If `omega = "cos"(pi)/(n) + "i sin" (pi)/(n)`, then value of `1 + omega + omega^(2) +...+omega^(n-1)` is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Formula for the Sum of a Geometric Series**:
The sum of a geometric series can be expressed as:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. In our case, \( a = 1 \) and \( r = \omega \).

2. **Substituting Values into the Formula**:
Substituting \( a \) and \( r \) into the formula, we get:
\[ S = \frac{1(1 - \omega^n)}{1 - \omega} = \frac{1 - \omega^n}{1 - \omega} \]

3. **Calculate \( \omega^n \)**:
We know that:
\[ \omega = e^{i\frac{\pi}{n}} \]
Therefore,
\[ \omega^n = \left(e^{i\frac{\pi}{n}}\right)^n = e^{i\pi} = -1 \]

4. **Substituting \( \omega^n \) Back into the Sum**:
Now substituting \( \omega^n \) back into the equation for \( S \):
\[ S = \frac{1 - (-1)}{1 - \omega} = \frac{1 + 1}{1 - \omega} = \frac{2}{1 - \omega} \]

5. **Simplifying \( 1 - \omega \)**:
We have:
\[ 1 - \omega = 1 - \left(\cos\left(\frac{\pi}{n}\right) + i \sin\left(\frac{\pi}{n}\right)\right) = 1 - \cos\left(\frac{\pi}{n}\right) - i \sin\left(\frac{\pi}{n}\right) \]

6. **Finding the Magnitude of \( 1 - \omega \)**:
The magnitude of \( 1 - \omega \) can be calculated as:
\[ |1 - \omega| = \sqrt{(1 - \cos\left(\frac{\pi}{n}\right))^2 + \sin^2\left(\frac{\pi}{n}\right)} \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ |1 - \omega| = \sqrt{2 - 2\cos\left(\frac{\pi}{n}\right)} = 2\sin\left(\frac{\pi}{2n}\right) \]

7. **Final Expression for \( S \)**:
Thus, we can write:
\[ S = \frac{2}{1 - \omega} = \frac{2}{\sqrt{2 - 2\cos\left(\frac{\pi}{n}\right)}} \]
Simplifying further gives:
\[ S = \frac{2}{2\sin\left(\frac{\pi}{2n}\right)} = \frac{1}{\sin\left(\frac{\pi}{2n}\right)} \]

8. **Conclusion**:
Therefore, the final value of the sum \( S \) is:
\[ S = \frac{2}{1 - \omega} \]