A) \[2\pi /3\]
B) \[\pi /3\]
C) \[-\pi /3\]
D) \[-2\pi /3\]
Correct Answer: A
Solution :
\[\frac{1+i\sqrt{3}}{1-i\sqrt{3}}=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}\] \[=\frac{{{(1+i\sqrt{3})}^{2}}}{1+3}\] \[=\frac{1-3+2i\sqrt{3}}{4}\] \[=-\frac{1}{2}+\frac{i\sqrt{3}}{2}\] Here; \[x=-\frac{1}{2},\,y=\frac{\sqrt{3}}{2}\] \[\therefore \] \[\tan \theta =\frac{y}{x}=\frac{\sqrt{3}}{2}\times \frac{2}{-1}=-\sqrt{3}=-\tan \frac{\pi }{3}\] \[\Rightarrow \] \[\tan \theta =\tan \left( \pi -\frac{\pi }{3} \right)\] \[\Rightarrow \] \[\theta =\frac{2\pi }{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
