I'm currently working on a problem that asks to map the unit quarter disk from quadrant I to the upper half-plane, and then to use this mapping to find a harmonic function on the quarter disk, taking values 1 on the part of the boundary with |z| = 1 and taking the value 0 on the rest of the boundary of the quarter disk.

I think I computed the correct conformal mapping.

First, I applied the mapping $z^4$. This maps the quarter disk to the unit disk.

Then, I used the standard mapping from the UHP to the unit disk, and computed its inverse. Now I have a mapping taking the quarter disk to the upper half plane, as required:

$$z \to \frac{i(1+z^4)}{1-z^4}$$

I'm having trouble finding the harmonic function, though. I think that the part of the boundary of the quarter disk that is supposed to take the value 1 is mapped to the negative real axis. Ok, then I apply the harmonic $Arg$ function to the above mapping, and I should get $\pi$ as the output, along the negative real axis. Now I scale by 1/$\pi$ to get my boundary value of 1. The function is

$$z \to \frac{1}{\pi}Arg [\frac{i(1+z^4)}{1-z^4}]$$

However, I don't know how to make the rest of the boundary of the quarter disk take the value 0.



$z^4$ maps the (open) quarter disk on the disk minus the interval $[0,1)$. You should start with $z^2$, which takes the quarter disk to the half disk, and then $(z+1)/(z-1)$.

  • $\begingroup$ Hi Professor Aguirre, so starting with $z^2$ instead, the z+1/z-1 mapping will then send the imaginary axis to the unit circle. So, keep with the same strategy and use the inverse mapping, i.e., mapping the unit disk to the upper half plane? Thanks @JulianAguirre, $\endgroup$ – user284193 Oct 26 '15 at 21:52
  • $\begingroup$ I'm a little confused about following the $z^2$ mapping with the (z+1)(z-1) mapping... $\endgroup$ – user284193 Oct 26 '15 at 21:58
  • $\begingroup$ It will send $-1$ to $0$ and $1$ to $\infty$. The half disk is transformed into a quadrant. $\endgroup$ – Julián Aguirre Oct 26 '15 at 22:15
  • $\begingroup$ Hi Professor Aguirre, I took a break for a few days and am now revisiting this problem. It wasn't coming to me, so I worked out all of the standard mappings for a bit just to see what goes on with each. Then I arrived at something that you had suggested: apply the $z^2$ mapping to get the upper half-disk. Then I found a mapping that maps quadrant III to the upper half-disk. So, I computed its inverse mapping, which is the mapping you gave: (z+1)/(z-1). $\endgroup$ – User001 Oct 29 '15 at 20:48
  • $\begingroup$ Now I have a mapping from the upper half-disk to quadrant III. From here, getting a half-plane seemed a bit problematic, so I first rotated this image 180 degrees counterclockwise, so that it is now in quadrant 1. Now applying the $z^2$ mapping once more gets me the upper half-plane. What do you think? And also, can you please elaborate a little bit on why the $z^4$ mapping doesn't get me the whole disk (from the quarter disk) and instead leaves out an interval [0,1)? Is this an issue of continuity / single-valuedness? Thanks @JuliánAguirre, $\endgroup$ – User001 Oct 29 '15 at 20:48

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