Complex Numbers And Quadratic Equations question 837
Question: If $ z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}, $ then $ {{(\bar{z})}^{100}} $ lies in [AMU 1999]
Options:
A) I quadrant
B) II quadrant
C) III quadrant
D) IV quadrant
Show Answer
Answer:
Correct Answer: C
Solution:
$ z=\frac{1+i\sqrt{3}}{\sqrt{3}+i} $
$ \Rightarrow z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i} $
$ \Rightarrow $ $ z=\frac{\sqrt{3}+3i-i+\sqrt{3}}{3+1} $ $ =\frac{2(\sqrt{3}+i)}{4} $
$ \Rightarrow $ $ z=\frac{\sqrt{3}+i}{2}=[ \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} ] $ Now $ \bar{z}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6} $
Þ $ {{(\bar{z})}^{100}}={{[ \cos \frac{\pi }{6}-i\sin \frac{\pi }{6} ]}^{100}} $
Þ $ {{(\bar{z})}^{100}}=\cos \frac{50\pi }{3}-i\sin \frac{50\pi }{3} $ $ =\cos \frac{2\pi }{3}-i\sin \frac{2\pi }{3} $ $ {{(\bar{z})}^{100}} $ lies in III quadrant.
BETA
