Let's go through each part of the problem step-by-step.

Problem Overview

We have the equation: $z^7 - 1 = 0$ This equation is a 7th root of unity, so it has 7 roots, given by: $z_k = e^{2\pi i k / 7} \quad \text{for } k = 0, 1, 2, \ldots, 6.$

Let $w = e^{2\pi i / 7}$, which is one of the roots with an argument between $0$ and $\frac{\pi}{3}$, as required.

Part (a): Find $w$ in Exponential Form

We have: $w = e^{2\pi i / 7}.$ Thus, $w$ in exponential form is simply: $w = e^{2\pi i / 7}.$

Part (b): Show that $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$

The sum $1 + w + w^2 + w^3 + w^4 + w^5 + w^6$ represents the sum of all the 7th roots of unity. By a known property of roots of unity, the sum of all distinct 7th roots of unity is zero. This can be shown as follows:

Since $w$ is a root of $z^7 - 1 = 0$, we know: $w^7 = 1.$ This implies that $1, w, w^2, w^3, w^4, w^5, w^6$ are all distinct roots of the equation $z^7 = 1$.

The sum of all roots of unity (for any $n$-th roots) is always zero, so: $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0.$

Part (c): Show that $w^2 + w^5 = 2 \cos \frac{4\pi}{7}$

To prove this, we need to express $w^2 + w^5$ in terms of cosine. Note that: $w^5 = e^{2\pi i \cdot 5 / 7} \quad \text{and} \quad w^2 = e^{2\pi i \cdot 2 / 7}.$ Thus, $w^2 + w^5 = e^{2\pi i \cdot 2 / 7} + e^{2\pi i \cdot 5 / 7}.$

Using Euler’s formula, $e^{i\theta} + e^{-i\theta} = 2 \cos \theta$, we can express the sum as: $w^2 + w^5 = 2 \cos \left( \frac{4\pi}{7} \right).$

Part (d): Deduce that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$

To show this, we use the property that the real parts of the non-trivial 7th roots of unity sum up to $-\frac{1}{2}$.

The cosines of the arguments of the 7th roots are: $\cos \frac{2\pi}{7}, \quad \cos \frac{4\pi}{7}, \quad \cos \frac{6\pi}{7}, \quad \cos \frac{8\pi}{7} = -\cos \frac{6\pi}{7}, \quad \cos \frac{10\pi}{7} = -\cos \frac{4\pi}{7}, \quad \cos \frac{12\pi}{7} = -\cos \frac{2\pi}{7}.$

Thus, we see that: $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{10\pi}{7} + \cos \frac{12\pi}{7} = 0.$ Grouping terms symmetrically gives: $2\left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} \right) = -1,$ so: $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}.$

Summary of Answers

1. (a) $w = e^{2\pi i / 7}$.
2. (b) $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0$.
3. (c) $w^2 + w^5 = 2 \cos \frac{4\pi}{7}$.
4. (d) $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$.

Would you like further explanations or more details on any of these steps?

Here are 5 related questions to deepen your understanding:

1. How would you derive the general formula for the sum of $n$-th roots of unity?
2. Can you show why the real parts of roots of unity for odd $n$ sum to $-\frac{1}{2}$?
3. How do we use the geometric interpretation of roots of unity to understand their symmetric properties?
4. Why does Euler's formula help simplify expressions involving roots of unity?
5. How does this problem relate to Fourier analysis or trigonometric identities?

Tip: Remember that the symmetry of roots of unity often helps simplify complex expressions involving sums and products of these roots.