The Mandelbrot set contains three regions (two bulbs and a cardioid) with periods of three. These regions each contain a fixed point which is a root of the expression $x^3 + 2x^2 + x + 1$.
The fixed point at the center of the period-three cardioid has the value $$-\frac{1}{3} \left(2 + \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}}\right)$$
This simplifies to $-ρ^2$, where $ρ$ is the plastic constant equal to $\sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 - \sqrt{69}}{18}}$.
The fixed points at the centers of the two period-three bulbs are $$-\frac{2}{3} + \frac{1 + i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 - i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$ and $$-\frac{2}{3} + \frac{1 - i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 + i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$
The real part of these two points simplifies to $\frac{ρ^2 - 2} {2}$. The imaginary parts are approximately $\pm$ 0.744861766619744236593170... $i$, but I can't figure out a closed-form expression for that. I'm hopeful that I can find something in terms of $ρ$, since that worked for the real part and for the cardioid point, but so far I've been stumped.
Given that I know the real root of the function $x^3 + 2x^2 + x + 1$ is $-ρ^2$, I can factor that out to get $(x+ρ^2)(x^2 + (2 - ρ^2) x + (2 - ρ^6 + 3ρ^4 - 3ρ^2))$.
Using the quadratic formula on that second term, I get $\frac{ρ^2 ± \sqrt{-4 + 8ρ^2 - 11ρ^4 + 4ρ^6} - 2} {2}$.
Because $ρ^3 = ρ + 1$ and $ρ^4 = ρ^2 + ρ$, the term inside the square root simplifies to $ρ^2 - 3ρ$. That is negative, so factoring out $ί$ gives $\sqrt{3ρ - ρ²} ί$.
Rearranging, that gives me the two complex roots as $\frac{ρ^2 - 2} {2} ± \frac{\sqrt{3ρ - ρ²}} {2} ί$.