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How can I map the unit disc onto the upper half plane?

I tried mapping $(1,i,-1)\rightarrow(1,0,\infty)$ using cross-ratio: $z\rightarrow \frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$, but didn't give me the right answer...

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Do you know how to conformally map the upper half plane to the unit disc? If so, you can take the inverse of this using matrix algebra.

An example of conformal map from the upper half plane to the unit disc is:

$$z \to \frac{z-i}{z+i}$$

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  • $\begingroup$ Nope, I don't know how to ''select'' the right triples $(z_2,z_3,z_4)$. I think that's my problem $\endgroup$
    – wilco
    Commented Oct 18, 2013 at 12:54
  • $\begingroup$ edited with an example. $\endgroup$ Commented Oct 18, 2013 at 12:58
  • $\begingroup$ What are the triples you chose? And why? $\endgroup$
    – wilco
    Commented Oct 18, 2013 at 13:00
  • $\begingroup$ I want to take the boundary of the upper half plane to the boundary of the disc, so I want everything on the vertical line to have norm $1$ in the image. I choose to send $i$ to $0$ as the new center of the circle, and since I want points on the horizontal line to go to norm $1$, this forces me to make the denominator $z+i$, since $-i$ is symmetric to $i$ across the line. $\endgroup$ Commented Oct 18, 2013 at 13:03
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    $\begingroup$ I learned it like this: the cross ratio which maps $(z_2,z_3,z_4)\rightarrow (1,0,\infty)$ is your axiom and making use of inverse and composition, I can map anything to anything. I think your method is better, but I don't ''see'' it happening. E.g. wat do you define as the boundary of the upper half plane? And what do you mean with ''the vertical line'' $\endgroup$
    – wilco
    Commented Oct 18, 2013 at 13:12

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