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Conformal Mapping Question about Mobius Maps, mapping 1 region to another

0

$begingroup$

I am trying to map the region G, > (2)^0.5 to the infinite vertical strip at x = +/- pi.

I have started by using a Mobius Map which sends the two common points of the circles to 0 and infinity (z=+/- 1), namely, using the map f(z)=(z-1)/(z+1). This maps the unit circle to the positive imaginary axis and should map the other circle to a circle. However, when I calculate the image of the other circle, I don't get a region that is one of the axes as one would expect...

If Mobius maps take circlines to circlines, why is this the case?

complex-analysis conformal-geometry mobius-transformation

share|cite|improve this question

edited Mar 26 at 18:12

MathematicianP

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Such a Mobius transform doesn't exist: the two boundary circles have two common points, while the two boundary lines have one common point ($infty$). You can find a mapping between the two given regions in the form $a ln((1 + z)/(1 - z)) + b$.
  $endgroup$
  – Maxim
  Mar 26 at 18:29
  

  
  
  
  


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  $begingroup$
  the two boundary lines can have 2 common points though - 0 and ∞?
  $endgroup$
  – MathematicianP
  Mar 26 at 19:42
  

  
  
  
  


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  $begingroup$
  Which straight lines are we talking about? The vertical strip $-pi < operatornameRe z < pi$ is bounded by the lines $operatornameRe z = pm pi$. They do not go through zero.
  $endgroup$
  – Maxim
  Mar 26 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh i see, but shouldn't the mobius map (z-1)/(z+1) take the two circles to circles? I don't think it does for |z+i|=(2)^0.5
  $endgroup$
  – MathematicianP
  Mar 26 at 20:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It does. $(z - 1)/(z + 1)$ maps $|z + i| = sqrt 2$ to the straight line that goes through zero at the same angle with the imaginary axis as the angle between the two circles at $1$.
  $endgroup$
  – Maxim
  Mar 26 at 20:18
  

  
  
  
  

add a comment | 

0

$begingroup$

I am trying to map the region G, > (2)^0.5 to the infinite vertical strip at x = +/- pi.

I have started by using a Mobius Map which sends the two common points of the circles to 0 and infinity (z=+/- 1), namely, using the map f(z)=(z-1)/(z+1). This maps the unit circle to the positive imaginary axis and should map the other circle to a circle. However, when I calculate the image of the other circle, I don't get a region that is one of the axes as one would expect...

If Mobius maps take circlines to circlines, why is this the case?

complex-analysis conformal-geometry mobius-transformation

share|cite|improve this question

edited Mar 26 at 18:12

MathematicianP

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Such a Mobius transform doesn't exist: the two boundary circles have two common points, while the two boundary lines have one common point ($infty$). You can find a mapping between the two given regions in the form $a ln((1 + z)/(1 - z)) + b$.
  $endgroup$
  – Maxim
  Mar 26 at 18:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the two boundary lines can have 2 common points though - 0 and ∞?
  $endgroup$
  – MathematicianP
  Mar 26 at 19:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which straight lines are we talking about? The vertical strip $-pi < operatornameRe z < pi$ is bounded by the lines $operatornameRe z = pm pi$. They do not go through zero.
  $endgroup$
  – Maxim
  Mar 26 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh i see, but shouldn't the mobius map (z-1)/(z+1) take the two circles to circles? I don't think it does for |z+i|=(2)^0.5
  $endgroup$
  – MathematicianP
  Mar 26 at 20:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It does. $(z - 1)/(z + 1)$ maps $|z + i| = sqrt 2$ to the straight line that goes through zero at the same angle with the imaginary axis as the angle between the two circles at $1$.
  $endgroup$
  – Maxim
  Mar 26 at 20:18
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to map the region G, > (2)^0.5 to the infinite vertical strip at x = +/- pi.

I have started by using a Mobius Map which sends the two common points of the circles to 0 and infinity (z=+/- 1), namely, using the map f(z)=(z-1)/(z+1). This maps the unit circle to the positive imaginary axis and should map the other circle to a circle. However, when I calculate the image of the other circle, I don't get a region that is one of the axes as one would expect...

If Mobius maps take circlines to circlines, why is this the case?

complex-analysis conformal-geometry mobius-transformation

share|cite|improve this question

edited Mar 26 at 18:12

MathematicianP

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

$endgroup$

share|cite|improve this question

edited Mar 26 at 18:12

MathematicianP

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

share|cite|improve this question

edited Mar 26 at 18:12

MathematicianP

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416

asked Mar 26 at 16:03

MathematicianPMathematicianP

3416