If `omega ne 1` is a cube root of unity, then `1, omega, omega^(2)`

Options

A. are vertices of an equilateral triangle
B. lie on a straight line
C. lie on a circle of radius `sqrt(3//2)`
D. None of these

Answer

To solve the problem, we need to show that the points \(1\), \(\omega\), and \(\omega^2\) form the vertices of an equilateral triangle in the complex plane, given that \(\omega\) is a cube root of unity and \(\omega \neq 1\).

### Step-by-Step Solution:

1. **Identify Cube Roots of Unity**:
The cube roots of unity are the solutions to the equation \(z^3 = 1\). They are given by:
\[ z = 1, \quad z = \omega, \quad z = \omega^2 \]
where \(\omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i\) and \(\omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i\).

2. **Calculate the Distances**:
To show that the triangle is equilateral, we need to calculate the distances between the points \(1\), \(\omega\), and \(\omega^2\).

- **Distance between \(1\) and \(\omega\)**:
\[ |1 - \omega| = |1 - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)| = |1 + \frac{1}{2} - \frac{\sqrt{3}}{2} i| = \left| \frac{3}{2} - \frac{\sqrt{3}}{2} i \right| \]
The modulus is:
\[ |1 - \omega| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \]

- **Distance between \(1\) and \(\omega^2\)**:
\[ |1 - \omega^2| = |1 - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)| = |1 + \frac{1}{2} + \frac{\sqrt{3}}{2} i| = \left| \frac{3}{2} + \frac{\sqrt{3}}{2} i \right| \]
The modulus is:
\[ |1 - \omega^2| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \]

- **Distance between \(\omega\) and \(\omega^2\)**:
\[ |\omega - \omega^2| = \left| \left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) \right| = \left| \frac{\sqrt{3}}{2} i + \frac{\sqrt{3}}{2} i \right| = \left| \sqrt{3} i \right| \]
The modulus is:
\[ |\omega - \omega^2| = \sqrt{3} \]

3. **Conclusion**:
Since all three distances are equal:
\[ |1 - \omega| = |1 - \omega^2| = |\omega - \omega^2| = \sqrt{3} \]
Therefore, the points \(1\), \(\omega\), and \(\omega^2\) form the vertices of an equilateral triangle.