Let `z=1+ai` be a complex number, `a > 0`,such that `z^3` is a real number. Then the sum `1+z+z^2+...+ z^11` is equal to:

Options

A. `-1250sqrt(3)i`
B. `1250sqrt(3)i`
C. `-1365sqrt(3)i`
D. `1365sqrt(3)i`

Explanation

To solve the problem, we need to find the sum \( S = 1 + z + z^2 + \ldots + z^{11} \) where \( z = 1 + ai \) and \( a > 0 \), given that \( z^3 \) is a real number.

### Step 1: Find \( z^3 \)
We start by calculating \( z^3 \):
\[ z = 1 + ai \]
Using the binomial theorem, we can expand \( z^3 \):
\[ z^3 = (1 + ai)^3 = 1^3 + 3(1^2)(ai) + 3(1)(ai)^2 + (ai)^3 \]
Calculating each term:
- \( 1^3 = 1 \)
- \( 3(1^2)(ai) = 3ai \)
- \( 3(1)(ai)^2 = 3a^2(-1) = -3a^2 \)
- \( (ai)^3 = a^3(-i) = -a^3i \)

Combining these, we have:
\[ z^3 = 1 + 3ai - 3a^2 - a^3i = (1 - 3a^2) + (3a - a^3)i \]

### Step 2: Set the imaginary part to zero
For \( z^3 \) to be a real number, the imaginary part must be zero:
\[ 3a - a^3 = 0 \]
Factoring out \( a \):
\[ a(3 - a^2) = 0 \]
This gives us two cases:
1. \( a = 0 \) (not valid since \( a > 0 \))
2. \( 3 - a^2 = 0 \Rightarrow a^2 = 3 \Rightarrow a = \sqrt{3} \)

### Step 3: Substitute \( a \) back into \( z \)
Now substituting \( a \) back into \( z \):
\[ z = 1 + \sqrt{3}i \]

### Step 4: Find the sum \( S = 1 + z + z^2 + \ldots + z^{11} \)
This is a geometric series with first term \( a = 1 \) and common ratio \( r = z \):
\[ S = \frac{1 - z^{12}}{1 - z} \]

### Step 5: Calculate \( z^{12} \)
First, we express \( z \) in polar form:
\[ z = 1 + \sqrt{3}i = 2 \left( \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \right) = 2e^{i\frac{\pi}{3}} \]
Thus,
\[ z^{12} = (2e^{i\frac{\pi}{3}})^{12} = 2^{12} e^{i4\pi} = 4096 \cdot 1 = 4096 \]

### Step 6: Substitute \( z^{12} \) back into \( S \)
Now substituting \( z^{12} \) into the sum:
\[ S = \frac{1 - 4096}{1 - (1 + \sqrt{3}i)} = \frac{-4095}{-\sqrt{3}i} = \frac{4095}{\sqrt{3}i} \]
Multiplying numerator and denominator by \( i \):
\[ S = \frac{4095i}{\sqrt{3}(-1)} = -\frac{4095i}{\sqrt{3}} \]

### Final Answer
Thus, the sum \( S \) is:
\[ S = -\frac{4095i}{\sqrt{3}} \]