. $|Z_{1}|-|Z_{2}|$
• C. $||Z_{1}|-|Z_{2}||$
• D. $0$

Answer 1

Answer: (C)

We have,
$|z_{1}- z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} - 2|z_{1}| |z_{2}|$ cos $(\theta_{1} - \theta_{2})$
where $\theta_{1}= $ arg $(Z_{1})$ and $\theta_{2} =$ arg $(z_{2}).$ Given,
arg $(z_{1}, - z_{2}) = 0$
$\Rightarrow |z_{1},- z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} -2 |z_{1}| |z_{2}|$
$= (|z_{1}|-|z_{1}|)^{2}$
$\Rightarrow |z_{1},-z_{2}| = ||z_{1}|-|z_{2}||.$