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Let $ \Omega = \{z \in \mathbb{C} : |z| > 1, z \notin \mathbb{R}_{< -1}, z \notin \mathbb{R}_{\ge 2}\}. $ Find a conformal map which maps the region $\Omega$ to the upper half plane.

I would want to know what I am supposed to do first. I tried to shift by $1$ to the right $(z+1)$ then used $1/z$ but I was not sure about how the points around $z=-1$ and $z=2$ moved by these two maps if it is a good way to start this problem.

Thank you in advance.

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To start, apply the map $f_1(z)=z+1/z$ which collapses the unit circle onto the segment $[-2,2]$; the exterior of the unit circle is mapped onto the exterior of the circle. Then you have a plane with two slits, and things should become clearer.

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    @YeonjooYoo You can see a plot [here](http://math.fullerton.edu/mathews/c2003/JoukowskiTransMod.html). Also, what you really want to know is where the slits $[-\infty,-1]$ and $[2,+\infty]$ go. Both lie on the real line, so it's really a question about real function $x+1/x$ which you can plot easily, e.g., by typing "plot x+1/x" into Google search. And by the way, 2+1/2 is 5/2. not 3/2.2012-06-28