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

Welcome, future engineers! Today, we are going to peel back the layers of a problem that might look like a daunting algebraic mess at first glance, but is actually a beautiful exercise in pattern recognition. We are dealing with the cube roots of unity, specifically $\omega$.

Imagine the complex plane. You have the number $1$, $\omega$, and ${\omega }^2$ sitting on the unit circle, perfectly spaced at $120^{\circ }$ intervals. They form an equilateral triangle, a symbol of perfect balance. Our task is to evaluate $(1+\omega {)}^7$ and map it to the form $A+B\omega$.

Before we touch the exponent, we need our master key. The sum of the cube roots of unity is always zero:

This isn't just an equation; it's a geometric statement. If you place these three vectors head-to-tail, they form a closed loop, returning you to the origin. This identity is our secret weapon.

Look at the expression $(1+\omega )$. Using our master key, we can rearrange it:

Suddenly, the binomial $(1+\omega )$ is replaced by a single term, $-{\omega }^2$. This is the kind of simplification that turns a ten-minute problem into a thirty-second victory.

Handling $(-{\omega }^2{)}^7$

Now, we substitute this back into our original expression:

Here is where the trap lies. We are raising a negative term to an odd power, $7$. Remember, a negative number raised to an odd power remains negative.

So, $(-{\omega }^2{)}^7$ becomes $-{\omega }^{2\cdot 7}$, which is $-{\omega }^{14}$. Do not let the negative sign vanish! It is a common mistake, but you are better than that.

Now we face $-{\omega }^{14}$. How do we handle such a high power? We invoke the second fundamental property: ${\omega }^3=1$.

This means the powers of $\omega$ are cyclic. Every time you hit a multiple of $3$ in the exponent, it resets to $1$. Let's divide $14$ by $3$:

This means ${\omega }^{14}=({\omega }^3{)}^4\cdot {\omega }^2$. Since ${\omega }^3=1$, this simplifies to $1^4\cdot {\omega }^2={\omega }^2$. Our expression $-{\omega }^{14}$ has now collapsed into $-{\omega }^2$.

We are almost there. We have $-{\omega }^2$, but the problem asks for the form $A+B\omega$. We need to get rid of the square.

We return to our master key one last time: $1+\omega +{\omega }^2=0$. This tells us that:

Look at that! The entire expression $(1+\omega {)}^7$ has simplified back to $1+\omega$. By comparing $1+\omega$ with $A+B\omega$, it is clear that $A=1$ and $B=1$.

The final result is the pair $(A,B)=(1,1)$. You see? By using the properties of the roots rather than brute-force expansion, we navigated through the complexity with ease. Keep this mindset for your JEE journey: look for the symmetry, use the identities, and the path will reveal itself.