Complex Numbers And Quadratic Equations question 225
Question: $ {{[ \frac{1+\cos (\pi /8)+i\sin (\pi /8)}{1+\cos (\pi /8)-i\sin (\pi /8)} ]}^{8}} $ is equal to [RPET 2001]
Options:
A) - 1
B) 0
C) 1
D) 2
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{[ \frac{1+\cos (\pi /8)+i\sin (\pi /8)}{1+\cos (\pi /8)-i\sin (\pi /8)} ]}^{8}} $ $ ={{[ \frac{2{{\cos }^{2}}( \pi /16 )+2i\sin ( \pi /16 )\cos ( \pi /16 )}{2{{\cos }^{2}}( \pi /16 )-2i\sin ( \pi /16 )\cos ( \pi /16 )} ]}^{8}} $ $ =\frac{{{[\cos ( \pi /16 )+i\sin ( \pi /16 )]}^{8}}}{{{[\cos ( \pi /16 )-i\sin ( \pi /16 )]}^{8}}} $ $ ={{[ \cos \frac{\pi }{16}+i\sin \frac{\pi }{16} ]}^{8}}{{[ \cos \frac{\pi }{16}+i\sin \frac{\pi }{16} ]}^{8}} $ $ ={{[\cos (\pi /16)+i\sin (\pi /16)]}^{16}} $ $ =\cos 16( \frac{\pi }{16} )+i\sin 16( \frac{\pi }{16} ) $ = $ \cos \pi =-1 $ .
BETA
