Complex Numbers And Quadratic Equations question 88
Question: If $ z=\frac{1-i\sqrt{3}}{1+i\sqrt{3}}, $ then $ arg(z)= $ [Roorkee 1990; UPSEAT 2004]
Options:
A) $ 60^{o} $
B) $ 120^{o} $
C) $ 240^{o} $
D) $ 300^{o} $
Show Answer
Answer:
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( \frac{1-i\sqrt{3}}{1+i\sqrt{3}} )=arg(1-i\sqrt{3})-arg(1+i\sqrt{3}) $ $ =-60^{o}-60^{o}=-120^{o} $ or $ 240^{o} $ .
BETA
