Complex Numbers And Quadratic Equations question 451
Question: The expression $ {{[ \frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}} ]}^{8}}= $
Options:
A) 1
B) -1
C) i
D) -i
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ Let\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}=z $
$ \Rightarrow {{[ \frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}} ]}^{8}} $ $ ={{( \frac{1+z}{1+\frac{1}{z}} )}^{8}} $ $ =z^{8}={{( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} )}^{8}} $ $ ={{( \cos ( \frac{\pi }{2}-\frac{\pi }{8} )+i\sin ( \frac{\pi }{2}-\frac{\pi }{8} ) )}^{8}} $ $ ={{( \cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8} )}^{8}} $ $ =\cos 3\pi =-1 $
BETA
