Find transformation that maps region between two circles to vertical strip












1












$begingroup$


I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.



I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.



All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(










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$endgroup$












  • $begingroup$
    Yes, I'll correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 20:48










  • $begingroup$
    Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
    $endgroup$
    – Song
    Jan 25 at 20:53
















1












$begingroup$


I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.



I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.



All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(










share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, I'll correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 20:48










  • $begingroup$
    Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
    $endgroup$
    – Song
    Jan 25 at 20:53














1












1








1





$begingroup$


I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.



I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.



All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(










share|cite|improve this question











$endgroup$




I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.



I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.



All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(







complex-analysis






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share|cite|improve this question













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edited Jan 25 at 20:48







AstlyDichrar

















asked Jan 25 at 20:33









AstlyDichrarAstlyDichrar

42248




42248












  • $begingroup$
    Yes, I'll correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 20:48










  • $begingroup$
    Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
    $endgroup$
    – Song
    Jan 25 at 20:53


















  • $begingroup$
    Yes, I'll correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 20:48










  • $begingroup$
    Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
    $endgroup$
    – Song
    Jan 25 at 20:53
















$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48




$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48












$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53




$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53










2 Answers
2






active

oldest

votes


















0












$begingroup$

HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 21:06










  • $begingroup$
    Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 21:09



















0












$begingroup$

Hint:



Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
      $endgroup$
      – AstlyDichrar
      Jan 25 at 21:06










    • $begingroup$
      Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
      $endgroup$
      – Ted Shifrin
      Jan 25 at 21:09
















    0












    $begingroup$

    HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
      $endgroup$
      – AstlyDichrar
      Jan 25 at 21:06










    • $begingroup$
      Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
      $endgroup$
      – Ted Shifrin
      Jan 25 at 21:09














    0












    0








    0





    $begingroup$

    HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?






    share|cite|improve this answer









    $endgroup$



    HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 25 at 20:54









    Ted ShifrinTed Shifrin

    64.4k44692




    64.4k44692












    • $begingroup$
      I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
      $endgroup$
      – AstlyDichrar
      Jan 25 at 21:06










    • $begingroup$
      Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
      $endgroup$
      – Ted Shifrin
      Jan 25 at 21:09


















    • $begingroup$
      I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
      $endgroup$
      – AstlyDichrar
      Jan 25 at 21:06










    • $begingroup$
      Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
      $endgroup$
      – Ted Shifrin
      Jan 25 at 21:09
















    $begingroup$
    I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 21:06




    $begingroup$
    I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
    $endgroup$
    – AstlyDichrar
    Jan 25 at 21:06












    $begingroup$
    Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 21:09




    $begingroup$
    Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 21:09











    0












    $begingroup$

    Hint:



    Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint:



      Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:



        Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.






        share|cite|improve this answer









        $endgroup$



        Hint:



        Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 20:54









        BernardBernard

        123k741117




        123k741117






























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