Schwarz Function of an Ellipse

Question

I want to find the Schwarz function of the ellipse define by $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0. $$ To do so, substitute $$ x = \frac{z+\bar{z}}{2}, \quad y = \frac{z - \bar{z}}{2i} $$ into the ellipse equation and solve for $\bar{z}$. With $c:c^2 = a^2 - b^2$ being the focal distance of the ellipse, one obtains $$ \bar{z} = \frac{(a^2+b^2)z \pm 2ab\sqrt{z^2-c^2}}{c^2}. $$ Now at this point, a few texts I have read discard the positive root solution and define the Schwarz function of the ellipse as $$ S(z) = \frac{(a^2+b^2)z - 2ab\sqrt{z^2-c^2}}{c^2} $$ and then use this to evaluate line integrals along the ellipse: $$ \oint_{\partial \Omega} \frac{\bar{z}}{z-\zeta}\, d\zeta = \oint_{\partial \Omega} \frac{S(z)}{z-\zeta}\, d\zeta. $$ I have a a question about this. Why can we discard the positive root solution? The Schwarz function is supposed to be such that $S(z) = \bar{z}$ everywhere on the ellipse but the negative root solution does not satisfy this; e.g. $S(-a) \neq -a$.

Answer 1

As you say, the Schwarz function is supposed to be such that $S(z)=\bar{z}$ everywhere on the ellipse, and in the example you give $S(-a)\neq -a$. The solution seems to be that we must choose the negative root if Re$(z)\geq 0$ and the positive root otherwise.