question_answer

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

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

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 \[\theta =2k\pi ,k\in Z.\]