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# Let $\;z_{1}\;$ and $\;z_{2}\;$ be two complex numbers such that $\;|z_{1}+z_{2}| = |z_{1}|+|z_{2}|\;$ Then show that $\;arg(z_{1}) - arg(z_{2}) =0$

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;4$

Answer : $\;0$
Explanation :
Let $\;z_{1}= r_{1}(cos \theta_{1} + i sin \theta_{1}) \;$ and $\;z_{2}= r_{2}(cos \theta_{2} + i sin \theta_{2})$
Where $\;r_{1} =| z_{1}|\; ,arg(z_{1}) = \theta_{1}\;. r_{2} = |z_{2}|\;, arg(z_{2}) = \theta_{2}$
$\;|z_{1}+z_{2}| = |z_{1}|+|z_{2}|\;$
$=|r_{1}(cos \theta_{1} + i sin \theta_{1}) + r_{2}(cos \theta_{2} + i sin \theta_{2})| = r_{1}+r_{2}$
$=r^{2}_{1} + r^{2}_{2} + 2r_{1} r_{2} cos(\theta_{1} - cos \theta_{2}) = (r_{1}+r_{2})^{2}$
i.e $cos (\theta_{1} - \theta_{2}) =1$
i.e $\; (\theta_{1} - \theta_{2})\; \; i\;.e\;. arg(z_{1}) = arg(z_{2})$