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If $z_{1} \: and \: z_{2}$ are two non-zero complex numbers such that $\left | z_{1}+z_{2} \right |= \left | z_{1} \right |+\left | z_{2} \right |,$ then arg $z_{1} \: - \: arg\: z_{2}$ is equal to :

Option 1)
$-\pi$

Option 2)
$\pi/2$

Option 3)
$-\pi/2$

Option 4)
$0$

Answers (2)

best_answer

As we learnt in

Property of Modulus of z(Complex Number) -

$|z_{1}+z_{2}|^{2}=\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}+z_{1}.\bar{z_{2}}+z_{2}.\bar{z_{1}}$

- wherein

|.| denotes modulus of z

$\bar{z}$ denotes conjugate of z

|z₁ + z₂| = |z₁| + |z₂|

|z₁ + z₂| = (|z₁| + |z₂|)²

$|z|^{2}+|z_{2}|^{2}+Re(z_{1}\overline{z_{2}})=|z_{1}|+|z_{2}|^{2}+2|z_{1}| |z_{2}|$

$\therefore\ \; Re(z_{1}\bar{z_{2}})=|z_{1}| |z_{2}|$

Now let $z_{1}=|z_{1}|(cos\theta_{1}+isin\theta_{1})$

$z_{2}=|z_{2}| (cos\theta_{2}+i sin\theta_{2})$

$|z_{1}| |z_{2}| cos(\theta_{1}-\theta_{2})=|z_{1}| |z_{2}|$

$\therefore\ \; cos(\theta_{1}-\theta_{2})=1$

$\therefore\ \; \theta_{1}-\theta_{2}=0$

$\therefore\ \; arg|z_{1}|-arg|z_{2}|=0$

Correct option is 4.

Option 1)

$-\pi$

This is an incorrect option.

Option 2)

$\pi/2$

This is an incorrect option.

Option 3)

$-\pi/2$

This is an incorrect option.

Option 4)

$0$

This is the correct option.

Posted by

divya.saini

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