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

Q: 2     z=-\sqrt{3}+i

Answers (1)

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Given the problem is
z=-\sqrt{3}+i
Now, let 
r\cos \theta = - \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1
Square and add both the sides 
r^2(\cos^2\theta +\sin^2\theta)= (-\sqrt3)^2+(1)^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 = -\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1
\cos \theta = -\frac{\sqrt3}{2} \ \ \ and \ \ \ \sin \theta =\frac{1}{2}
Since values of   \cos \theta  is negative and  value \sin \theta is positive and  we know that this is the case in  II quadrant
Therefore,
Argument = \left ( \pi - \frac{\pi}{6} \right )= \frac{5\pi}{6}
Therefore, the argument  is

 \frac{5\pi}{6}

Posted by

Gautam harsolia

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