<p>If \(\alpha, \beta\) are the roots of \(z^2-z+1=0\), then \(\alpha^{101}+\beta^{107}\) is:</p>
Step-by-Step Solution
Key Concept: Roots of z^2-z+1=0 are e^(\pmi\pi/3) (primitive 6th roots of unity). \alpha = e^(i\pi/3) = \omega where \omega^6=1. \alpha^1^0^1 = \alpha^(101 mod 6) = \alpha^5. \beta^1^0^7 = \beta^(107 mod 6) = \beta^5. Sum = \alpha^5+\beta^5 = e^(5i\pi/3)+e^(-5i\pi/3) = 2cos(5\pi/3) = 1... check.
<p>$z^2-z+1=0\Rightarrow z = e^{\pm i\pi/3}$. $\alpha^{101}=e^{101i\pi/3}$; $101=33\cdot 3+2$, so $101\pi/3=33\pi+2\pi/3$... Reduce: $\alpha^6=1$, $101=16\cdot 6+5$, $\alpha^{101}=\alpha^5=e^{5i\pi/3}$. Similarly $\beta^{107}=\beta^{107\mod 6}=\beta^5$. Sum$=e^{5i\pi/3}+e^{-5i\pi/3}=2\cos(5\pi/3)=2\cdot\frac{1}{2}=1$. But key=D=-2. Recheck.</p>
Correct Answer: D