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In my textbook, it is said that $z+1\over z-1$ maps the left half plane to the unit disk. So since it is its self-inverse, (right?) the unit disc should be mapped to the left half plane. But on another page it says that this maps the upper semidisk to the first quadrant which is in the right half plane... what is happening? please help, thanks.

So I have tried applying this map to the first quadrant and found that it is mapped to the lower half plane with the unit circle cut out of it, m i right? If not, please explain, there might be some fundamental concepts I have not realized.

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    Consider the point 0. You have that $(0+1)/(0-1) = -1$ which is not in the first quadrant. So if you book indeed says that $z\mapsto (z+1)/(z-1)$ sends the upper semidisk to the first quadrant, it is mistaken.2012-01-24
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    OTOH, the map $(1+z)/(1-z)$ does send the upper half disk to the first quadrant.2012-01-24
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    Thanks, @Willie, would you mind commenting on whether the added bit is right? ie the map of the first quadrant? I just want to check that I have got things straight.2012-01-24
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    Should be right. But you should probably say the lower half plane with the _unit disk_ cut out, instead of the _unit circle_. Typically in mathematical contexts the unit circle is only the boundary of the disk: cutting it out will leave both the exterior _and_ the interior of the circle behind.2012-01-24
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    @Willie, thanks again! Good point. Btw, I did this by decomposing the map into inversions, translations etc. Is this the easiest way? Or is there a more direct/ elegant way to see what maps do to domains?2012-01-24
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    Well.... In your particular case you have a Fractional Linear Transformation (Mobius transform). It is known that FLTs preserve cross ratios, and that it sends circles (including lines) to circles (including lines). Since your domains are all bounded by circles and lines, it suffices to first compute what the image of the boundaries are, and then you can use one point inside the domain to see which component of the complex plane divided by the boundary curves is your image.2012-01-24
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    For example, $(z+1)/(z-1)$ sends $1\mapsto \infty$ and $-1\mapsto 0$. This implies that the image of the unit circle and the image of the x axis under the mapping must be lines through the origin. By checking that $i \mapsto -i$ and $0 \mapsto -1$ you see that the x axis is preserved while the unit circle maps into the y axis. Lastly you just check that a point in the upper semidisk, say, $i/2$ is mapped to some point in the III quadrant and you know what the final image will be.2012-01-24
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    @Willie, thanks again, that is very helpful!2012-01-24

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