The problem asks for the value of $\omega^{100} + \omega^{101} + \omega^{102}$, where $\omega$ is a complex cube root of unity. Let’s solve this step by step.

Properties of Cube Roots of Unity

The complex cube roots of unity are the solutions to the equation:

$\omega^3 = 1$

These roots are:

• $1$ (real cube root)
• $\omega$, where $\omega = e^{2\pi i / 3}$
• $\omega^2 = e^{4\pi i / 3}$

These roots satisfy the following relations: $1 + \omega + \omega^2 = 0$

Simplifying Powers of $\omega$

Since $\omega^3 = 1$, powers of $\omega$ repeat every 3 steps. Therefore:

• $\omega^{100} = \omega^{100 \mod 3} = \omega^1 = \omega$
• $\omega^{101} = \omega^{101 \mod 3} = \omega^2$
• $\omega^{102} = \omega^{102 \mod 3} = \omega^0 = 1$

Now, substitute these values into the expression: $\omega^{100} + \omega^{101} + \omega^{102} = \omega + \omega^2 + 1$

Using the property $\omega + \omega^2 + 1 = 0$, we find: $\omega^{100} + \omega^{101} + \omega^{102} = 0$

Final Answer:

The value of $\omega^{100} + \omega^{101} + \omega^{102}$ is $0$, which corresponds to option c).

Would you like more details or have any other questions?

Here are 5 related questions to expand on this concept:

1. What are the roots of unity for other powers, such as the 4th or 5th roots of unity?
2. How can the properties of cube roots of unity be applied to solve equations in number theory?
3. Can you derive the general expression for the $n$-th roots of unity?
4. How are roots of unity represented geometrically in the complex plane?
5. How does the symmetry of roots of unity help simplify polynomial equations?

Tip: The property $1 + \omega + \omega^2 = 0$ is a key identity when working with cube roots of unity, and it can greatly simplify many complex number problems.