A) \[\frac{2\pi }{3}\]
B) \[\frac{\pi }{3}\]
C) \[-\frac{\pi }{3}\]
D) \[-\frac{2\pi }{3}\]
Correct Answer: D
Solution :
Let \[z=-1-i\sqrt{3}\] then\[\alpha ={{\tan }^{-1}}\left| \,\frac{b}{a}\, \right|={{\tan }^{-1}}\left| \,-\frac{\sqrt{3}}{1}\, \right|=\frac{\pi }{3}\] Clearly, z is in III quadrant. Therefore argument \[\theta =-(\pi -\alpha )=-(\pi -\pi /3)=\frac{-2\pi }{3}\].
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