Question

If \(\alpha,\beta\in C\) are the distinct roots of the equation \(x^2-x+1=0\), then \(\alpha^{101}+\beta^{107}\) is equal to :

• A. 2
• B. -1
• C. 0
• D. 1

Hint:

The roots are \(e^{\pm i\pi/3}\), so \(\alpha^6=\beta^6=1\). Reduce the exponents modulo 6.

Answer 1

The Correct Option is D

\(x^2-x+1=0\Rightarrow x=\frac{1\pm i\sqrt3}{2}=e^{\pm i\pi/3}\). Let \(\alpha=e^{i\pi/3}\), \(\beta=e^{-i\pi/3}\).
\(\alpha^6=\beta^6=1\).
\(101=6\cdot16+5\Rightarrow\alpha^{101}=\alpha^5=e^{5i\pi/3}\).
\(107=6\cdot17+5\Rightarrow\beta^{107}=\beta^5=e^{-5i\pi/3}\).
Sum \(=2\cos\frac{5\pi}{3}=2\cdot\frac12=1\).
Answer: (4) 1