The value of `sum_(r=1)^(8)(sin(2rpi)/9+icos(2rpi)/9)`, is

Options

A. `-1`
B. 1
C. i
D. `-i`

Answer

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

\[ S = \sum_{r=1}^{8} \left( \sin\left(\frac{2\pi r}{9}\right) + i \cos\left(\frac{2\pi r}{9}\right) \right) \]

### Step 1: Rewrite the sum
We can rewrite the sine and cosine terms using Euler's formula. According to Euler's formula, we have:

\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]

Thus, we can express our sum as:

\[ S = \sum_{r=1}^{8} \left( i \cos\left(\frac{2\pi r}{9}\right) + \sin\left(\frac{2\pi r}{9}\right) \right) \]

### Step 2: Change the index of summation
To simplify the computation, we can change the index of summation to start from \( r=0 \) to \( r=8 \) and then subtract the term at \( r=0 \):

\[ S = \sum_{r=0}^{8} \left( \sin\left(\frac{2\pi r}{9}\right) + i \cos\left(\frac{2\pi r}{9}\right) \right) - \left( \sin(0) + i \cos(0) \right) \]

### Step 3: Evaluate the term at \( r=0 \)
Calculating the term at \( r=0 \):

\[ \sin(0) = 0, \quad \cos(0) = 1 \implies \text{So, } \sin(0) + i \cos(0) = 0 + i(1) = i \]

### Step 4: Sum the series
Now we need to evaluate:

\[ \sum_{r=0}^{8} \left( \sin\left(\frac{2\pi r}{9}\right) + i \cos\left(\frac{2\pi r}{9}\right) \right) \]

Using the property of complex exponentials, we can write:

\[ \sin\left(\frac{2\pi r}{9}\right) + i \cos\left(\frac{2\pi r}{9}\right) = i \left( \cos\left(\frac{2\pi r}{9}\right) - i \sin\left(\frac{2\pi r}{9}\right) \right) = i e^{-i\frac{2\pi r}{9}} \]

Thus, our sum becomes:

\[ S = i \sum_{r=0}^{8} e^{-i\frac{2\pi r}{9}} - i \]

### Step 5: Evaluate the geometric series
The sum \( \sum_{r=0}^{8} e^{-i\frac{2\pi r}{9}} \) is a geometric series with first term \( a = 1 \) and common ratio \( r = e^{-i\frac{2\pi}{9}} \):

\[ \sum_{r=0}^{n-1} ar^r = \frac{1 - r^n}{1 - r} = \frac{1 - e^{-i\frac{2\pi \cdot 9}{9}}}{1 - e^{-i\frac{2\pi}{9}}} = \frac{1 - e^{-2\pi i}}{1 - e^{-i\frac{2\pi}{9}}} \]

Since \( e^{-2\pi i} = 1 \):

\[ \sum_{r=0}^{8} e^{-i\frac{2\pi r}{9}} = \frac{1 - 1}{1 - e^{-i\frac{2\pi}{9}}} = 0 \]

### Step 6: Final result
Substituting back into our equation for \( S \):

\[ S = i \cdot 0 - i = -i \]

Thus, the final value of the sum is:

\[ \boxed{-i} \]