Animated Solution for Mathematics - Complex Numbers: Let $\alpha$ be a root of the equation $1+x^2+x^4=0$. Then the value of ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$ is equal to:

Imagine you are standing before a massive, intimidating algebraic expression: ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$. It looks like a mountain, but in the world of JEE Advanced, every mountain has a path to the summit.

Our journey begins with the equation $x^4+x^2+1=0$. To reveal its secret, we use the technique of completing the square by adding and subtracting $x^2$:

Suddenly, the expression transforms into $(x^2+1{)}^2-x^2=0$. This is the key that unlocks the door.

With the difference of squares identity, $A^2-B^2=(A-B)(A+B)$, we can split our equation into two elegant quadratic factors:

This gives us two distinct paths: $x^2+x+1=0$ and $x^2-x+1=0$.

If $\alpha$ is a root of the first, it is a complex cube root of unity, meaning ${\alpha }^3=1$. If $\alpha$ is a root of the second, it is a cube root of $-1$, meaning ${\alpha }^3=-1$. This is the fundamental realization that simplifies our massive exponents.

Now, look at our target expression: ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$. Notice that all the exponents are multiples of $3$.

We can rewrite them as:

This is where the magic happens. Whether ${\alpha }^3=1$ or ${\alpha }^3=-1$, the expression collapses into a simple arithmetic problem.

If ${\alpha }^3=1$, we get:

If ${\alpha }^3=-1$, we get:

In both cases, the result is the same. The complexity vanishes, leaving us with the elegant final answer of 1.