Möbius Transformation Example












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?










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?










    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?










      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






      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











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991372%2fm%25c3%25b6bius-transformation-example%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991372%2fm%25c3%25b6bius-transformation-example%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅