The equation to solve is:

$z^3 - 1 = 0$

Step 1: Rewrite the equation

Rewriting the equation, we have:

$z^3 = 1$

This represents the complex cube roots of unity.

Step 2: Represent 1 in polar form

The number $1$ can be represented in polar form as:

$1 = e^{i 2\pi k}, \quad k \in \mathbb{Z}$

Here, $k$ represents the possible angles as multiples of $2\pi$. For cube roots, we divide the angle $2\pi k$ by $3$.

Step 3: General solution

The cube roots of unity are given by:

$z = e^{i \frac{2\pi k}{3}}, \quad k = 0, 1, 2$

Step 4: Write each solution explicitly

1. For $k = 0$: $z_0 = e^{i \frac{2\pi \cdot 0}{3}} = e^{i 0} = 1$

2. For $k = 1$: $z_1 = e^{i \frac{2\pi \cdot 1}{3}} = e^{i \frac{2\pi}{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$

3. For $k = 2$: $z_2 = e^{i \frac{2\pi \cdot 2}{3}} = e^{i \frac{4\pi}{3}} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$

Final Solutions

The three cube roots of unity are:

$z_0 = 1, \quad z_1 = -\frac{1}{2} + i \frac{\sqrt{3}}{2}, \quad z_2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}.$

Visualization Tip

These roots are evenly spaced on the unit circle in the complex plane, at angles $0$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$.

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Would you like further clarification or a detailed geometric representation of the roots?

Here are some related questions for deeper understanding:

1. What is the significance of cube roots of unity in mathematics?
2. How do the cube roots of unity relate to polynomial factorization?
3. Can we derive the same roots using factorization of $z^3 - 1$?
4. What are the properties of the cube roots of unity?
5. How can we visualize these roots in the Argand plane?

Tip: The sum of all cube roots of unity is always zero, which is a useful property!