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If $z_1\:\:and\:\:z_2$ are non zero complex numbers such that $|z_1+z_2|=|z_1|+|z_2|$, then $arg z_1-arg z_2=?$

$\begin{array}{1 1}(A) \;- \pi \\(B)\;-\frac{\pi}{2} \\(C)\;0 \\(D)\;\large\frac{\pi}{2}\end{array}$

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Toolbox:
  • $cos\theta_1cos\theta_2+sin\theta_1sin\theta_2=cos(\theta_1-\theta_2)$
Let $z_1=r_1(cos\theta_1+isin\theta_1)\:\:and$
$z_2=r_2(cos\theta_2+isin\theta_2)$
$\Rightarrow\:|z_1|=r_1,\:|z_2|=r_2,\:argz_1=\theta_1\:\:and\:\:argz_2=\theta_2$
$z_1+z_2=(r_1cos\theta_1+r_2cos\theta_2)+i(r_1sin\theta_1+r_2sin\theta_2)$
$\Rightarrow\:|z_1+z_2|=\sqrt {(r_1cos\theta_1+r_2cos\theta_2)^2+(r_1sin\theta_1+r_2sin\theta_2)^2}$
$=\sqrt {r_1^2+r_2^2+2r_1r_2(cos\theta_1cos\theta_2+sin\theta_1sin\theta_2)}$
$=\sqrt {r_1^2+r_2^2+2r_1r_2(cos(\theta_1-\theta_2))}$
If $|z_1+z_2|=|z_1|+|z_2|,$ then
$=\sqrt {r_1^2+r_2^2+2r_1r_2(cos(\theta_1-\theta_2))}=r_1+r_2$
$\Rightarrow\:cos(\theta_1-\theta_2)=1$
$\Rightarrow\:\theta_1-\theta_2=0$
$\Rightarrow\:argz_1-argz_2=0$
answered Jul 12, 2013 by rvidyagovindarajan_1
 

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