If `omega` is a complex cube root of unity, then value of expression `cos[{(1-omega)(1-omega^2) +...+(10-omega) (10-omega^2)}pi/900]`

Options

A. `-1`
B. 0
C. 1
D. `sqrt(3//2)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \cos\left[\frac{(1-\omega)(1-\omega^2) + (2-\omega)(2-\omega^2) + \ldots + (10-\omega)(10-\omega^2)}{900}\pi\right] \]

where \(\omega\) is a complex cube root of unity.

### Step 1: Understanding the properties of \(\omega\)
The complex cube roots of unity satisfy the following properties:
- \(1 + \omega + \omega^2 = 0\)
- \(\omega^3 = 1\)

From the first property, we can express \(\omega + \omega^2 = -1\).

### Step 2: Simplifying the expression
We can rewrite each term in the summation:
\[ (a - \omega)(a - \omega^2) = a^2 - (a(\omega + \omega^2)) + \omega \cdot \omega^2 \]
Using the properties of \(\omega\):
- \(\omega \cdot \omega^2 = \omega^3 = 1\)
- \(\omega + \omega^2 = -1\)

Thus, we have:
\[ (a - \omega)(a - \omega^2) = a^2 + a + 1 \]

### Step 3: Summing the terms from 1 to 10
Now we need to sum this expression for \(a\) from 1 to 10:
\[ \sum_{a=1}^{10} (a^2 + a + 1) \]

This can be separated into three sums:
\[ \sum_{a=1}^{10} a^2 + \sum_{a=1}^{10} a + \sum_{a=1}^{10} 1 \]

### Step 4: Calculating the individual sums
1. **Sum of squares**:
\[ \sum_{a=1}^{n} a^2 = \frac{n(n+1)(2n+1)}{6} \]
For \(n = 10\):
\[ \sum_{a=1}^{10} a^2 = \frac{10(11)(21)}{6} = 385 \]

2. **Sum of first \(n\) natural numbers**:
\[ \sum_{a=1}^{n} a = \frac{n(n+1)}{2} \]
For \(n = 10\):
\[ \sum_{a=1}^{10} a = \frac{10(11)}{2} = 55 \]

3. **Sum of 1's**:
\[ \sum_{a=1}^{10} 1 = 10 \]

### Step 5: Combining the sums
Now we combine the sums:
\[ \sum_{a=1}^{10} (a^2 + a + 1) = 385 + 55 + 10 = 450 \]

### Step 6: Plugging into the cosine expression
Now we substitute back into our cosine expression:
\[ \cos\left[\frac{450\pi}{900}\right] = \cos\left[\frac{\pi}{2}\right] \]

### Step 7: Evaluating the cosine
We know that:
\[ \cos\left[\frac{\pi}{2}\right] = 0 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{0} \]