# Find the modulus and the arguments of each of the complex numbers.Q: 1       $z=-1-i\sqrt{3}$

G Gautam harsolia

Given the problem is
$z=-1-i\sqrt{3}$
Now, let
$r\cos \theta = - 1 \ \ \ and \ \ \ r\sin \theta = -\sqrt3$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-\sqrt3)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+3$
$r^2 =4$
$r= 2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta = -1 \ \ \ and \ \ \ 2\sin \theta = -\sqrt3$
$\cos \theta = -\frac{1}{2} \ \ \ and \ \ \ \sin \theta = -\frac{\sqrt3}{2}$
Since, both the values of   $\cos \theta \ and \ \sin \theta$  is negative and we know that they are negative in III quadrant
Therefore,
Argument = $-\left ( \pi - \frac{\pi}{3} \right )= - \frac{2\pi}{3}$
Therefore, the argument  is

$- \frac{2\pi}{3}$

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