We have, $$\arg z_1 = \frac{k\,\pi}3, \quad z_1 = \left(\tfrac{1+\sqrt{-3}}{2}\right)^k\tag1$$ $$\arg z_2=\frac{k\,\pi}3, \quad z_2 = \left( B\Big(\color{blue}{\tfrac{-5+8\,\sqrt{-11}}{27}};\,\tfrac12,\tfrac13\Big)\right)^k \tag2$$ for $k=1,2,3$ and incomplete beta function $B(z;a,b)$. The second with $k=1$ is by V. Reshetnikov. But that cannot be an isolated result.
Reshetnikov states (without giving details) that $(2)$ is equivalent to, $$B\left(\frac19;\,\frac16,\frac13\right) = \frac12\,B\left(\frac16,\frac13\right) =\frac{1}{2\sqrt{\pi}}\,\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)\tag3$$ Thus, $(3)$ is related to, $$I\left(\color{blue}{\frac19};\,\frac16,\frac13\right)=\frac12 \tag4$$ with regularized beta function $I(z;a,b)$. The equation $$I\left(z;\,\frac16,\frac13\right)=\frac1n \tag5$$ is quite easy to solve. For example, the smallest real root of, $$-1 + 99 z - 243 z^2 + 81 z^3= 0\tag6$$ will yield $n=3$. More specifically, $$I\left(\frac19\Big(1-4\sin\big(\tfrac{\pi}{18}\big)\Big)^2;\,\frac16,\frac13 \right)=\frac13\quad\quad\tag7$$ $$\quad I\left(\frac13\Big(1-\frac{\sqrt2}{\sqrt[4]3}\Big)^2;\,\frac16,\frac13\right) =\frac14\quad\tag8$$ and so on.
Q: But how do we get from $\color{blue}{\frac19}$ of $(4)$ to $\color{blue}{\frac{-5+8\,\sqrt{-11}}{27}}$ of $(2)$? (I assume a hypergeometric transformation is involved.)
If the answer can be found, then perhaps we can use an expression $P(z)$ of a root $z$ of $(6)$ to find some, $$\arg\,B\left( P(z);\,\frac12,\frac13\right)=\beta\,\pi\tag7$$ where $\beta$ is a rational/algebraic number?
