Complex Numbers And Quadratic Equations question 213
Question: If $ (\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )…….. $ $ (\cos n\theta +i\sin n\theta )=1 $ , then the value of $ \theta $ is[Karnataka CET 1992; Kurukshetra CEE 2002]
Options:
A) $ 4m\pi $
B) $ \frac{2m\pi }{n(n+1)} $
C) $ \frac{4m\pi }{n(n+1)} $
D) $ \frac{m\pi }{n(n+1)} $
Show Answer
Answer:
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 ( \frac{n(n+1)}{2}\theta )+i\sin ( \frac{n(n+1)}{2}\theta )=1 $ $ \cos ( \frac{n(n+1)}{2}\theta )=1\text{ and }\sin ( \frac{n(n+1)}{2}\theta )=0 $
Þ $ \frac{n(n+1)}{2}\theta =2m\pi \Rightarrow \theta =\frac{4m\pi }{n(n+1)}, $ where $ m\in I. $
BETA
