Let `omega ne 1` be cube root of unity. Suppose `(1 + omega)^(2023) = A + B omega` where A, B `in` R, then `|A + Bi|^(2)` is equal to ______________.

Answer

To solve the problem, we need to find the value of \(|A + Bi|^2\) where \((1 + \omega)^{2023} = A + B\omega\) and \(\omega\) is a cube root of unity.

### Step-by-Step Solution:

1. **Understanding Cube Roots of Unity**:
The cube roots of unity are given by the equation \(x^3 = 1\). The roots are \(1, \omega, \omega^2\) where \(\omega = e^{2\pi i/3}\) and \(\omega^2 = e^{-2\pi i/3}\). They satisfy the relation:
\[ 1 + \omega + \omega^2 = 0 \]
Hence, we can express \(\omega^2\) as:
\[ \omega^2 = -1 - \omega \]

2. **Expressing \(1 + \omega\)**:
We know that:
\[ 1 + \omega = -\omega^2 \]

3. **Calculating \((1 + \omega)^{2023}\)**:
We can rewrite:
\[ (1 + \omega)^{2023} = (-\omega^2)^{2023} = (-1)^{2023} (\omega^2)^{2023} \]
Since \(2023\) is odd, \((-1)^{2023} = -1\). Thus,
\[ (1 + \omega)^{2023} = -(\omega^2)^{2023} \]

4. **Using the properties of \(\omega\)**:
We know that \(\omega^3 = 1\), so:
\[ (\omega^2)^{2023} = \omega^{4046} = \omega^{4046 \mod 3} = \omega^{1} = \omega \]
Therefore,
\[ (1 + \omega)^{2023} = -\omega \]

5. **Setting up the equation**:
We have:
\[ (1 + \omega)^{2023} = A + B\omega \]
Equating both sides gives:
\[ -\omega = A + B\omega \]
This implies:
\[ A = 0 \quad \text{and} \quad B = -1 \]

6. **Finding \(|A + Bi|^2\)**:
Now we need to calculate:
\[ |A + Bi|^2 = |0 - i|^2 = | -i |^2 = |i|^2 \]
The modulus of \(i\) is \(1\), so:
\[ |i|^2 = 1^2 = 1 \]

### Final Answer:
\[ |A + Bi|^2 = 1 \]