Complex Numbers And Quadratic Equations question 207
Question: $ \frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}} $ is equal to [MNR 1985; UPSEAT 2000]
Options:
A) $ \cos \theta -i\sin \theta $
B) $ \cos 9\theta -i\sin 9\theta $
C) $ \sin \theta -i\cos \theta $
D) $ \sin 9\theta -i\cos 9\theta $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}=\frac{{{(\cos +i\sin \theta )}^{4}}}{i^{5}{{( \frac{1}{i}\sin \theta +\cos \theta )}^{5}}} $ $ =\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{i{{(\cos \theta -i\sin \theta )}^{5}}}=\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{i{{(\cos \theta +i\sin \theta )}^{-5}}} $ (By property) $ =\frac{1}{i}{{(\cos \theta +i\sin \theta )}^{9}}=\sin 9\theta -i\cos 9\theta $ .
BETA
