A) \[\frac{\pi }{3}\]
B) \[-\frac{\pi }{3}\]
C) \[\frac{2\pi }{3}\]
D) \[-\frac{2\pi }{3}\]
Correct Answer: C
Solution :
Here \[-1+\sqrt{-3}=r{{e}^{i\theta }}\]Þ \[-1+i\sqrt{3}=r{{e}^{i\theta }}\] \[=r\cos \theta +ir\sin \theta \] Equating real and imaginary parts, we get \[r\cos \theta =-1\]and \[r\sin \theta =\sqrt{3}\] Hence\[\tan \theta =-\sqrt{3}\,\,\Rightarrow \tan \theta =\tan \frac{2\pi }{3}\].Hence \[\theta =\frac{2\pi }{3}\].
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