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The question here is not so much to find the mapping, but exactly how one goes about to find the correct map.

I have $\{z\in\mathbb{C}: -\pi/4<\arg(z)<\pi/2\}$, which requires to be turned into a quarter disk in the lower left half plane.

Now most of the steps I understand, however what I do not understand is how to get the sector into an easier plane to work with (half plane, or whatever else).

As such the process in order to get through this first map is of much more interest to me than the actual solution. I would be very grateful if someone could explain to me how this particular type of mapping works.

Ronan

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  • $\begingroup$ Would specify, you wanna get something like that? Transformation $\endgroup$
    – Kaster
    Apr 24, 2013 at 23:52
  • $\begingroup$ Yes, that is apparently what I want to do first. Thank you for the illustration Kaster. $\endgroup$
    – Ronan
    Apr 25, 2013 at 0:01

1 Answer 1

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The easiest ways to deal with stuff like this (corners) is to use logarithms or fractional powers of $z$ (which use logarithms). So, for instance, taking a log here gives you a rectangle stretching to infinity on the left, which you can play around with, while squaring this gives you three quarters of the unit circle.

If you can get the shape to be a quadrant, you can square it to get a half plane.

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  • $\begingroup$ Ok, so then the 'true' first step for this would be to create a full quadrant sized 'field' (not sure what the right term is). In which case one would do something like: $\{z->z*a\}$ Where a would be a given constant? With a quick try I would assume $a=\exp(\pi/3)$ $\endgroup$
    – Ronan
    Apr 24, 2013 at 23:54
  • $\begingroup$ that should be $a=\exp(i*2*pi/3)$. I took the total angle between both rays (=$3*\pi/4$), set ($3*\pi/4 *x=\pi/2$) yielding ($2*\pi/3$). But at this point I have the feeling that I have no idea what I am doing... $\endgroup$
    – Ronan
    Apr 25, 2013 at 0:22
  • $\begingroup$ Try using a combination of rotations and $z^{4/3}$ to get an upper half disk, and then use the standard map of the disk to a half plane to get a quarter plane. Then use the map $z^2$ to get a full half plane. $\endgroup$ Apr 25, 2013 at 0:45
  • $\begingroup$ Alright, so I understand where the $z^{4/3}$ arrives, I had gotten that earlier by setting the wanted total angle to $\pi$. I guess, my confusion arises from why that isn't enough. In my mind, performing the map $z->z^{3/4}$ is sufficient to create the upper half disk. Or does that only create that appropriate 'distance'? $\endgroup$
    – Ronan
    Apr 25, 2013 at 0:57
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    $\begingroup$ so then from what I gather there is no need for rotation? Once you use the {$z^{3/4}=w$} map, you get the full upper half plane. From there you could do, say, {$w->(w+-i)/(w+1)$} to get it into a unit disk (and thus move on with your life :p). I guess I had not noticed that the $z^{3/4}$ mapped to upper half plane and not upper half disk. $\endgroup$
    – Ronan
    Apr 25, 2013 at 1:54

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