Let $$n$$ denote the number of solutions of the equation $$z^2 + 3\bar{z} = 0$$, where $$z$$ is a complex number. Then the value of $$\sum_{k=0}^{\infty} \frac{1}{n^k}$$ is equal to

We need to find the number of solutions of $$z^2 + 3\bar{z} = 0$$, where $$z$$ is a complex number.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$. The equation becomes $$(x + iy)^2 + 3(x - iy) = 0$$.

Expanding: $$x^2 + 2ixy - y^2 + 3x - 3iy = 0$$.

Separating real and imaginary parts: $$(x^2 - y^2 + 3x) + i(2xy - 3y) = 0$$.

For this to equal zero, both real and imaginary parts must be zero:

Real part: $$x^2 - y^2 + 3x = 0$$ $$-(1)$$

Imaginary part: $$2xy - 3y = 0$$ $$-(2)$$

From equation $$(2)$$: $$y(2x - 3) = 0$$, so either $$y = 0$$ or $$x = \frac{3}{2}$$.

Case 1: $$y = 0$$. Substituting into $$(1)$$: $$x^2 + 3x = 0$$, so $$x(x + 3) = 0$$, giving $$x = 0$$ or $$x = -3$$. This gives solutions $$z = 0$$ and $$z = -3$$.

Case 2: $$x = \frac{3}{2}$$. Substituting into $$(1)$$: $$\frac{9}{4} - y^2 + \frac{9}{2} = 0$$, so $$y^2 = \frac{9}{4} + \frac{9}{2} = \frac{27}{4}$$, giving $$y = \pm\frac{3\sqrt{3}}{2}$$. This gives two more solutions: $$z = \frac{3}{2} \pm \frac{3\sqrt{3}}{2}i$$.

So the total number of solutions is $$n = 4$$.

Now we compute $$\sum_{k=0}^{\infty} \frac{1}{n^k} = \sum_{k=0}^{\infty} \frac{1}{4^k} = \sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k$$.

This is an infinite geometric series with first term $$1$$ and common ratio $$r = \frac{1}{4}$$. Since $$|r| < 1$$, the sum is $$\frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$.

The answer is $$\frac{4}{3}$$, which is Option B.