If `a_1,a_2 ...a_n` are nth roots of unity then`1/(1-a_1) +1/(1-a_2)+1/(1-a_3)..+1/(1-a_n)` is equal to

Options

A. `(n-1)/2`
B. `n/2`
C. `(2^(n)-1)/(2)`
D. none of these

Explanation

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

\[ S = \frac{1}{1 - a_1} + \frac{1}{1 - a_2} + \frac{1}{1 - a_3} + \ldots + \frac{1}{1 - a_n} \]

where \( a_1, a_2, \ldots, a_n \) are the \( n \)th roots of unity. The \( n \)th roots of unity are given by:

\[ a_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]

### Step 1: Rewrite the expression

We can rewrite \( S \) as:

\[ S = \sum_{k=0}^{n-1} \frac{1}{1 - a_k} \]

### Step 2: Find a common denominator

The common denominator for the terms in \( S \) is:

\[ \prod_{k=0}^{n-1} (1 - a_k) \]

Thus, we can express \( S \) as:

\[ S = \frac{\sum_{k=0}^{n-1} \prod_{j \neq k} (1 - a_j)}{\prod_{j=0}^{n-1} (1 - a_j)} \]

### Step 3: Evaluate the denominator

The denominator \( \prod_{j=0}^{n-1} (1 - a_j) \) can be evaluated using the fact that \( a_j \) are the roots of the polynomial \( x^n - 1 = 0 \). Therefore, we can express this as:

\[ \prod_{j=0}^{n-1} (1 - a_j) = P(1) \quad \text{where } P(x) = x^n - 1 \]

Calculating \( P(1) \):

\[ P(1) = 1^n - 1 = 0 \]

This means that the product \( \prod_{j=0}^{n-1} (1 - a_j) \) is equal to \( n \) because:

\[ P(x) = (x - 1)(x - a_1)(x - a_2) \ldots (x - a_{n-1}) \]

### Step 4: Evaluate the numerator

Now we need to evaluate the numerator:

\[ \sum_{k=0}^{n-1} \prod_{j \neq k} (1 - a_j) \]

Using the fact that \( \prod_{j \neq k} (1 - a_j) \) is the polynomial evaluated at \( 1 \) excluding the term \( (1 - a_k) \):

\[ \sum_{k=0}^{n-1} \prod_{j \neq k} (1 - a_j) = n \cdot \prod_{j=0}^{n-1} (1 - a_j) \]

### Step 5: Combine the results

Thus, we have:

\[ S = \frac{n \cdot \prod_{j=0}^{n-1} (1 - a_j)}{\prod_{j=0}^{n-1} (1 - a_j)} = n \]

### Final Result

Thus, the final result is:

\[ S = \frac{n(n-1)}{2} \]