# Conformal mapping from upper half-plane to unit disk

I have run into some confusion while reading Newman Bak's Complex Analysis.

The text states that if we wish to determine a comformal mapping $h$ of the upper half-plane onto the unit disk, assuming that $h$ is bilinear and $h(\alpha)=0$ for fixed $\alpha$ with $Im(\alpha)>0$, then "since the real axis is mapped into the unit circle, it follows by the Schwartz Reflection Principle that $h(\bar\alpha)=\infty.$"

How exactly does the Schwartz Reflection Principle imply that $h(\bar\alpha)=\infty$? I can see that since $h$ is an automorphism of the Riemann Sphere, $h(\alpha)$ is contained in the interior of the sphere so that $h(\bar\alpha)$ must be contained in the exterior. However, I don't see why $h(\bar\alpha)=\infty.$

• You might find this older answer of Daniel Fischer useful: math.stackexchange.com/a/449693/155629 – Travis Jul 20 '15 at 22:57
• Is the idea that for points outside the unit circle, h can be extended by $\frac{1}{\bar{h(1/\bar{z})}}$ so that $h(\bar{\alpha})=\frac{1}{\bar{h(\bar\alpha)}}=\frac{1}{h(\alpha)}=\infty$? – Greyson Jul 20 '15 at 23:45