Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\alpha$ and $\beta$ be the roots of $x^{2}+x+1=0 .$ If $n$ be a positive integer, then $\alpha^{n}+\beta^{n}$ is
Options:
Solution:
2143 Upvotes
Verified Answer
The correct answer is:
$2 \cos \frac{2 n \pi}{3}$
We have, $x^{2}+x+1=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2}$
$\Rightarrow \quad \alpha=\frac{-1+\sqrt{3} i}{2}$ and $\beta=\frac{-1-\sqrt{3} i}{2}$
or $\quad \alpha=e^{\frac{i2\pi}{3}}$ and $\beta=e^{\frac{-2\pi i}{3}}$
$\therefore$
$\alpha^{n}+\beta^{n}=e^{\frac{2 n i t}{3}}+e^{\frac{-2\pi i}{3}}$
$=2\left(\frac{e^{\frac{2n\pi i}{3}}+e^{\frac{-2n\pi i}{3}}}{2}\right)$
$=2 \cos \left(\frac{2 n \pi}{3}\right)$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2}$
$\Rightarrow \quad \alpha=\frac{-1+\sqrt{3} i}{2}$ and $\beta=\frac{-1-\sqrt{3} i}{2}$
or $\quad \alpha=e^{\frac{i2\pi}{3}}$ and $\beta=e^{\frac{-2\pi i}{3}}$
$\therefore$
$\alpha^{n}+\beta^{n}=e^{\frac{2 n i t}{3}}+e^{\frac{-2\pi i}{3}}$
$=2\left(\frac{e^{\frac{2n\pi i}{3}}+e^{\frac{-2n\pi i}{3}}}{2}\right)$
$=2 \cos \left(\frac{2 n \pi}{3}\right)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.