Animated Solution for Mathematics - Complex Numbers: If $\omega$ is an imaginary cube root of unity, then $(1+\omega -{\omega }^2{)}^7$ equals

Welcome, students! Today we are tackling a classic algebra problem involving the cube roots of unity. Let's visualize this on the complex plane.

We have our three roots: $1$, $\omega$, and ${\omega }^2$, sitting perfectly on the unit circle, forming an equilateral triangle. These are not just numbers; they are vectors that hold a beautiful, hidden balance.

Before we jump into the calculation, let's recall the two golden rules of these roots: 1. Their sum is always zero: $1+\omega +{\omega }^2=0$. 2. The cyclic property: ${\omega }^3=1$.

These two properties are our main tools. Now look at the equation: $(1+\omega -{\omega }^2{)}^7$. Expanding this directly would be a nightmare, but we can simplify the interior first.

Using our sum property, we can easily isolate $1+\omega$. By shifting ${\omega }^2$ to the other side, we find that:

Geometrically, this is the vector sum of $1$ and $\omega$. Let's substitute this value back into our original expression.

We replace the $(1+\omega )$ part with $-{\omega }^2$. Inside the bracket, we now have:

The whole expression is now reduced to a single term raised to the power of $7$:

When raising a product to a power, the exponent applies to each factor individually. We distribute the power of $7$ to the $-2$ and to the ${\omega }^2$:

We must now simplify ${\omega }^{14}$ using the property ${\omega }^3=1$. We can express the exponent $14$ as $12+2$:

Since ${\omega }^3=1$, this simplifies to:

Putting it all together, our final answer is:

The key takeaway here is to always look for ways to use the sum property to reduce terms before dealing with large exponents.