Complex Numbers And Quadratic Equations question 680
Question: The equation whose roots are the $ n^{th} $ power of the roots of the equation $ x^{2}-2xcos\theta +1=0 $ is given by
Options:
A) $ x^{2}+2x\cos n\theta +1=0 $
B) $ x^{2}-2x\cos n\theta +1=0 $
C) $ x^{2}-2xsinn\theta +1=0 $
D) $ x^{2}+2xsinn\theta +1=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
The roots of the given equation are $ x=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}=\cos \theta \pm i\sin \theta $ Let $ \alpha =cos\theta +isin\theta \And \beta = cos \theta -i sin \theta $ Then $ {{\alpha }^{n}} = cosn\theta +isin n\theta $ $ {{\beta }^{n}} = cos n\theta - i sin n\theta $ [Using De Moivre Theorem] $ {{\alpha }^{n}}+{{\beta }^{n}}=2cosn\theta and{{\alpha }^{n}}\cdot {{\beta }^{n}}=1 $
$ \therefore $ The required equation is $ x^{2}-2x\cos n\theta +1=0 $
BETA
