A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
\[{{x}^{n}}=1=(\cos 0+i\sin {{0}^{o}})=\cos 2r\pi +i\sin 2r\pi ={{e}^{i2r\pi }}\] Þ \[x={{e}^{i(2r\pi /n)}},r=0,\,1,\,2\]??.. Obviously the roots are\[1,{{e}^{2\pi i/n}},{{e}^{4\pi i/n}}\]...... which are obviously in G.P. with common ratio\[{{e}^{2\pi i/n}}\].
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