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The wedge is given by $0< \arg(z) < \alpha \pi$, for $0 < \alpha <1$.

If I use the complex logarithm, principal branch $-\pi < 0 \le \pi$, then

$$z \mapsto w = \operatorname{Log}(z)$$ $$ = \ln|z|+ i\arg(z)$$ $$= u+iv$$

so that $0<v<\alpha\pi$, which shows that the wedge maps onto a semi-infinite, horizontal half-strip, with height ranging from $0$ to $\alpha \pi$.

What can I do next? Is the Logarithm mapping even a good start? I usually like it, when I see a circular region that I have to start with, but not for very great reasons other than that I know what the mapping does to a circular region...

Thanks,

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    $\begingroup$ Mapping a strip to the unit circle sounds like a job for a linear fractional transform, but I don't have the means to check that at the moment. $\endgroup$ – IPoiler Sep 29 '15 at 1:42
  • $\begingroup$ Thanks for your comment -- perhaps go to a half-plane ...and then onto the unit disk by simple symmetry arguments for LFTs. I'll work on this now. Thanks, @IPoiler. $\endgroup$ – User001 Sep 29 '15 at 1:52
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This is a standard problem in beginning complex variable, as I learned from Ahlfors. First you apply the function $z\mapsto z^{1/a}$. It’s pretty clear that $a$ must be not too big, I guess less than $2$, and positive, I’m sure. You see that this sends your wedge to the upper half plane. Then any good map like $z\mapsto (iz+1)/(z+i)$ will get the UHP to the disk. To define $z^{1/a}$, you have to use a suitable branch of the logarithm, and that’s why you want $0<a<2$.

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  • $\begingroup$ Wow... @Lubin. That power map is too clever. What if I were stuck at the horizontal half-strip, with height 0 to $\alpha \pi$? I was stuck on this half strip after applying Log initially to the wedge. I'm hoping to proceed from my naive approach, in case there are other clever mappings that I do not get a chance to see and memorize. What do you think? Is it just an overall bad idea to start with the Log mapping? I am currently trying to go from the half-strip to a half-plane, but am not sure how...thanks, $\endgroup$ – User001 Sep 29 '15 at 2:07
  • $\begingroup$ Perhaps the $z^2$ mapping to double the angle, going from a max argument of pi/2 to pi ... so either way...we need a power mapping, I think. $\endgroup$ – User001 Sep 29 '15 at 2:12
  • $\begingroup$ Actually, that doesn't quite work... $\endgroup$ – User001 Sep 29 '15 at 2:15
  • $\begingroup$ I will stick with your suggestion. Thanks so much @Lubin. Have a great night :-) $\endgroup$ – User001 Sep 29 '15 at 2:18
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    $\begingroup$ Yes that branch of the logarithm would be fine. I would have chosen $0\le\arg(z)<2\pi$, but the effect is the same. $\endgroup$ – Lubin Sep 29 '15 at 18:03

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