Do either (a) or (b).
Suppose \(f\) is meromorphic on \(\mathbb{D}\), continuous on \(\overline{\mathbb{D}}\) except at finitely many points of \(\mathbb{D}\), and real on \(\mathbb{T}\). Prove that \(f\) is a rational function.
Let \(\Omega\subset\mathbb{C}\) be the region inside the unit circle \(|z|=1\) and outside the circle \(\bigl|z-\frac14\bigr|=\frac14\). Find a one-to-one conformal map from \(\Omega\) onto an annulus \(r<|z|<1\) for the appropriate value of \(r\).
Solution
We solve part (a). Since \(f\) is meromorphic in \(\mathbb D\), it has only finitely many poles there: otherwise the poles would have an accumulation point in the compact closed disk after excluding a thin boundary collar, while continuity on the boundary makes \(f\) bounded near \(\mathbb T\).
Define on the exterior disk \[ F(z)=\overline{f(1/\overline z)},\qquad |z|>1. \] This is meromorphic there. On \(|z|=1\), the hypothesis that \(f\) is real gives \[ F(z)=\overline{f(z)}=f(z). \] Schwarz reflection across the unit circle therefore glues \(f\) and \(F\) to a meromorphic function on the Riemann sphere. A meromorphic function on the compact Riemann sphere is rational. Hence the original \(f\) is the restriction of a rational function.
For completeness, part (b) can also be made explicit. The two boundary circles belong to a coaxial family with limiting points \[ \alpha=\frac{1-\sqrt3}{4},\qquad \beta=\frac{1+\sqrt3}{4}. \] Thus \(T(z)=(z-\alpha)/(z-\beta)\) maps them to concentric circles. Multiplying \(T\) by a constant normalizes the outer image circle to radius \(1\); the inner radius is the constant ratio of the two resulting moduli. This gives the required conformal annulus map.