Möbius Transformation Example












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I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?






complex-analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 9 '18 at 13:33









      John SmithJohn Smith

      678




      678






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          Is $z mapsto frac{az+b}{z+1}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:40






          • 1




            $begingroup$
            Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:44












          • $begingroup$
            So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:47






          • 1




            $begingroup$
            No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:50












          • $begingroup$
            Ah ups. Okay thank you!
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:50



















          1












          $begingroup$

          A mobius transformation may be given as a function
          $$
          f(z) = frac{az+b}{cz + d}
          $$
          with some niceness restrictions on $a, b, c, d$.



          If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does this imply that $a = 0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:43





















          1












          $begingroup$

          When working with a Möbius transformation
          $$f(x) = frac{ax+b}{cx+d}
          $$
          you are allowed to do just a little bit of "infinity arithmetic". For example:
          $$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
          $$
          which equals $infty$ when $c=0$.



          And then a simple computation gives
          $$f^{-1}(x) = frac{da - b}{-cx+a}
          $$
          and so
          $$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
          $$
          which, again, equals $infty$ when $c=0$.



          You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            I come up with an alternate answer where you can easily get rid of the limits.
            Let the transformation be
            $w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:



            $0=frac{2ai+b}{2ci+d},
            infty =frac{-a+b}{-c+d} $             ..We get $-c+d =0$
            And the equation is $b=-4id$



            Now solve for $a,b,c,d$. We get
            $a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
            Put these values of $a,b,c,d$ in the transformation and simplify we get
            $w=frac{z+2/i}{z/2+1/2} $.
            And hence $w= frac{2z-4i}{z+1}$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
              $endgroup$
              – Henry
              Nov 9 '18 at 14:22










            • $begingroup$
              Very good advice!
              $endgroup$
              – John Smith
              Nov 9 '18 at 14:25











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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Is $z mapsto frac{az+b}{z+1}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:40






            • 1




              $begingroup$
              Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:44












            • $begingroup$
              So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:47






            • 1




              $begingroup$
              No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:50












            • $begingroup$
              Ah ups. Okay thank you!
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:50
















            1












            $begingroup$

            Is $z mapsto frac{az+b}{z+1}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:40






            • 1




              $begingroup$
              Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:44












            • $begingroup$
              So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:47






            • 1




              $begingroup$
              No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:50












            • $begingroup$
              Ah ups. Okay thank you!
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:50














            1












            1








            1





            $begingroup$

            Is $z mapsto frac{az+b}{z+1}$






            share|cite|improve this answer









            $endgroup$



            Is $z mapsto frac{az+b}{z+1}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 9 '18 at 13:37









            Richard MartinRichard Martin

            1,61318




            1,61318












            • $begingroup$
              Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:40






            • 1




              $begingroup$
              Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:44












            • $begingroup$
              So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:47






            • 1




              $begingroup$
              No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:50












            • $begingroup$
              Ah ups. Okay thank you!
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:50


















            • $begingroup$
              Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:40






            • 1




              $begingroup$
              Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:44












            • $begingroup$
              So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:47






            • 1




              $begingroup$
              No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
              $endgroup$
              – Richard Martin
              Nov 9 '18 at 13:50












            • $begingroup$
              Ah ups. Okay thank you!
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:50
















            $begingroup$
            Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:40




            $begingroup$
            Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:40




            1




            1




            $begingroup$
            Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:44






            $begingroup$
            Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:44














            $begingroup$
            So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:47




            $begingroup$
            So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:47




            1




            1




            $begingroup$
            No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:50






            $begingroup$
            No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
            $endgroup$
            – Richard Martin
            Nov 9 '18 at 13:50














            $begingroup$
            Ah ups. Okay thank you!
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:50




            $begingroup$
            Ah ups. Okay thank you!
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:50











            1












            $begingroup$

            A mobius transformation may be given as a function
            $$
            f(z) = frac{az+b}{cz + d}
            $$
            with some niceness restrictions on $a, b, c, d$.



            If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Does this imply that $a = 0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:43


















            1












            $begingroup$

            A mobius transformation may be given as a function
            $$
            f(z) = frac{az+b}{cz + d}
            $$
            with some niceness restrictions on $a, b, c, d$.



            If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Does this imply that $a = 0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:43
















            1












            1








            1





            $begingroup$

            A mobius transformation may be given as a function
            $$
            f(z) = frac{az+b}{cz + d}
            $$
            with some niceness restrictions on $a, b, c, d$.



            If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.






            share|cite|improve this answer









            $endgroup$



            A mobius transformation may be given as a function
            $$
            f(z) = frac{az+b}{cz + d}
            $$
            with some niceness restrictions on $a, b, c, d$.



            If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 9 '18 at 13:39









            ArthurArthur

            116k7116199




            116k7116199












            • $begingroup$
              Does this imply that $a = 0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:43




















            • $begingroup$
              Does this imply that $a = 0$?
              $endgroup$
              – John Smith
              Nov 9 '18 at 13:43


















            $begingroup$
            Does this imply that $a = 0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:43






            $begingroup$
            Does this imply that $a = 0$?
            $endgroup$
            – John Smith
            Nov 9 '18 at 13:43













            1












            $begingroup$

            When working with a Möbius transformation
            $$f(x) = frac{ax+b}{cx+d}
            $$
            you are allowed to do just a little bit of "infinity arithmetic". For example:
            $$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
            $$
            which equals $infty$ when $c=0$.



            And then a simple computation gives
            $$f^{-1}(x) = frac{da - b}{-cx+a}
            $$
            and so
            $$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
            $$
            which, again, equals $infty$ when $c=0$.



            You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              When working with a Möbius transformation
              $$f(x) = frac{ax+b}{cx+d}
              $$
              you are allowed to do just a little bit of "infinity arithmetic". For example:
              $$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
              $$
              which equals $infty$ when $c=0$.



              And then a simple computation gives
              $$f^{-1}(x) = frac{da - b}{-cx+a}
              $$
              and so
              $$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
              $$
              which, again, equals $infty$ when $c=0$.



              You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                When working with a Möbius transformation
                $$f(x) = frac{ax+b}{cx+d}
                $$
                you are allowed to do just a little bit of "infinity arithmetic". For example:
                $$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
                $$
                which equals $infty$ when $c=0$.



                And then a simple computation gives
                $$f^{-1}(x) = frac{da - b}{-cx+a}
                $$
                and so
                $$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
                $$
                which, again, equals $infty$ when $c=0$.



                You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.






                share|cite|improve this answer









                $endgroup$



                When working with a Möbius transformation
                $$f(x) = frac{ax+b}{cx+d}
                $$
                you are allowed to do just a little bit of "infinity arithmetic". For example:
                $$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
                $$
                which equals $infty$ when $c=0$.



                And then a simple computation gives
                $$f^{-1}(x) = frac{da - b}{-cx+a}
                $$
                and so
                $$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
                $$
                which, again, equals $infty$ when $c=0$.



                You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 9 '18 at 13:45









                Lee MosherLee Mosher

                49.8k33686




                49.8k33686























                    1












                    $begingroup$

                    I come up with an alternate answer where you can easily get rid of the limits.
                    Let the transformation be
                    $w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:



                    $0=frac{2ai+b}{2ci+d},
                    infty =frac{-a+b}{-c+d} $                     ..We get $-c+d =0$
                    And the equation is $b=-4id$



                    Now solve for $a,b,c,d$. We get
                    $a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
                    Put these values of $a,b,c,d$ in the transformation and simplify we get
                    $w=frac{z+2/i}{z/2+1/2} $.
                    And hence $w= frac{2z-4i}{z+1}$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                      $endgroup$
                      – Henry
                      Nov 9 '18 at 14:22










                    • $begingroup$
                      Very good advice!
                      $endgroup$
                      – John Smith
                      Nov 9 '18 at 14:25
















                    1












                    $begingroup$

                    I come up with an alternate answer where you can easily get rid of the limits.
                    Let the transformation be
                    $w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:



                    $0=frac{2ai+b}{2ci+d},
                    infty =frac{-a+b}{-c+d} $                     ..We get $-c+d =0$
                    And the equation is $b=-4id$



                    Now solve for $a,b,c,d$. We get
                    $a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
                    Put these values of $a,b,c,d$ in the transformation and simplify we get
                    $w=frac{z+2/i}{z/2+1/2} $.
                    And hence $w= frac{2z-4i}{z+1}$.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                      $endgroup$
                      – Henry
                      Nov 9 '18 at 14:22










                    • $begingroup$
                      Very good advice!
                      $endgroup$
                      – John Smith
                      Nov 9 '18 at 14:25














                    1












                    1








                    1





                    $begingroup$

                    I come up with an alternate answer where you can easily get rid of the limits.
                    Let the transformation be
                    $w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:



                    $0=frac{2ai+b}{2ci+d},
                    infty =frac{-a+b}{-c+d} $                     ..We get $-c+d =0$
                    And the equation is $b=-4id$



                    Now solve for $a,b,c,d$. We get
                    $a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
                    Put these values of $a,b,c,d$ in the transformation and simplify we get
                    $w=frac{z+2/i}{z/2+1/2} $.
                    And hence $w= frac{2z-4i}{z+1}$.






                    share|cite|improve this answer











                    $endgroup$



                    I come up with an alternate answer where you can easily get rid of the limits.
                    Let the transformation be
                    $w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:



                    $0=frac{2ai+b}{2ci+d},
                    infty =frac{-a+b}{-c+d} $                     ..We get $-c+d =0$
                    And the equation is $b=-4id$



                    Now solve for $a,b,c,d$. We get
                    $a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
                    Put these values of $a,b,c,d$ in the transformation and simplify we get
                    $w=frac{z+2/i}{z/2+1/2} $.
                    And hence $w= frac{2z-4i}{z+1}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 9 at 15:43









                    Thomas Shelby

                    3,7092525




                    3,7092525










                    answered Nov 9 '18 at 14:20









                    HenryHenry

                    327




                    327








                    • 1




                      $begingroup$
                      And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                      $endgroup$
                      – Henry
                      Nov 9 '18 at 14:22










                    • $begingroup$
                      Very good advice!
                      $endgroup$
                      – John Smith
                      Nov 9 '18 at 14:25














                    • 1




                      $begingroup$
                      And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                      $endgroup$
                      – Henry
                      Nov 9 '18 at 14:22










                    • $begingroup$
                      Very good advice!
                      $endgroup$
                      – John Smith
                      Nov 9 '18 at 14:25








                    1




                    1




                    $begingroup$
                    And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                    $endgroup$
                    – Henry
                    Nov 9 '18 at 14:22




                    $begingroup$
                    And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
                    $endgroup$
                    – Henry
                    Nov 9 '18 at 14:22












                    $begingroup$
                    Very good advice!
                    $endgroup$
                    – John Smith
                    Nov 9 '18 at 14:25




                    $begingroup$
                    Very good advice!
                    $endgroup$
                    – John Smith
                    Nov 9 '18 at 14:25


















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