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find all the mobius transformation that map the units disc$ D$ onto the left half plane

$H^-$ = {$w \in C : Re w <0 $}

My attempts : I know that all the Möbius transformations can be written on the form

$$w=f(z)=\frac{az+b}{cz+d}$$

here i don't know how can i find all the mobius transformation .

as im not getting in my heads pliz help me

José Carlos Santos
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jasmine
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    (1) Mobius transformations send (generalized) circles to (generalized) circles. ("Generalized" circles include straight lines in the plane.) (2) Mobius transformations act sharply 3-transitively - that is, they are determined by where they send any 3 points. – anon Mar 21 '18 at 10:12
  • @anon,,,im not understanding,,ur 2) and 3) ..lines – jasmine Mar 21 '18 at 10:13
  • I only had (1) and (2), not (3). Do you mean you (a) don't understand what I'm saying or (b) don't understand how to apply what I'm saying to your problem? – anon Mar 21 '18 at 10:15
  • (2) Mobius transformations act sharply 3-transitively - that is, they are determined by where they send any 3 points......this line not getting in my head – jasmine Mar 21 '18 at 10:16
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    If you know $(u,v,w)$ and $(f(u),f(v),f(w))$, then $f$ is uniquely determined. (You can't have two different Mobius transformations that send $(u,v,w)$ to the same tuple.) So I'll add a part (3): the cross ratio $$f(z)=\frac{z-u}{z-v}\frac{w-v}{w-u} $$ is a Mobius transformation that maps the ordered triple $(u,v,w)$ to $(0,\infty,1)$. Note $0,\infty,1$ are all on the real axis. Thus, this maps the unit disk to either the upper or lower half-planes (since it maps the unit circle to the real axis). Can you go from there? – anon Mar 21 '18 at 10:18

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Hint: It's a standard result that the Möbius transformations which map the unit disk onto itself are those of the form$$z\mapsto\omega\frac{z-a}{1-\overline az},$$with $|\omega|=1$ (there's a proof here). Now, find one Möbius which maps the left half plane onto the unit disk and you're don (well, almost).

José Carlos Santos
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  • if i take a = -1 which -1<0 then $f(z) = \frac { z-1} {1-z}$ is the required mobius transformation....Is its corrects ??? @ jose carlos sir – jasmine Mar 21 '18 at 10:26
  • @lomber No. That's the constant function $-1$. – José Carlos Santos Mar 21 '18 at 10:27
  • @ Jose carlos sir,,,i was thinking but .. im not getting that .... mobius which maps the left half plane onto the unit disk – jasmine Mar 21 '18 at 10:39
  • @lomber Take$$f(z)=\frac{z+1}{z-1}.$$ – José Carlos Santos Mar 21 '18 at 10:43
  • carlos sir..but according to ur given answer in above that is $f(z) = \frac {z-a}{1-az} $ that mean if i put a= -1 as im getting $ f(z) = \frac{z+1} {1+z}$.....but u have wriiten $f(z) = \frac {z+1}{z-1}$ – jasmine Mar 21 '18 at 10:52
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    @lomber The functions that I mentioned in my answer are the Möbius transformations which map the open unito disk onto itself. – José Carlos Santos Mar 21 '18 at 10:53
  • carlos sir ..one last doubts .... how u take $ f(z) = \frac {z+1}{z-1}$??? is there any procedure and how can u claim that$ f(z) = \frac {z+1}{z-1}$ maps the left half plane onto the unit disk ? – jasmine Mar 21 '18 at 11:16
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    @lomber Sure. The elements of the left half plane are those numbers $z$ which are closer to $-1$ than to $1$. In other words, $|z+1|<|z-1|$. But this means that $\left|\frac{z+1}{z-1}\right|<1$. – José Carlos Santos Mar 21 '18 at 11:18
  • ok,,,,thanks a lots @jose carlos sir ...now i got its – jasmine Mar 21 '18 at 11:19