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If $z_1$ and $z_2$ are two non zero complex numbers such that $\mid z_1+z_2\mid=\mid z_1\mid+\mid z_2\mid$ then $arg(z_1)-arg(z_2)$ is equal to

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

1 Answer

$z_1=r_1(\cos \theta_1+i\sin \theta_1)$
$z_2=r_2(\cos \theta_2+i\sin \theta_2)$
$|z_1+z_2|^2=r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$
$\Rightarrow (r_1+r_2)^2+2r_1r_2[\cos(\theta_1-\theta_2)-1]$
Now $|z_1+z_2|=|z_1|+|z_2|$
$\iff \cos(\theta_1-\theta_2)=1$
$\iff \theta_1-\theta_2=0$
Hence (D) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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