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
 = 0, f'(z_0) > 0)
(i.e., real and positive). Also suppose that
is symmetric with respect to the real axis.
Claim:
 = \overline{f(z)})
.
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
 = f^{-1} \circ \rho_K \circ f \circ \rho_\mathbb{R}(z) )
which maps from
to
. Since
and
is open, we can get
 \subseteq \Omega \cap \mathbb{R})
. If
)
then
 = w)
because
 = w)
and
 \in f(\mathbb{R})\Rightarrow \rho_K(f(w)) = f(w))
. So
 = f^{-1} \circ f(w) = w)
. 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.
 = z)
. So
 = \rho_K \circ f(z) \Rightarrow f(\overline{z}) = \rho_K(f(z)))
.
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
 ) \subseteq \mathbb{R})
, 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
.
