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Find conformal mapping that maps set $A=\{(x,y)\in\mathbb{R}^2 : \operatorname{y\le 0}\}$ to unit disk. I know that such a mapping exists from Riemann Theorem.

Note: I don't want full answer. I expect only some starting point hint.

Thanks

Edit:

I can use rotation by $\frac{-\pi}{2}$ on map:

$$ z \mapsto \frac{z-1}{z+1}$$

So my map looks like that: $$z \mapsto e^{-i\frac{\pi}{2}} \frac{z-1}{z+1}$$

Edit 2:

$$z \mapsto \frac{e^{-i\frac{\pi }{2}}z-1}{e^{-i\frac{\pi }{2}}z+1}$$

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  • $\begingroup$ Doesn't it still require some improvement? Substituting -i gives infinity, obviously outside the unit disc. $\endgroup$ – D777 Feb 21 at 14:34
  • $\begingroup$ Just realized, you have rotated the plane by 90 degrees in the wrong direction. $\endgroup$ – D777 Feb 21 at 14:43
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A map from the complex half-plane of numbers with positive real part to the unit disc is given by $$ z \mapsto \frac{z-1}{z+1}$$ For details on this see an earlier question Mapping half-plane to unit disk?

You should be able to modify this for you purpose; note that your set corresponds to the half-plane with negative imaginary part.

Your attempt is good in that you can indeed compose this map with a rotation. However, note that you have to rotate first and then use the map given above. (You did it the other way round.)

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  • $\begingroup$ I edited my post $\endgroup$ – luka5z Jul 10 '15 at 15:04
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    $\begingroup$ This is almost fine. I expanded my post and also corrected an error; but in fact you understood it the way I should have written it. $\endgroup$ – quid Jul 10 '15 at 15:14
  • $\begingroup$ I edited my post again. $\endgroup$ – luka5z Jul 10 '15 at 15:17
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    $\begingroup$ It is $e^{i \pi/2} z$ not $e^{i \pi z/2} $. Also note that $e^{i \pi/2}=-i$ which allows for some siplification. $\endgroup$ – quid Jul 10 '15 at 15:19
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    $\begingroup$ I see you just correct the first thing. Yes this is fine. You can further simplfy using $e^{i\pi/2}=-i$. $\endgroup$ – quid Jul 10 '15 at 15:20

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