How can I show that the Joukowsky transform, $J(z)= z+\frac{1}{z}$
maps the set $R =\{(x,y)\in R^2: x^2+y^2>1, y>0\}$
conformally onto the upper half-plane?
It's clear that the parts of the boundary that are on the negative / positive real axis are mapped to the real axis in the $w$-plane.
Also, the part of the boundary that is the upper half of the unit circle, I notice that under the mapping $Z+\frac{1}{z}$, for $z=e^{i\theta}$, I get $$J(e^{i\theta}) = e^{i\theta} + e^{-i\theta}$$
$$= z + z*$$
$= 2Re(z)$, i.e., this part of the boundary also gets mapped to the real axis.
I'd like to use some sort of orientation preserving argument now and conclude that since the domain is to the left of a point traveling along the boundary from left to right, the point traces out the real line on the $w$-plane from negative to positive, and hence the domain must also be to the left of the real line, i.e., the upper half-plane. However, this is not in the form of a Mobius transformation, $\frac{az+b}{cz+d}$, so I don't think we have this kind of orientation argument to use.
And, the $2Re(z)$ also doesn't seem to give the orientation that I mentioned above and am hoping for - that the boundary of the set $R$, from left to right, gets mapped to the real line, from left to right, so that the domain is to the left of the real line; that is, the upper half-plane.
Any ideas would be greatly appreciated. Thanks in advance,
Edit: Or perhaps a connectedness argument is better?
Also, I can note that $J(\infty)=\infty$. So, no points in the complex plane $C$ get mapped to infinity.
One more thing: some test points easily show that these points from the set $R$ map to the upper half-plane in the $w$-plane.
I just need to find a way to conclude that the image is, in fact, the whole upper half-plane.