Let `omega ne 1`, be a cube root of unity, and `f : I rarr C` be defined by `f(n) = 1 + omega^(n) + omega^(2n)`, then range of f is

Options

A. {0}
B. {0, 3}
C. {0, 1, 3}
D. {0, 1}

Answer

To find the range of the function \( f(n) = 1 + \omega^n + \omega^{2n} \), where \( \omega \) is a cube root of unity (specifically, \( \omega \neq 1 \)), we can follow these steps:

### Step 1: Understand the properties of cube roots of unity
The cube roots of unity are the solutions to the equation \( x^3 = 1 \). They are given by:
- \( 1 \)
- \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \)
- \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \)

These roots satisfy the relation:
\[ 1 + \omega + \omega^2 = 0 \]

### Step 2: Evaluate \( f(n) \) for different integer values of \( n \)
We will compute \( f(n) \) for \( n = 0, 1, 2, 3, \ldots \) to observe any patterns.

1. **For \( n = 0 \)**:
\[ f(0) = 1 + \omega^0 + \omega^{0} = 1 + 1 + 1 = 3 \]

2. **For \( n = 1 \)**:
\[ f(1) = 1 + \omega + \omega^2 = 1 + \omega + \omega^2 = 1 + 0 = 0 \]

3. **For \( n = 2 \)**:
\[ f(2) = 1 + \omega^2 + \omega^{4} = 1 + \omega^2 + \omega = 1 + 0 = 0 \]
(since \( \omega^4 = \omega \))

4. **For \( n = 3 \)**:
\[ f(3) = 1 + \omega^3 + \omega^{6} = 1 + 1 + 1 = 3 \]

5. **For \( n = 4 \)**:
\[ f(4) = 1 + \omega^4 + \omega^{8} = 1 + \omega + \omega^2 = 1 + 0 = 0 \]

6. **For \( n = 5 \)**:
\[ f(5) = 1 + \omega^5 + \omega^{10} = 1 + \omega^2 + \omega = 1 + 0 = 0 \]

7. **For \( n = 6 \)**:
\[ f(6) = 1 + \omega^6 + \omega^{12} = 1 + 1 + 1 = 3 \]

### Step 3: Identify the pattern
From the calculations above, we can see that:
- \( f(n) = 3 \) when \( n \) is a multiple of 3 (i.e., \( n = 0, 3, 6, \ldots \))
- \( f(n) = 0 \) when \( n \) is not a multiple of 3 (i.e., \( n = 1, 2, 4, 5, 7, 8, \ldots \))

### Conclusion: Determine the range of \( f(n) \)
The range of the function \( f(n) \) is therefore:
\[ \{0, 3\} \]