If `w=cos""(pi)/(n)+isin""(pi)/(n)` then value of `1+w+w^(2)+.......+w^(n-1)` is :

Options

A. `1+i`
B. `1+itan((pi)/(n))`
C. `1+icot((pi)/(2n))`
D. None of these

Answer

To solve the problem, we need to find the value of the expression \( S_n = 1 + w + w^2 + \ldots + w^{n-1} \) where \( w = \cos\left(\frac{\pi}{n}\right) + i \sin\left(\frac{\pi}{n}\right) \).

### Step-by-step Solution:

1. **Express \( w \) in Exponential Form**:
\[ w = e^{i \frac{\pi}{n}} \]

2. **Recognize the Series as a Geometric Series**:
The expression \( S_n \) can be recognized as a geometric series with the first term \( a = 1 \), common ratio \( r = w \), and \( n \) terms.
\[ S_n = 1 + w + w^2 + \ldots + w^{n-1} \]

3. **Use the Formula for the Sum of a Geometric Series**:
The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Substituting \( a = 1 \) and \( r = w \):
\[ S_n = \frac{1 - w^n}{1 - w} \]

4. **Calculate \( w^n \)**:
Since \( w = e^{i \frac{\pi}{n}} \):
\[ w^n = \left(e^{i \frac{\pi}{n}}\right)^n = e^{i \pi} = -1 \]

5. **Substitute \( w^n \) into the Sum**:
Now substituting \( w^n = -1 \) into the expression for \( S_n \):
\[ S_n = \frac{1 - (-1)}{1 - w} = \frac{1 + 1}{1 - w} = \frac{2}{1 - w} \]

6. **Simplify \( 1 - w \)**:
We know that:
\[ w = \cos\left(\frac{\pi}{n}\right) + i \sin\left(\frac{\pi}{n}\right) \]
Therefore:
\[ 1 - w = 1 - \left(\cos\left(\frac{\pi}{n}\right) + i \sin\left(\frac{\pi}{n}\right)\right) = 1 - \cos\left(\frac{\pi}{n}\right) - i \sin\left(\frac{\pi}{n}\right) \]

7. **Final Expression for \( S_n \)**:
Thus, we have:
\[ S_n = \frac{2}{1 - \cos\left(\frac{\pi}{n}\right) - i \sin\left(\frac{\pi}{n}\right)} \]