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

We are given the complex number $$z = 1 + ai$$ with $$a > 0$$ and the information that $$z^3$$ is a real number. Our task is to evaluate the sum $$1 + z + z^2 + \ldots + z^{11}$$.

First, we express $$z$$ in polar (modulus-argument) form. We have

$$|z| = \sqrt{1^2 + a^2} = \sqrt{1 + a^2},$$

and, because the real part is positive, the principal argument is

$$\theta = \tan^{-1} a.$$

Thus

$$z = |z|(\cos\theta + i\sin\theta) = \sqrt{1 + a^2}\,(\cos\theta + i\sin\theta).$$

Now, using De Moivre’s theorem,

$$z^3 = |z|^3 \bigl(\cos(3\theta) + i\sin(3\theta)\bigr).$$

For $$z^3$$ to be real, its imaginary part must vanish. That happens precisely when

$$\sin(3\theta) = 0 \quad\Longrightarrow\quad 3\theta = n\pi \quad(n\in\mathbb Z).$$

Because $$\theta = \tan^{-1}a$$ lies strictly between 0 and $$\dfrac{\pi}{2}$$ (since $$a > 0$$), the only admissible multiple is

$$3\theta = \pi \quad\Longrightarrow\quad \theta = \dfrac{\pi}{3}.$$

Substituting $$\theta = \dfrac{\pi}{3}$$ into $$\tan\theta = a$$ yields

$$a = \tan\!\left(\dfrac{\pi}{3}\right) = \sqrt3.$$

Hence

$$z = 1 + \sqrt3\,i.$$

Its modulus is

$$|z| = \sqrt{1 + (\sqrt3)^2} = \sqrt4 = 2,$$

so the polar form is

$$z = 2\bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\bigr).$$

We now turn to the desired sum

$$S = 1 + z + z^2 + \cdots + z^{11}.$$

This is a finite geometric progression with common ratio $$z\neq1$$, so we use the standard formula

$$\text{(Finite GP sum)}\qquad 1 + r + r^2 + \cdots + r^{n} = \dfrac{r^{n+1} - 1}{r - 1}.$$

Here $$r = z$$ and $$n = 11$$, whence

$$S = \dfrac{z^{12} - 1}{z - 1}.$$

We evaluate the numerator $$z^{12}$$ via De Moivre’s theorem again:

$$z^{12} = 2^{12}\,\Bigl(\cos\bigl(12\cdot\dfrac{\pi}{3}\bigr) + i\sin\bigl(12\cdot\dfrac{\pi}{3}\bigr)\Bigr) = 2^{12}\,\bigl(\cos4\pi + i\sin4\pi\bigr).$$

Because $$\cos4\pi = 1$$ and $$\sin4\pi = 0$$, we obtain

$$z^{12} = 2^{12} \times 1 = 4096.$$

Next, the denominator is straightforward:

$$z - 1 = (1 + \sqrt3\,i) - 1 = \sqrt3\,i.$$

Substituting these into the formula for $$S$$, we get

$$S = \dfrac{4096 - 1}{\sqrt3\,i} = \dfrac{4095}{\sqrt3\,i}.$$

To simplify, we recall the reciprocal of $$i$$:

$$\frac{1}{i} = -i.$$

Using this fact,

$$S = 4095 \cdot \frac{1}{\sqrt3}\cdot\frac{1}{i} = 4095 \cdot \frac{1}{\sqrt3} \cdot (-i) = -\frac{4095}{\sqrt3}\,i.$$

Finally, we rationalize the denominator in the coefficient:

$$\frac{4095}{\sqrt3} = \frac{4095\sqrt3}{3} = 1365\sqrt3.$$

Thus

$$S = -1365\sqrt3\,i.$$

Hence, the correct answer is Option B.