Determinants Matrices Question 55
Question: If $ a=\cos \theta +i\sin \theta , $
$ b=\cos 2\theta -i\sin 2\theta , $
$ c=\cos 3\theta +i\sin 3\theta $ and if , then
Options:
A) $ \theta =2k\pi ,k\in Z $
B) $ \theta =(2k+1)\pi ,k\in Z $
C) $ \theta =(4k+1)\pi ,k\in Z $
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ =(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) $
$ =\frac{1}{2}(a+b+c)[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]=0 $
$ \Rightarrow a+b+c=0 $ or $ a=b=c $ If $ a+b+c=0 $ , we have $ \cos \theta +\cos 2\theta +\cos 3\theta =0 $ and $ \sin \theta -sin2\theta +sin3\theta =0 $ or $ \cos 2\theta (2cos\theta +1)=0 $ and $ \sin 2\theta (1-2cos\theta )=0 $ …(1) Which is not possible as $ \cos 2\theta =0 $ , gives $ \sin 2\theta \ne 0,cos\theta \ne 1/2. $ and $ \cos \theta =-1/2 $ gives $ \sin 2\theta \ne 0, $
$ \cos \theta \ne 1/2. $ therefore, Eq. (1) does not hold simultaneously. Therefore, $ a+b+c\ne 0 $
$ \Rightarrow a=b=c $
$ \therefore {e^{i\theta }}={e^{-2i\theta }}={e^{3i\theta }} $ Which is satisfied only by $ {e^{i\theta }}=1, $ i.e., $ \cos \theta =1,\sin \theta =0; $ so $ \theta =2k\pi ,k\in Z. $
BETA
