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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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1 Answer 1

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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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  • $\begingroup$ Thank you for the answer. I am still not sure how the point outside of the disk are mapped. Because when I try to plug in poins on the vertical line with Re(z)=2,I am not sure how the points are mapped by z+1/z. In particular z=2,it is mapped to 3/2 but I do not see any pattern of the points on the line. $\endgroup$ Commented Jun 28, 2012 at 18:08
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    $\begingroup$ @YeonjooYoo You can see a plot here. 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. $\endgroup$ Commented Jun 28, 2012 at 18:52

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