A) \[60{}^\circ \]
B) \[120{}^\circ \]
C) \[-120{}^\circ \]
D) \[-135{}^\circ \]
Correct Answer: D
Solution :
\[\frac{1-i\sqrt{3}}{1+i\sqrt{3}}=\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}\] \[=\frac{{{(1-i\sqrt{3})}^{2}}}{1-{{(i\sqrt{3})}^{2}}}=\frac{1-3-2\sqrt{3}i}{1+3}\] \[=\frac{-2-2\sqrt{3}i}{4}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\] \[\therefore \] \[\arg =-\pi +{{\tan }^{-1}}(1)\] \[=-\pi +\frac{\pi }{4}=\frac{-3\pi }{4}\]
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