Animated Solution for Mathematics - Complex Numbers: If $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^4+x^3+x^2+x+1=0$, then ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$ is equal to.

Imagine you are standing before the equation $x^4+x^3+x^2+x+1=0$. At first glance, it looks like a standard, albeit slightly intimidating, quartic equation.

But for the trained eye of a JEE aspirant, this is not just a polynomial; it is a beautiful, symmetric structure waiting to be unlocked. This equation is a geometric progression with five terms. Recognizing this is your first step toward mastery.

We want to find the sum of the $2021^{st}$ powers of its roots. If we try to solve for the roots directly using the quartic formula, we will be lost in a sea of algebra.

Instead, let us use a classic, elegant trick. We multiply both sides of the equation by $(x-1)$. We must be careful: we can only do this if $x eq 1$.

If we plug $x=1$ into the original equation, we get $5=0$, which is clearly false. Thus, $x$ is not $1$, and we are free to proceed. The product $(x-1)(x^4+x^3+x^2+x+1)$ collapses perfectly into:

This is the magic of the geometric series!

Now, we have $x^5=1$. The solutions to this equation are the $5^{th}$ roots of unity. These roots are not just random numbers; they are points on the complex plane that form a perfect regular pentagon inscribed in the unit circle.

One of these roots is the real number $1$. The other four—$\alpha ,\beta ,\gamma ,\delta$—are the roots of our original quartic equation.

Because the sum of all $n^{th}$ roots of unity is always zero, we know that $1+\alpha +\beta +\gamma +\delta =0$. This immediately tells us that the sum of our four roots is:

This is a powerful piece of information that we will hold onto.

Now, we face the target expression: ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$. A power of $2021$ seems impossible to compute, but remember the property of the $5^{th}$ roots of unity: ${\alpha }^5=1$.

This means the powers of $\alpha$ cycle every five steps. To simplify ${\alpha }^{2021}$, we divide $2021$ by $5$. We find that $2021=5\times 404+1$.

Therefore, ${\alpha }^{2021}=({\alpha }^5{)}^{404}\cdot {\alpha }^1$. Since ${\alpha }^5=1$, this simplifies beautifully to $1^{404}\cdot \alpha$, which is just $\alpha$.

The same logic applies to $\beta ,\gamma ,$ and $\delta$. Each term in our sum reduces to its base root.

Our complex expression ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$ has now transformed into the simple sum $\alpha +\beta +\gamma +\delta$. We already know this sum is $-1$.

What started as a terrifying problem involving high-degree polynomials and massive exponents has been reduced to a simple, elegant result. The final answer is -1.

This is the beauty of complex numbers and the roots of unity—they turn chaos into order. Keep this perspective, and no JEE problem will ever be able to intimidate you again.