Let `z = a("cos"(pi)/(5) + "i sin" (pi)/(5)),a in R, |a| lt 1`, then `S = z^(2015) + z^(2016) + z^(2017) + ...` equals

Options

A. `(a^(2015))/(z-1)`
B. `(a^(2015))/(1-z)`
C. `(a^(2015))/(1-a)`
D. `(a^(2015))/(a-1)`

Answer

To solve the problem, we start with the complex number given by:

\[ z = a \left( \cos\left(\frac{\pi}{5}\right) + i \sin\left(\frac{\pi}{5}\right) \right) \]

where \( a \in \mathbb{R} \) and \( |a| < 1 \). We need to find the sum:

\[ S = z^{2015} + z^{2016} + z^{2017} + \ldots \]

### Step 1: Identify the series as a geometric series

The series \( S \) can be recognized as an infinite geometric series where the first term \( a_1 = z^{2015} \) and the common ratio \( r = z \).

### Step 2: Use the formula for the sum of an infinite geometric series

The formula for the sum of an infinite geometric series is given by:

\[ S = \frac{a_1}{1 - r} \]

where \( |r| < 1 \). Here, we have:

\[ S = \frac{z^{2015}}{1 - z} \]

### Step 3: Substitute \( z \) into the formula

Now, we need to express \( z^{2015} \):

\[ z^{2015} = \left( a \left( \cos\left(\frac{\pi}{5}\right) + i \sin\left(\frac{\pi}{5}\right) \right) \right)^{2015} \]

Using Euler's formula, we can write:

\[ z = a e^{i \frac{\pi}{5}} \]

Thus,

\[ z^{2015} = a^{2015} e^{i \frac{2015 \pi}{5}} = a^{2015} e^{i 403 \pi} = a^{2015} e^{i \pi} = -a^{2015} \]

### Step 4: Substitute \( z^{2015} \) back into the sum formula

Now we substitute \( z^{2015} \) back into the sum \( S \):

\[ S = \frac{-a^{2015}}{1 - z} \]

### Step 5: Calculate \( 1 - z \)

We have:

\[ 1 - z = 1 - a \left( \cos\left(\frac{\pi}{5}\right) + i \sin\left(\frac{\pi}{5}\right) \right) \]

### Final Expression for \( S \)

Thus, the final expression for the sum \( S \) is:

\[ S = \frac{-a^{2015}}{1 - a \left( \cos\left(\frac{\pi}{5}\right) + i \sin\left(\frac{\pi}{5}\right) \right)} \]