A) \[-1+i\sqrt{3}\] or \[1-i\sqrt{3}\]
B) \[1\pm i\sqrt{3}\]
C) \[\sqrt{3}-i\] or \[1-i\sqrt{3}\]
D) \[-1\pm i\sqrt{3}\]
Correct Answer: A
Solution :
Let \[z=\sqrt{3}+i\] \[\therefore \] \[\arg (z)={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\] \[={{30}^{o}}\] For making a right angled triangle OPQ, point Q either in II nd quadrant or IVth quadrant. If the point Q is in IInd quadrant, then we take \[\theta ={{120}^{o}}\] \[\therefore \] \[\tan {{120}^{o}}=-\cot {{30}^{o}}\] \[=\frac{\sqrt{3}}{-1}\] \[\therefore \] Point Q is \[(-1,\sqrt{3)}\] and if the point Q is in IVth quadrant, then we take \[\theta =-{{60}^{o}}\] \[\therefore \] \[\tan (-{{60}^{o}})=-\tan {{60}^{o}}\] \[=-\frac{1}{\sqrt{3}}\] \[\therefore \] Point Q is \[(1,-\sqrt{3})\] Hence, option [a] is correct.You need to login to perform this action.
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