Complex Numbers And Quadratic Equations question 283
Question: Let $ {\omega_{n}}=\cos ( \frac{2\pi }{n} )+i\sin ( \frac{2\pi }{n} ),i^{2}=-1 $ , then $ (x+y{\omega_3}+z{\omega_3}^{2}) $ $ (x+y{\omega_3}^{2}+z{\omega_3}) $ is equal to [AMU 2001]
Options:
A) 0
B) $ x^{2}+y^{2}+z^{2} $
C) $ x^{2}+y^{2}+z^{2}-yz-zx-xy $ $ $
D) $ x^{2}+y^{2}+z^{2}+yz+zx+xy $
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Answer:
Correct Answer: C
Solution:
$ {\omega_{n}}=\cos ( \frac{2\pi }{n} )+i\sin ( \frac{2\pi }{n} ) $
$ \Rightarrow {\omega_3}=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+\frac{i\sqrt{3}}{2}=\omega $ and $ \omega _3^{2}={{( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} )}^{2}}=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3} $ $ =-\frac{1}{2}-\frac{i\sqrt{3}}{2}={{\omega }^{2}}. $
$ \therefore (x+y{\omega_3}+z\omega _3^{2})(x+y\omega _3^{2}+z{\omega_3}) $ $ =(x+y\omega +z{{\omega }^{2}})(x+y{{\omega }^{2}}+z\omega ) $ $ =x^{2}+y^{2}+z^{2}-xy-yz-zx $ .
BETA
