Complex Numbers And Quadratic Equations question 216
The value of expression $ ( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} ) $ $ ( \cos \frac{\pi }{2^{2}}+i\sin \frac{\pi }{2^{2}} ) $ ……..to $ n $ is [Kurukshetra CEE 1998]
Options:
A) $ -1 $
B) $ 1 $
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Answer:
Correct Answer: A
Solution:
$ ( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} )( \cos \frac{\pi }{2^{2}}+i\sin \frac{\pi }{2^{2}} )….. $ to $ \infty $ $ =\cos ( \frac{\pi }{2}+\frac{\pi }{2^{2}}+….. )+i\sin ( \frac{\pi }{2}+\frac{\pi }{2^{2}}+…. ) $ $ =\cos \frac{\pi }{2}( 1+\frac{1}{2}+\frac{1}{2^{3}}+….. )+i\sin \frac{\pi }{2}( 1+\frac{1}{2}+\frac{1}{2^{3}}+….. ) $ $ =\cos \frac{\pi }{2}( \frac{1}{1-\frac{1}{2}} )+i\sin \frac{\pi }{2}( \frac{1}{1-\frac{1}{2}} )=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2} =-1 $
BETA
