Animated Solution for Mathematics - Complex Numbers: Let $\alpha =\frac{-1+i\sqrt{3}}{2}$ and $\beta =\frac{-1-i\sqrt{3}}{2},i=\sqrt{-1}$. If $(7-7\alpha +9\beta {)}^{20}+(9+7\alpha -7\beta {)}^{20}+(-7+9\alpha +7\beta {)}^{20}+(14+7\alpha +7\beta {)}^{20}=m^{10}$, then $m$ is .........

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of high-power binomial expansions. You see a term raised to the power of 20, and your instinct might be to panic.

But in the world of JEE Advanced, panic is the enemy. Insight is your weapon. Let us embark on this journey to simplify the complex.

Unmasking the Cube Roots

Look at the definitions provided: $\alpha =\frac{-1+i\sqrt{3}}{2}$ and $\beta =\frac{-1-i\sqrt{3}}{2}$. If you have spent time with complex numbers, these should look like old friends.

These are the complex cube roots of unity. Specifically, $\alpha =\omega$ and $\beta ={\omega }^2$.

Why does this matter? Because $\omega$ is not just a number; it is a geometric operator. It obeys two sacred laws:

The moment you replace $\alpha$ and $\beta$ with $\omega$ and ${\omega }^2$, the problem stops being about arithmetic and starts being about symmetry. We are no longer dealing with messy fractions; we are dealing with the elegant structure of the roots of unity.

The Fourth Term - An Anchor in the Storm

Before we tackle the intimidating first three terms, let us look at the fourth term: $(14+7\alpha +7\beta {)}^{20}$. Let us call this $T_4$.

Substituting our new knowledge, we get:

Remember the second law? $\omega +{\omega }^2=-1$. Suddenly, the expression inside the parenthesis becomes $14+7(-1)$, which is simply $7$.

Thus, $T_4=7^{20}$. This is our anchor. We have successfully tamed the most intimidating-looking term by simply recognizing the identity of $\omega$.

The Cyclic Dance

Now, let us define the bases of the first three terms as $z_1$, $z_2$, and $z_3$:

Do you see the cyclic shift in the coefficients? It is almost like a dance. Let us test the relationship between them. What happens if we multiply $z_1$ by $\omega$?

Since ${\omega }^3=1$, this becomes $7\omega -7{\omega }^2+9$, which is exactly $z_2$!

Similarly, if you multiply $z_1$ by ${\omega }^2$, you will find that $z_1{\omega }^2=z_3$. This is the core insight. We have proven that $z_2=z_1\omega$ and $z_3=z_1{\omega }^2$. We are not dealing with three independent terms; we are dealing with one term ($z_1$) and its rotations in the complex plane.

The Grand Collapse

Now, let us bring it all together. The sum of the first three terms is $S=z_1^{20}+z_2^{20}+z_3^{20}$. Substituting our cyclic relationships, we get:

Now, we must reduce the powers of $\omega$. Since ${\omega }^3=1$, we divide the exponents by 3.

For ${\omega }^{20}$, $20=3\times 6+2$, so ${\omega }^{20}={\omega }^2$. For ${\omega }^{40}$, $40=3\times 13+1$, so ${\omega }^{40}=\omega$.

The bracket becomes $(1+{\omega }^2+\omega )$. And what is that? It is zero! The entire sum of the first three terms vanishes into thin air.

The Final Victory

We are left with a beautifully simple equation:

To solve for $m$, we rewrite $7^{20}$ as $(7^2{)}^{10}$, which is $49^{10}$. Comparing $49^{10}=m^{10}$, we find $m=49$.

See? The complexity was a mask. By identifying the underlying structure, we turned a terrifying expression into a simple exercise in properties. Keep this mindset, and no problem will ever be too big for you. You are a JEE warrior—now go forth and conquer.