Faber polynomials

Remarks

The map $\Phi$ exists and is unique by the Riemann mapping theorem, since $\widehat{\C} \setminus K$ is simply connected and has more than one boundary point. The constant $c$ is the logarithmic capacity of $K$. The polynomial $F_n$ has degree $n$ and leading coefficient $c^{-n}$. Example: for the disc $\abs{z - a} \le r$, $\Phi(z) = (z - a)/r$ and $F_n(z) = ((z - a)/r)^n$. Example: for $K = [-1, 1]$, $\Phi(z) = z + \sqrt{z^2 - 1}$, and $\Phi^n + \Phi^{-n} = 2 T_n$ with $T_n$ the Chebyshev polynomial; since $\Phi^{-n} \to 0$ at $\infty$, $F_n = 2 T_n$ for $n \ge 1$. Non-example: the union of two disjoint closed discs does not qualify, because its complement in $\widehat{\C}$ is not simply connected, so there is no map $\Phi$. With $\Psi = \Phi^{-1}$, the Faber polynomials have the generating function $\Psi'(w)/(\Psi(w) - z) = \sum_{n \ge 0} F_n(z) w^{-n-1}$ for $\abs{w}$ large.