Complex Numbers And Quadratic Equations question 230
Question: $ {{( \frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } )}^{n}} $ = [Kerala (Engg.) 2002]
Options:
A) $ \cos ( \frac{n\pi }{2}-n\theta )+i\sin ( \frac{n\pi }{2}-n\theta ) $
B) $ \cos ( \frac{n\pi }{2}+n\theta )+i\sin ( \frac{n\pi }{2}+n\theta ) $
C) $ \sin ( \frac{n\pi }{2}-n\theta )+i\cos ( \frac{n\pi }{2}-n\theta ) $
D) $ \cos n( \frac{\pi }{2}+2\theta )+i\sin n( \frac{\pi }{2}+2\theta ) $
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Answer:
Correct Answer: A
Solution:
$ {{( \frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } )}^{n}}={{( \frac{1+\cos \alpha +i\sin \alpha }{1+\cos \alpha -i\sin \alpha } )}^{n}} $ $ ( $ where $ \alpha =\frac{\pi }{2}-\theta ) $ $ ={{( \frac{2{{\cos }^{2}}\frac{\alpha }{2}+2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}-2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} )}^{n}} $ $ ={{( \frac{\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}-i\sin \frac{\alpha }{2}} )}^{n}} $ $ ={{( \frac{cis( \frac{\alpha }{2} )}{cis( -\frac{\alpha }{2} )} )}^{n}} $ $ ={{{ cis( \frac{\alpha }{2}+\frac{\alpha }{2} ) }}^{n}}=cis(n\alpha ) $ = $ cisn( \frac{\pi }{2}-\theta )=cis( \frac{n\pi }{2}-n\theta ) $ $ =\cos ( \frac{n\pi }{2}-n\theta )+i\sin ( \frac{n\pi }{2}-n\theta ) $ .
BETA
