I have an exercise and an approach to it's solution with a gap in it. The exercise comes from Ahlfors Complex Analysis text in Ch. 6 on Conformal mappings.

Suppose that

is a Riemann map (conformal) from a simply connected domain

to the open unit disk

with

(i.e., real and positive). Also suppose that

is symmetric with respect to the real axis.

Claim:

.

Scratch Work: For the moment assume

is a mobius map and recall that mobius tranforms map Riemann circles to Riemann circles (including "lines"). Define

be a reflection across the real axis and

be the reflection about the circle

. Form the function

which maps from

to

. Since

and

is open, we can get

. If

then

because

and

. So

. Therefore, we have that

is analytic (since conformal) and equals the identity map on

. Then, analytic extension gives us that

is the identity map, i.e.

. So

.

So there is the first gap; I don't see how to find that

as claimed. If it is true, it must be that

so that

, but I'm not seeing it.

When we consider that

is a general conformal map, there is a similar gap. In this case it is not necessarily true that the real line maps to a Riemann "circle". We have the Carathedory-Osgood extension that gives a homeomorphic extension

, so the boundary of

maps to the boundary of

. But, again, I don't see why the segment of

should map into a line (or specifically into the real axis) in the disk so that we can conclude that the reflection

is in fact

.