A) \[{{60}^{o}}\]
B) \[{{120}^{o}}\]
C) \[{{240}^{o}}\]
D) \[{{300}^{o}}\]
Correct Answer: C
Solution :
If \[z=\frac{1-i\sqrt{3}}{1+i\sqrt{3}}=\frac{(1-i\sqrt{3})(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\] \[=\frac{1-3-2i\sqrt{3}}{1+3}=\frac{-2-2i\sqrt{3}}{4}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}\] Thus \[arg(z)={{\tan }^{-1}}\frac{y}{x}={{\tan }^{-1}}\sqrt{3}=\frac{\pi }{3}={{60}^{o.}}\] Since the complex number lies in III quadrant, therefore \[arg\,(z)\] is \[{{180}^{o}}\] + \[{{60}^{o}}={{240}^{o}}\] Aliter: \[arg\left( \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \right)=arg(1-i\sqrt{3})-arg(1+i\sqrt{3})\] \[=-{{60}^{o}}-{{60}^{o}}=-{{120}^{o}}\]or\[{{240}^{o}}\].
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