Animated Solution for Mathematics - Complex Numbers: If $\omega (\ne 1)$ is a cube root of unity and $(1+\omega {)}^7=A+B\omega$ then $A$ and $B$ are respectively

Imagine you are standing on the complex plane. You are looking at the unit circle, and suddenly, three points appear, perfectly spaced at $120^{\circ }$ from each other. These are the cube roots of unity: $1$, $\omega$, and ${\omega }^2$.

They are not just numbers; they are the building blocks of rotational symmetry in the complex world. When you face a problem like $(1+\omega {)}^7=A+B\omega$, your first instinct might be to expand it using the Binomial Theorem.

Resist that urge! In the JEE, the path of least resistance is usually the path of deepest insight. Let us embark on a journey to solve this not by brute force, but by understanding the geometric soul of these numbers.

The most profound property of these roots is that they sum to zero:

Geometrically, this means if you place these three vectors head-to-tail, you end up exactly where you started—at the origin. This is our master key.

Look at the term inside our parenthesis: $1+\omega$. If we rearrange our identity, we see that $1+\omega =-{\omega }^2$.

Suddenly, the expression $(1+\omega {)}^7$ transforms into $(-{\omega }^2{)}^7$. We have replaced a binomial with a single term. This is the beauty of complex algebra; we simplify by substitution, not by expansion.

Now, we face $(-{\omega }^2{)}^7$. Do not let the negative sign or the exponent intimidate you. We apply the power rule:

Since $7$ is an odd number, $(-1{)}^7$ remains $-1$. Now, we are left with $-{\omega }^{14}$.

Here is where the second property of unity roots comes into play: ${\omega }^3=1$. This is the cyclic nature of $\omega$. Every time you multiply by ${\omega }^3$, you are essentially multiplying by $1$. It is a loop.

To reduce ${\omega }^{14}$, we divide the exponent $14$ by $3$. We find that $14=3\times 4+2$.

This means ${\omega }^{14}=({\omega }^3{)}^4\cdot {\omega }^2$. Since ${\omega }^3=1$, the term $({\omega }^3{)}^4$ becomes $1^4$, which is just $1$. We are left with $-{\omega }^2$.

We have arrived at $-{\omega }^2$. But wait, the question demands the answer in the form $A+B\omega$. We are currently at a quadratic term, and we need a linear one.

We return to our trusty identity: $1+\omega +{\omega }^2=0$. If we isolate the $-{\omega }^2$ term, we see that:

This is a beautiful moment of closure. We substitute this back into our expression, and we get $1+\omega$.

Now, compare this to the required form $A+B\omega$. It is immediate: $A=1$ and $B=1$.

You have successfully navigated the trap, utilized the geometric properties, and arrived at the solution with elegance. Remember, in mathematics, the most complex-looking problems often yield to the simplest truths.