Rational Convex Hull of Hartog's Triangle

Question

Here is a fun problem I have been thinking about of late:

Show that the rationally convex hull of the closed Hartog's triangle $$\{ z \in \mathbb{C}^2 : \left| z_1 \right| \leq \left| z_2 \right| \leq 1 \}$$ is the closed bidisk.

Answer 1

The rationally convex hull of this compact set is the closed bidisk. Indeed, suppose $p$ is a polynomial that has no root on the closed Hartogs triangle. By continuity, $p$ has no zero in an open neighbourhood of the closed triangle. For each fixed $z_1$, the function $1/p$ has a Laurent series expansion in $z_2$ in the annulus $0 < \left| z_1 \right| < \left| z_2 \right| < 1$. When $z_1$ is close to zero, the series becomes a Maclaurin series that converges in a full disk where $\left| z_2 \right| < 1$. The Laurent series coefficients are represented by integrals of the form $$c_k = \frac{1}{2\pi i} \int_{\Omega} \frac{1}{w^{k+1} p(z_1,w)}dw,$$ which shows us that they depend holomorphically on $z_1$. Since the coefficients with negative indices vanish for $z_1$ in a neighbouhood of $0$, they vanish identically. Hence the Laurent series for $1/p$ is a Maclaurin series even when $z_1$ is far away from $0$. So, the polynomial $p$ cannot have any roots in the bidisk.