Complex Numbers And Quadratic Equations question 256
Question: If $ \omega $ is a complex cube root of unity, then for positive integral value of $ n $ , the product of $ \omega .{{\omega }^{2}}.{{\omega }^{3}}……..{{\omega }^{n}} $ , will be [Roorkee 1991]
Options:
A) $ \frac{1-i\sqrt{3}}{2} $
B) $ -\frac{1-i\sqrt{3}}{2} $
C) 1
D) (b) and (c) both
Show Answer
Answer:
Correct Answer: D
Solution:
The product is given by $ \omega .{{\omega }^{2}}.{{\omega }^{3}}…..{{\omega }^{n}}={{\omega }^{1+2+3+……+n}}={{\omega }^{n(n+1)/2}} $ On putting $ n=1,2,3,….., $ we get $ ={{\omega }^{1(1+1)/2}}=\omega ,{{\omega }^{2(2+1)/2}}={{\omega }^{3}}=1,…..{{\omega }^{4(5)/2}}={{\omega }^{10}}=\omega $ Hence it gives the values 1 and $ \omega $ only.
BETA
