A) \[\frac{\pi }{3}+2n\pi ,n\in I\]
B) \[n\pi +\frac{\pi }{6},n\in I\]
C) \[n\pi -\frac{\pi }{3},n\in I\]
D) \[2n\pi -\frac{\pi }{3},n\in I\]
Correct Answer: B
Solution :
We have \[\sqrt{3}+i=(a+ib)(c+id)\] \[\therefore ac-bd=\sqrt{3}\]and \[ad+bc=1\] Now tan-1\[\left( \frac{b}{a} \right)+{{\tan }^{-1}}\left( \frac{d}{c} \right)\] \[\begin{align} & ={{\tan }^{-1}}\left( \frac{\frac{b}{a}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}} \right)={{\tan }^{-1}}\left( \frac{bc+ad}{ac-bd} \right)={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right) \\ & \\ \end{align}\] \[=n\pi +\frac{\pi }{6},n\in I\]
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