Animated Solution for Mathematics - Complex Numbers: If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then ${\alpha }^{2009}+{\beta }^{2009}=$

Welcome, fellow traveler on the JEE journey. Today, we are going to tackle a problem that, at first glance, looks like a brute-force nightmare. You see an equation, $x^2-x+1=0$, and you are asked to find the sum of its roots raised to the power of $2009$.

Your instinct might be to panic. How on earth are you supposed to calculate ${\alpha }^{2009}$? Are you going to multiply the root by itself two thousand times?

Of course not. In the world of JEE Advanced, we don't calculate; we observe, we simplify, and we conquer.

The Quadratic Trap

Let us start by looking at the equation $x^2-x+1=0$. If you were to blindly apply the quadratic formula,

you would find the roots to be $x=\frac{1\pm i\sqrt{3}}{2}$.

Now, you could stop here. You could try to use De Moivre's Theorem, converting these into polar form: $r(\cos \theta +i\sin \theta )$. You would find the modulus $r=1$ and the argument $\theta =\frac{\pi }{3}$.

Then, raising this to the power of $2009$ would involve calculating $\cos (\frac{2009\pi }{3})+i\sin (\frac{2009\pi }{3})$. It is a valid path, but it is a long, winding road filled with potential for arithmetic errors. Is there a better way? Always.

The Bridge to Unity

This is where the true JEE aspirant distinguishes themselves. We look at $x^2-x+1=0$ and we see a ghost. We see the ghost of the identity $x^3+1=(x+1)(x^2-x+1)$.

If $x^2-x+1=0$, then $x^3+1=0$, which implies $x^3=-1$. This is the key! The roots of our equation are not just random complex numbers; they are related to the cube roots of unity.

Recall that the cube roots of unity, denoted by $\omega$ and ${\omega }^2$, satisfy the equation $x^2+x+1=0$. Our equation is slightly different, but the symmetry is identical. By comparing our roots $x=\frac{1\pm i\sqrt{3}}{2}$ with the standard cube roots of unity $\omega =\frac{-1+i\sqrt{3}}{2}$ and ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$, we realize a beautiful connection: our roots $\alpha$ and $\beta$ are simply $-{\omega }^2$ and $-\omega$.

The Power of Reduction

Now, watch the magic happen. We need to evaluate ${\alpha }^{2009}+{\beta }^{2009}$. Substituting our new expressions, we get:

Since $2009$ is an odd number, the negative sign survives. We can factor it out:

This is where the cyclic nature of $\omega$ saves us. We know that ${\omega }^3=1$. This means that any power of $\omega$ is simply $\omega$ raised to the remainder of the exponent when divided by $3$.

Let us perform the division:

For $4018$: $4018=3\times 1339+1$. Thus, ${\omega }^{4018}={\omega }^1=\omega$.

For $2009$: $2009=3\times 669+2$. Thus, ${\omega }^{2009}={\omega }^2$.

Our expression has now collapsed from a terrifying power of $2009$ into a simple, elegant sum:

The Final Collapse

We are at the finish line. We recall the fundamental property of the cube roots of unity: $1+\omega +{\omega }^2=0$. This implies that $\omega +{\omega }^2=-1$.

Substituting this into our expression, we get:

And there it is. The final answer is 1.

Look at what we just did. We took a problem that seemed to require massive computation and reduced it to a simple identity. This is the essence of physics and mathematics in the JEE.

It is rarely about how fast you can calculate; it is about how clearly you can see the underlying structure. Whenever you encounter high powers in a complex number problem, do not reach for the calculator. Reach for the properties of unity.

Look for the cyclic nature. Look for the symmetry. You have the tools; you just need to trust your intuition. Keep practicing, keep observing, and keep falling in love with the elegance of the solution. You are doing great.