A) \[4m\pi \]
B) \[\frac{2m\pi }{n(n+1)}\]
C) \[\frac{4m\pi }{n(n+1)}\]
D) \[\frac{m\pi }{n(n+1)}\]
Correct Answer: C
Solution :
We have \[(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )\] ......\[(\cos n\theta +i\sin n\theta )=1\] Þ \[\cos (\theta +2\theta +3\theta +...+n\theta )+i\sin (\theta +2\theta +.+n\theta )=1\] Þ \[\cos \left( \frac{n(n+1)}{2}\theta \right)+i\sin \left( \frac{n(n+1)}{2}\theta \right)=1\] \[\cos \left( \frac{n\,(n+1)}{2}\theta \right)\,=\,1\text{ and }\sin \left( \frac{n(n+1)}{2}\theta \right)\,=0\] Þ \[\frac{n(n+1)}{2}\theta =2m\pi \Rightarrow \theta =\frac{4m\pi }{n(n+1)},\]where \[m\in I.\]You need to login to perform this action.
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