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I am trying to find a complex mapping from the left-hand side of the plane to the unit disk, such that $$\lvert z_1\rvert\gt\lvert z_2\rvert\iff \lvert f(z_1)\rvert\gt\lvert f(z_2)\rvert$$ I wasn't sure what to call this property, so the title is a bit weird. According to this answer

there is a holomorphic bijection from the open unit disk onto a region $U$ if and only if $U$ is simply connected and the complement of $U$ (in the Riemann sphere) has at least two points.

The Möbius transformation $w=\frac{z+1}{z-1}$ maps the left side into the unit disk, but it doesn't satisfy that property. I tried some other forms of the bilinear transformation without any luck, and I suspect that such transformation may not exist. Is my suspicion correct?

edm
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polfosol
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1 Answers1

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In the title you didn't say 'onto' but in the question you have put this extra condition. Since $f$ is onto there exists $z_2$ such that $w_2=f(z_2)=0$. But then $|z_1|<|z_2|$ implies $|w_1|<0$, a contradiction.

Kavi Rama Murthy
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  • I don't get what you mean. Please ignore my abuse of words. The condition restricts $f$ to the ones that $f(0)=0$ and I don't see any contradiction – polfosol Oct 16 '18 at 11:54
  • @polfosol 0 is not in the domain of the function (left half plane)... – Albert Oct 16 '18 at 11:58
  • @Glougloubarbaki does it make any difference if I change the domain to left half of the plane including the imaginary axis? – polfosol Oct 16 '18 at 12:01
  • @polfosol kind of, because what's a holomorphic function on a set that is not open? but I suspect that in any case there are no such functions even without any condition like onto or $f(0)=0$ – Albert Oct 16 '18 at 12:03
  • @Glougloubarbaki I suspected that too. But I am not sure yet – polfosol Oct 16 '18 at 12:07