The problem asks us to find the value of the product $(1 + \omega)(1 + \omega^2)(1 + \omega^3)\dots(1 + \omega^{3n})$, where $\omega$ is an imaginary cube root of unity.

Step 1: Understanding Cube Roots of Unity

The cube roots of unity are the solutions to the equation $x^3 = 1$. These roots are $1$, $\omega$, and $\omega^2$, where:

• $\omega = e^{2\pi i / 3}$
• $\omega^2 = e^{4\pi i / 3}$

The roots satisfy the following properties:

• $\omega^3 = 1$
• $1 + \omega + \omega^2 = 0$

Step 2: Simplifying the Expression

The product given in the problem is: $(1 + \omega)(1 + \omega^2)(1 + \omega^3)\dots(1 + \omega^{3n})$

Since $\omega^3 = 1$, the powers of $\omega$ repeat every three terms. Therefore, the terms inside the product can be grouped as: $(1 + \omega)(1 + \omega^2)(1 + 1) \times (1 + \omega)(1 + \omega^2)(1 + 1) \times \dots \text{ (n times)}$

Each group consists of three terms: $(1 + \omega)(1 + \omega^2)(1 + 1)$

We can simplify the product of each group: $(1 + \omega)(1 + \omega^2)(1 + 1)$

Step 3: Calculating the Value of Each Group

Let's calculate $(1 + \omega)(1 + \omega^2)(1 + 1)$:

• $1 + 1 = 2$
• $(1 + \omega)(1 + \omega^2)$

Using the identity $1 + \omega + \omega^2 = 0$: $1 + \omega = -\omega^2 \quad \text{and} \quad 1 + \omega^2 = -\omega$ So, $(1 + \omega)(1 + \omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$

Thus, each group simplifies to: $(1 + \omega)(1 + \omega^2)(1 + 1) = 1 \times 2 = 2$

Step 4: Calculating the Entire Product

Since there are $n$ such groups, the entire product is: $2^n$

Final Answer:

The value of the product is $2^n$, corresponding to option b. $2^n$.

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Do you want any more details or have any questions? Here are some related questions you might find interesting:

1. How can the properties of roots of unity be applied in other polynomial equations?
2. What are the geometric interpretations of the cube roots of unity?
3. How does this problem change if we consider fourth roots of unity?
4. Can this problem be extended to a general $k$-th root of unity?
5. How would you prove the identity $1 + \omega + \omega^2 = 0$ without using the explicit form of $\omega$?

Tip: Understanding the properties of roots of unity is crucial when dealing with complex numbers, as they often simplify polynomial and trigonometric expressions.