Finding a conformal map from this domain into the unit disc

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Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.



Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.



I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?



Help is appreciated.







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  • Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
    – quasi
    Aug 19 at 9:23










  • In our book we defined a conformal map as 'holomorphic bijection'.
    – Kamil
    Aug 19 at 9:26










  • What book are you using?
    – quasi
    Aug 19 at 9:27










  • 'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
    – Kamil
    Aug 19 at 9:29










  • It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
    – quasi
    Aug 19 at 9:43















up vote
1
down vote

favorite












Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.



Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.



I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?



Help is appreciated.







share|cite|improve this question




















  • Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
    – quasi
    Aug 19 at 9:23










  • In our book we defined a conformal map as 'holomorphic bijection'.
    – Kamil
    Aug 19 at 9:26










  • What book are you using?
    – quasi
    Aug 19 at 9:27










  • 'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
    – Kamil
    Aug 19 at 9:29










  • It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
    – quasi
    Aug 19 at 9:43













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.



Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.



I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?



Help is appreciated.







share|cite|improve this question












Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.



Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.



I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?



Help is appreciated.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 19 at 9:12









Kamil

1,90421237




1,90421237











  • Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
    – quasi
    Aug 19 at 9:23










  • In our book we defined a conformal map as 'holomorphic bijection'.
    – Kamil
    Aug 19 at 9:26










  • What book are you using?
    – quasi
    Aug 19 at 9:27










  • 'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
    – Kamil
    Aug 19 at 9:29










  • It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
    – quasi
    Aug 19 at 9:43

















  • Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
    – quasi
    Aug 19 at 9:23










  • In our book we defined a conformal map as 'holomorphic bijection'.
    – Kamil
    Aug 19 at 9:26










  • What book are you using?
    – quasi
    Aug 19 at 9:27










  • 'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
    – Kamil
    Aug 19 at 9:29










  • It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
    – quasi
    Aug 19 at 9:43
















Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23




Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23












In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26




In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26












What book are you using?
– quasi
Aug 19 at 9:27




What book are you using?
– quasi
Aug 19 at 9:27












'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29




'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29












It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43





It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43











2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.






share|cite|improve this answer




















  • Great. Could you explain to me how you arrived at this map? What was your logic?
    – Kamil
    Aug 19 at 11:02










  • what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
    – Nosrati
    Aug 19 at 11:06










  • The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
    – Kamil
    Aug 19 at 11:39











  • yes, it is.....
    – Nosrati
    Aug 19 at 11:45

















up vote
-1
down vote













It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form






share|cite|improve this answer




















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Hint: With conformal mapping
    $$w=left(dfracz+1z-1right)^2$$
    you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.






    share|cite|improve this answer




















    • Great. Could you explain to me how you arrived at this map? What was your logic?
      – Kamil
      Aug 19 at 11:02










    • what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
      – Nosrati
      Aug 19 at 11:06










    • The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
      – Kamil
      Aug 19 at 11:39











    • yes, it is.....
      – Nosrati
      Aug 19 at 11:45














    up vote
    1
    down vote



    accepted










    Hint: With conformal mapping
    $$w=left(dfracz+1z-1right)^2$$
    you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.






    share|cite|improve this answer




















    • Great. Could you explain to me how you arrived at this map? What was your logic?
      – Kamil
      Aug 19 at 11:02










    • what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
      – Nosrati
      Aug 19 at 11:06










    • The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
      – Kamil
      Aug 19 at 11:39











    • yes, it is.....
      – Nosrati
      Aug 19 at 11:45












    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Hint: With conformal mapping
    $$w=left(dfracz+1z-1right)^2$$
    you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.






    share|cite|improve this answer












    Hint: With conformal mapping
    $$w=left(dfracz+1z-1right)^2$$
    you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 19 at 9:44









    Nosrati

    20.7k41644




    20.7k41644











    • Great. Could you explain to me how you arrived at this map? What was your logic?
      – Kamil
      Aug 19 at 11:02










    • what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
      – Nosrati
      Aug 19 at 11:06










    • The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
      – Kamil
      Aug 19 at 11:39











    • yes, it is.....
      – Nosrati
      Aug 19 at 11:45
















    • Great. Could you explain to me how you arrived at this map? What was your logic?
      – Kamil
      Aug 19 at 11:02










    • what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
      – Nosrati
      Aug 19 at 11:06










    • The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
      – Kamil
      Aug 19 at 11:39











    • yes, it is.....
      – Nosrati
      Aug 19 at 11:45















    Great. Could you explain to me how you arrived at this map? What was your logic?
    – Kamil
    Aug 19 at 11:02




    Great. Could you explain to me how you arrived at this map? What was your logic?
    – Kamil
    Aug 19 at 11:02












    what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
    – Nosrati
    Aug 19 at 11:06




    what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
    – Nosrati
    Aug 19 at 11:06












    The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
    – Kamil
    Aug 19 at 11:39





    The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
    – Kamil
    Aug 19 at 11:39













    yes, it is.....
    – Nosrati
    Aug 19 at 11:45




    yes, it is.....
    – Nosrati
    Aug 19 at 11:45










    up vote
    -1
    down vote













    It seems that your region $A$ is the limit as $R to + infty$ of
    $$
    A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
    $$
    I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form






    share|cite|improve this answer
























      up vote
      -1
      down vote













      It seems that your region $A$ is the limit as $R to + infty$ of
      $$
      A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
      $$
      I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form






      share|cite|improve this answer






















        up vote
        -1
        down vote










        up vote
        -1
        down vote









        It seems that your region $A$ is the limit as $R to + infty$ of
        $$
        A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
        $$
        I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form






        share|cite|improve this answer












        It seems that your region $A$ is the limit as $R to + infty$ of
        $$
        A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
        $$
        I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 19 at 9:26









        Laurent

        127




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