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Image of Cayley Transform

0

$begingroup$

I know that the Cayley-Transform $varphi(z)=fracz-iz+i$ maps stripes like $zinmathbb C: 0<Im(z)<b$ onto certain "sickle" looking domains. For example, Cayley maps $zinmathbb C: 0<Im(z)<pi$ onto $D_1/2(1/2)$ (the Disk of radius $1/2$ centered at $1/2$).

I am interested in how one could parameterize the image of the general stripe as mentioned above. If I'm not mistaken it is of the form $D_a(1-a)$ for some $ainmathbb R^+$ that depends on $b$. Alternatively, I would be interested in the image of the line $zinmathbb C: Im(z)=b$ under the Calyey-Transform, which must be some circle line centered at the real axis. I have already tried plugging in some $x+icdot b$ into Cayley to find $a$ but without any result so far.

I thought this would be some elementary knowledge but so far I cannot seem to find this result anywhere so that's why I am asking around here. Does anyone happen to know?

complex-analysis

share|cite|improve this question

asked Mar 26 at 17:11

RedLanternRedLantern

517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the Cayley-Transform is a special case of a Mobius Transform, which map circles/lines to circles/lines. To express your image region, use the fact Mobius Transforms are determined by where three points are mapped to get the radius and center of the image circle(s)/line(s) and parameterize from there
  $endgroup$
  – Brevan Ellefsen
  Mar 27 at 8:58
  

  
  
  
  

add a comment | 

0

$begingroup$

I know that the Cayley-Transform $varphi(z)=fracz-iz+i$ maps stripes like $zinmathbb C: 0<Im(z)<b$ onto certain "sickle" looking domains. For example, Cayley maps $zinmathbb C: 0<Im(z)<pi$ onto $D_1/2(1/2)$ (the Disk of radius $1/2$ centered at $1/2$).

I am interested in how one could parameterize the image of the general stripe as mentioned above. If I'm not mistaken it is of the form $D_a(1-a)$ for some $ainmathbb R^+$ that depends on $b$. Alternatively, I would be interested in the image of the line $zinmathbb C: Im(z)=b$ under the Calyey-Transform, which must be some circle line centered at the real axis. I have already tried plugging in some $x+icdot b$ into Cayley to find $a$ but without any result so far.

I thought this would be some elementary knowledge but so far I cannot seem to find this result anywhere so that's why I am asking around here. Does anyone happen to know?

complex-analysis

share|cite|improve this question

asked Mar 26 at 17:11

RedLanternRedLantern

517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the Cayley-Transform is a special case of a Mobius Transform, which map circles/lines to circles/lines. To express your image region, use the fact Mobius Transforms are determined by where three points are mapped to get the radius and center of the image circle(s)/line(s) and parameterize from there
  $endgroup$
  – Brevan Ellefsen
  Mar 27 at 8:58
  

  
  
  
  

add a comment | 

$begingroup$

I know that the Cayley-Transform $varphi(z)=fracz-iz+i$ maps stripes like $zinmathbb C: 0<Im(z)<b$ onto certain "sickle" looking domains. For example, Cayley maps $zinmathbb C: 0<Im(z)<pi$ onto $D_1/2(1/2)$ (the Disk of radius $1/2$ centered at $1/2$).

I am interested in how one could parameterize the image of the general stripe as mentioned above. If I'm not mistaken it is of the form $D_a(1-a)$ for some $ainmathbb R^+$ that depends on $b$. Alternatively, I would be interested in the image of the line $zinmathbb C: Im(z)=b$ under the Calyey-Transform, which must be some circle line centered at the real axis. I have already tried plugging in some $x+icdot b$ into Cayley to find $a$ but without any result so far.

I thought this would be some elementary knowledge but so far I cannot seem to find this result anywhere so that's why I am asking around here. Does anyone happen to know?

complex-analysis

share|cite|improve this question

asked Mar 26 at 17:11

RedLanternRedLantern

517

$endgroup$

share|cite|improve this question

asked Mar 26 at 17:11

RedLanternRedLantern

517

share|cite|improve this question

asked Mar 26 at 17:11

RedLanternRedLantern

517

asked Mar 26 at 17:11

RedLanternRedLantern

517

asked Mar 26 at 17:11

RedLanternRedLantern

517