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The conformal mapping $z\mapsto \frac{1}{z}$ doesn't map onto the unit disc, but it maps to the unit disc minus the interval $(0,1)\subset\mathbb{R}$. I tried using the fact that the circle is perpendicular to the ray and that they have a common point on their boundary, but I am not sure if one point is enough. I am not sure how to proceed because I am not sure how the boundary of the image below will be mapped to the boundary of the unit disc.

I'm not looking for an entire answer, just a way to deal (or get rid of) with this ray and possibly a technique that would apply to a larger class of problems.

enter image description here

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First apply $\sqrt z$, the branch defined on $\mathbb{C}\setminus[0,\infty)$: $\sqrt{|z|i^{i\theta}}=\sqrt{|z|}e^{i\theta/2}$. This will take $D$ into $\{|z|>1,\text{Im}(z)>0\}$. Now apply $(z+1)/(z-1)$. I hope you can follow from here.

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  • $\begingroup$ Can you comment on how the conformal map $\sqrt z$ works? I understand how $z^2$ works - intuitively, I double the angle. For example the first quadrant maps to the second, and so on. But if given a region and the $\sqrt z$ map, how do I know which part to get rid of? If I had to guess, the image of $U$ under the $\sqrt z$ map would be $\lbrace \vert z \vert >1, \text{Im}z>0 \rbrace$ instead of the one you gave. $\endgroup$ – The Substitute Feb 15 '15 at 13:27
  • $\begingroup$ Oops! I have edited my answer, explaining tha branch used. $\sqrt z$ halfs the angle. $\endgroup$ – Julián Aguirre Feb 15 '15 at 15:24

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