maximality for the slit mapping in the previous paragraph , so that > 0. The two inequalities for are contradictory. The proof of the uniqueness of the conformal parallel slit transformation is given in and . Applying the inverse of the Joukowsky transform to the horizontal slit domain, it can be assumed that is a domain bounded by the unit circle and contains analytic arcs and isolated points (the images of other the inverse of the Joukowsky transform under the other parallel horizontal slits). Thus, taking a fixed in , there is a univalent mapping = = with image a horizontal slit domain. Suppose that is another uniformizer with = . The images under or of each have a fixed y-coordinate so are horizontal segments. On the other hand = is holomorphic in . If it is constant, then it must be identically zero since = 0. Suppose is non-constant. Then by assumption are all horizontal lines. If is not in one of these lines, Cauchy's argument principle shows that the number of solutions of = in is zero (any will eventually be encircled by contours in close to the 's). This contradicts the fact that the non-constant holomorphic function is an open mapping. Sketch proof via Dirichlet problem Given U and a point z0 in U, we want to construct a function f which maps U to the unit disk and z0 to 0. For this sketch, we will assume that U is bounded and its boundary is smooth, much like Riemann did. Write where g = u + iv is some (to be determined) holomorphic function with real part u and imaginary part v. It is then clear that z0 is the only zero of f. We require |f(z)| = 1 for z ∈ ∂U, so we need on the boundary. Since u is the real part of a holomorphic function, we know that u is necessarily a harmonic function; i.e., it satisfies Laplace's equation. The question then becomes: does a real-valued harmonic function u exist that is defined on all of U and has the given boundary condition? The positive answer is provided by the Dirichlet principle. Once the existence of u has been established, the Cauchy–Riemann equations for the holomorphic function g allow us to find v (this argument depends on the assumption that U be simply connected). Once u and v have been constructed, one