Animated Solution for Mathematics - Complex Numbers: If $\alpha ,\beta \in \mathbb{C}$ are the distinct roots, of the equation $x^2-x+1=0$, then ${\alpha }^{101}+{\beta }^{107}$ is equal to :

We begin with the given quadratic equation:

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we substitute $a=1$, $b=-1$, and $c=1$. This yields:

These values represent our roots, $\alpha$ and $\beta$. They are not merely random complex numbers; they are intrinsically linked to the cube roots of unity.

Recall the definition of the complex cube roots of unity, where $\omega =\frac{-1+i\sqrt{3}}{2}$ and ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$. If you examine our roots $\frac{1\pm i\sqrt{3}}{2}$ closely, you will notice they are exactly $-{\omega }^2$ and $-\omega$.

This identification is the critical "Aha!" moment. By recognizing our roots as $-\omega$ and $-{\omega }^2$, we unlock the two fundamental properties of $\omega$:

Now, we address the expression ${\alpha }^{101}+{\beta }^{107}$. Substituting our identified values, we obtain:

Because $101$ and $107$ are odd, the negative signs persist, resulting in:

We now apply the property ${\omega }^3=1$ by dividing the exponents by $3$. For the first term, $101=3\times 33+2$, so ${\omega }^{101}={\omega }^2$. For the second term, $214=3\times 71+1$, so ${\omega }^{214}=\omega$.

Our expression simplifies to:

Recalling our second property, $1+\omega +{\omega }^2=0$, we know that ${\omega }^2+\omega =-1$. Substituting this into our expression, we get:

The massive powers have vanished, leaving behind a simple, beautiful integer. The final result is 1.