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  4. The argument of (1+i√3/1-i√3) is

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Q. The argument of $ \frac{1+i\sqrt{3}}{1-i\sqrt{3}} $ is

J & K CETJ & K CET 2006

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} $