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

Answer 1

Answer: (D)

1

Answer 2

The roots of the equation $x^2 - x + 1 = 0$ are $\alpha$ and $\beta$. Using the quadratic formula, the roots are $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$. Let $\alpha = \frac{1 + i\sqrt{3}}{2}$ and $\beta = \frac{1 - i\sqrt{3}}{2}$.

These roots are related to the 6th roots of unity. Specifically, they are $e^{i\pi/3}$ and $e^{-i\pi/3}$. Therefore, $\alpha^6 = 1$ and $\beta^6 = 1$.

To simplify $\alpha^{101}$ and $\beta^{107}$, we consider the exponents modulo 6: $101 \equiv 5 \pmod 6$ $107 \equiv 5 \pmod 6$

So, $\alpha^{101} = \alpha^5$ and $\beta^{107} = \beta^5$.

We know that $\alpha^5 = \alpha^{-1}$ and $\beta^5 = \beta^{-1}$ since $\alpha^6 = 1$ and $\beta^6 = 1$. Also, from the equation $x^2 - x + 1 = 0$, we have $1 = x - x^2$. If $x = \alpha$, then $1 = \alpha - \alpha^2$. If $x = \beta$, then $1 = \beta - \beta^2$.

Alternatively, we can use the property that if $\omega$ is a complex cube root of $-1$, then $\omega^2 - \omega + 1 = 0$. The roots of $x^2 - x + 1 = 0$ are indeed related to cube roots of $-1$. Let's consider $\alpha^5$. Since $\alpha^6 = 1$, $\alpha^5 = \alpha^{-1}$. From $x^2 - x + 1 = 0$, dividing by $x$ (since $x \neq 0$), we get $x - 1 + \frac{1}{x} = 0$, so $\frac{1}{x} = 1 - x$. Thus, $\alpha^{-1} = 1 - \alpha$ and $\beta^{-1} = 1 - \beta$. So, $\alpha^5 = 1 - \alpha$ and $\beta^5 = 1 - \beta$.

Therefore, $\alpha^{101} + \beta^{107} = \alpha^5 + \beta^5 = (1 - \alpha) + (1 - \beta) = 2 - (\alpha + \beta)$.

From Vieta's formulas for the equation $x^2 - x + 1 = 0$, the sum of the roots $\alpha + \beta = -(-1)/1 = 1$.

Substituting this value: $\alpha^{101} + \beta^{107} = 2 - 1 = 1$.