Find the modulus and the arguments of each of the complex numbers.

Q: 1       z=-1-i\sqrt{3}

Answers (1)
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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