If `omega ` is the imaginary cube roots of unity, then the number of pair of integers (a,b) such that `|aomega + b| = 1` is ______.

Explanation

To solve the problem, we need to find the number of ordered pairs of integers \((a, b)\) such that \(|a\omega + b| = 1\), where \(\omega\) is a non-real cube root of unity.

### Step-by-Step Solution:

1. **Understanding the Cube Roots of Unity**:
The cube roots of unity are the solutions to the equation \(x^3 = 1\). They are given by:
\[ 1, \quad \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \]
Here, \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

2. **Setting Up the Equation**:
We start with the equation:
\[ |a\omega + b| = 1 \]
Squaring both sides gives:
\[ |a\omega + b|^2 = 1^2 \]
This implies:
\[ (a\omega + b)(a\overline{\omega} + b) = 1 \]
Since \(\overline{\omega} = \omega^2\), we rewrite this as:
\[ (a\omega + b)(a\omega^2 + b) = 1 \]

3. **Expanding the Left Side**:
Expanding the left-hand side:
\[ a^2(\omega \cdot \omega^2) + ab(\omega + \omega^2) + b^2 = 1 \]
Since \(\omega \cdot \omega^2 = 1\) and \(\omega + \omega^2 = -1\), we have:
\[ a^2 + ab(-1) + b^2 = 1 \]
Simplifying gives:
\[ a^2 - ab + b^2 = 1 \]

4. **Rearranging the Equation**:
We can rearrange this equation as:
\[ a^2 - ab + b^2 - 1 = 0 \]

5. **Finding Integer Solutions**:
We can treat this as a quadratic in \(a\):
\[ a^2 - ab + (b^2 - 1) = 0 \]
The discriminant of this quadratic must be a perfect square for \(a\) to be an integer:
\[ D = b^2 - 4(b^2 - 1) = -3b^2 + 4 \]
Setting \(D = k^2\) for some integer \(k\):
\[ 4 - 3b^2 = k^2 \implies 3b^2 + k^2 = 4 \]

6. **Finding Integer Pairs**:
We need to find integer pairs \((b, k)\) such that \(3b^2 + k^2 = 4\):
- If \(b = 0\), then \(k^2 = 4 \Rightarrow k = \pm 2\).
- If \(b = 1\), then \(3 + k^2 = 4 \Rightarrow k^2 = 1 \Rightarrow k = \pm 1\).
- If \(b = -1\), it gives the same results as \(b = 1\).
- If \(b = 2\) or \(|b| > 2\), then \(3b^2 > 4\), which is not possible.

The valid pairs are:
- For \(b = 0\): \(k = 2\) or \(-2\) gives \(a^2 = 1\) leading to \((1, 0)\) and \((-1, 0)\).
- For \(b = 1\): \(k = 1\) or \(-1\) gives \(a^2 = 1\) leading to \((1, 1)\) and \((-1, 1)\).
- For \(b = -1\): similar to \(b = 1\) gives \((1, -1)\) and \((-1, -1)\).

7. **Counting Ordered Pairs**:
The pairs we found are:
- \((1, 0)\), \((-1, 0)\)
- \((1, 1)\), \((-1, 1)\)
- \((1, -1)\), \((-1, -1)\)
- \((0, 1)\), \((0, -1)\)
- \((1, 0)\), \((0, 1)\)

Thus, the total number of ordered pairs \((a, b)\) is \(6\).

### Final Answer:
The number of pairs of integers \((a, b)\) such that \(|a\omega + b| = 1\) is **6**.