The cube roots of unity lie on a circle

Options

A. `z = 1 `
B. `| z - 1| = 1 `
C. ` | z + 1| = 1 `
D. ` | z - omega| = 1 `

Answer

To solve the problem of showing that the cube roots of unity lie on a circle, we can follow these steps:

### Step 1: Define the Cube Roots of Unity
The cube roots of unity are the solutions to the equation \( z^3 = 1 \). These solutions can be expressed as:
\[ z = 1, \quad z = \omega = e^{2\pi i / 3}, \quad z = \omega^2 = e^{4\pi i / 3} \]
Where \( \omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \) and \( \omega^2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \).

### Step 2: Write the Roots in Complex Form
The cube roots of unity can be represented in the complex plane as:
- \( z_0 = 1 \) (which is \( 1 + 0i \))
- \( z_1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \)
- \( z_2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \)

### Step 3: Find the Magnitude of Each Root
To show that these points lie on a circle, we need to find the magnitude of each root:
1. For \( z_0 \):
\[ |z_0| = |1| = 1 \]
2. For \( z_1 \):
\[ |z_1| = \left| -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
3. For \( z_2 \):
\[ |z_2| = \left| -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

### Step 4: Conclude the Location of the Roots
Since the magnitude of each cube root of unity is 1, we can conclude that all three points lie on a circle with radius 1 centered at the origin (0, 0) in the complex plane.

### Step 5: Equation of the Circle
The equation of a circle in the complex plane is given by:
\[ |z| = r \]
Where \( r \) is the radius. In this case, since the radius is 1, the equation of the circle is:
\[ |z| = 1 \]

### Conclusion
Thus, we have shown that the cube roots of unity lie on a circle of radius 1 centered at the origin.

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