Show that $ad-bc ne 0$ for a composition of two Mobius transformations.












0














I could prove that a composition of two Mobius transformations is again a Mobius transformation. Let be $T(z)=dfrac{a_1z+b_1}{c_1z+d_1}, (a_1d_1-c_1b_1 ne 0)$ and $S(z)=dfrac{a_2z+b_2}{c_2z+d_2}, (a_2d_2-c_2b_2 ne 0)$, so $S[T(z)]=dfrac{a_3z+b_3}{c_3z+d_3}$. But I can't prove whether $a_3d_3-c_3b_3 ne 0)$. The book says $a_3d_3-c_3b_3 = (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ but in order that to hold also must $c_2b_2(c_1b_1+a_1d_1)=0$ hold to make $a_3d_3-c_3b_3 - (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ to be zero. I did calculations thrice! Where am I doing wrong?






complex-analysis proof-verification mobius-transformation







share|cite|improve this question




















  • 1




    Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
    – zwim
    Nov 20 '18 at 21:39










  • @zwim, it solve my problem. thanks a lot :)
    – 72D
    Nov 20 '18 at 22:09
















0














I could prove that a composition of two Mobius transformations is again a Mobius transformation. Let be $T(z)=dfrac{a_1z+b_1}{c_1z+d_1}, (a_1d_1-c_1b_1 ne 0)$ and $S(z)=dfrac{a_2z+b_2}{c_2z+d_2}, (a_2d_2-c_2b_2 ne 0)$, so $S[T(z)]=dfrac{a_3z+b_3}{c_3z+d_3}$. But I can't prove whether $a_3d_3-c_3b_3 ne 0)$. The book says $a_3d_3-c_3b_3 = (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ but in order that to hold also must $c_2b_2(c_1b_1+a_1d_1)=0$ hold to make $a_3d_3-c_3b_3 - (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ to be zero. I did calculations thrice! Where am I doing wrong?






complex-analysis proof-verification mobius-transformation







share|cite|improve this question




















  • 1




    Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
    – zwim
    Nov 20 '18 at 21:39










  • @zwim, it solve my problem. thanks a lot :)
    – 72D
    Nov 20 '18 at 22:09














0












0








0







I could prove that a composition of two Mobius transformations is again a Mobius transformation. Let be $T(z)=dfrac{a_1z+b_1}{c_1z+d_1}, (a_1d_1-c_1b_1 ne 0)$ and $S(z)=dfrac{a_2z+b_2}{c_2z+d_2}, (a_2d_2-c_2b_2 ne 0)$, so $S[T(z)]=dfrac{a_3z+b_3}{c_3z+d_3}$. But I can't prove whether $a_3d_3-c_3b_3 ne 0)$. The book says $a_3d_3-c_3b_3 = (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ but in order that to hold also must $c_2b_2(c_1b_1+a_1d_1)=0$ hold to make $a_3d_3-c_3b_3 - (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ to be zero. I did calculations thrice! Where am I doing wrong?






complex-analysis proof-verification mobius-transformation







share|cite|improve this question















I could prove that a composition of two Mobius transformations is again a Mobius transformation. Let be $T(z)=dfrac{a_1z+b_1}{c_1z+d_1}, (a_1d_1-c_1b_1 ne 0)$ and $S(z)=dfrac{a_2z+b_2}{c_2z+d_2}, (a_2d_2-c_2b_2 ne 0)$, so $S[T(z)]=dfrac{a_3z+b_3}{c_3z+d_3}$. But I can't prove whether $a_3d_3-c_3b_3 ne 0)$. The book says $a_3d_3-c_3b_3 = (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ but in order that to hold also must $c_2b_2(c_1b_1+a_1d_1)=0$ hold to make $a_3d_3-c_3b_3 - (a_1d_1-c_1b_1)(a_2d_2-c_2b_2)$ to be zero. I did calculations thrice! Where am I doing wrong?






complex-analysis proof-verification mobius-transformation




complex-analysis proof-verification mobius-transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 21:25

























asked Nov 20 '18 at 21:20









72D

572116




572116








  • 1




    Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
    – zwim
    Nov 20 '18 at 21:39










  • @zwim, it solve my problem. thanks a lot :)
    – 72D
    Nov 20 '18 at 22:09














  • 1




    Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
    – zwim
    Nov 20 '18 at 21:39










  • @zwim, it solve my problem. thanks a lot :)
    – 72D
    Nov 20 '18 at 22:09








1




1




Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
– zwim
Nov 20 '18 at 21:39




Probably a calculation error anyway. bpaste.net/show/156c07aa4363 check all the steps there.
– zwim
Nov 20 '18 at 21:39












@zwim, it solve my problem. thanks a lot :)
– 72D
Nov 20 '18 at 22:09




@zwim, it solve my problem. thanks a lot :)
– 72D
Nov 20 '18 at 22:09










2 Answers
2






active

oldest

votes


















0














Note thatbegin{align}Sbigl(T(z)bigr)&=frac{a_2frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2frac{a_1z+b_1}{c_1z+d_1}+d_2}\&=frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\&=frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}end{align}



So,begin{align}begin{vmatrix}a_3&b_3\c_3&d_3end{vmatrix}&=begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\c_2a_1+d_2c_1&c_2b_1+d_2b_1end{vmatrix}\&=detleft(begin{bmatrix}a_2&b_2\c_2&d_2end{bmatrix}.begin{bmatrix}a_1&b_1\c_1&d_1end{bmatrix}right)\&=begin{vmatrix}a_2&b_2\c_2&d_2end{vmatrix}.begin{vmatrix}a_1&b_1\c_1&d_1end{vmatrix}\&neq0.end{align}






share|cite|improve this answer























  • In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
    – 72D
    Nov 20 '18 at 21:37








  • 1




    I have expanded and corrected my answer. Is it clear now?
    – José Carlos Santos
    Nov 20 '18 at 22:06



















0














There is a natural identification of these transformations with that of $2times 2$ matrices with non-zero determinant:



$$frac{az+b}{cz+d}longleftrightarrow begin{pmatrix}a&b\
c&d
end{pmatrix}.$$



You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.






share|cite|improve this answer





















    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%2f3006920%2fshow-that-ad-bc-ne-0-for-a-composition-of-two-mobius-transformations%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Note thatbegin{align}Sbigl(T(z)bigr)&=frac{a_2frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2frac{a_1z+b_1}{c_1z+d_1}+d_2}\&=frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\&=frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}end{align}



    So,begin{align}begin{vmatrix}a_3&b_3\c_3&d_3end{vmatrix}&=begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\c_2a_1+d_2c_1&c_2b_1+d_2b_1end{vmatrix}\&=detleft(begin{bmatrix}a_2&b_2\c_2&d_2end{bmatrix}.begin{bmatrix}a_1&b_1\c_1&d_1end{bmatrix}right)\&=begin{vmatrix}a_2&b_2\c_2&d_2end{vmatrix}.begin{vmatrix}a_1&b_1\c_1&d_1end{vmatrix}\&neq0.end{align}






    share|cite|improve this answer























    • In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
      – 72D
      Nov 20 '18 at 21:37








    • 1




      I have expanded and corrected my answer. Is it clear now?
      – José Carlos Santos
      Nov 20 '18 at 22:06
















    0














    Note thatbegin{align}Sbigl(T(z)bigr)&=frac{a_2frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2frac{a_1z+b_1}{c_1z+d_1}+d_2}\&=frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\&=frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}end{align}



    So,begin{align}begin{vmatrix}a_3&b_3\c_3&d_3end{vmatrix}&=begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\c_2a_1+d_2c_1&c_2b_1+d_2b_1end{vmatrix}\&=detleft(begin{bmatrix}a_2&b_2\c_2&d_2end{bmatrix}.begin{bmatrix}a_1&b_1\c_1&d_1end{bmatrix}right)\&=begin{vmatrix}a_2&b_2\c_2&d_2end{vmatrix}.begin{vmatrix}a_1&b_1\c_1&d_1end{vmatrix}\&neq0.end{align}






    share|cite|improve this answer























    • In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
      – 72D
      Nov 20 '18 at 21:37








    • 1




      I have expanded and corrected my answer. Is it clear now?
      – José Carlos Santos
      Nov 20 '18 at 22:06














    0












    0








    0






    Note thatbegin{align}Sbigl(T(z)bigr)&=frac{a_2frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2frac{a_1z+b_1}{c_1z+d_1}+d_2}\&=frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\&=frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}end{align}



    So,begin{align}begin{vmatrix}a_3&b_3\c_3&d_3end{vmatrix}&=begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\c_2a_1+d_2c_1&c_2b_1+d_2b_1end{vmatrix}\&=detleft(begin{bmatrix}a_2&b_2\c_2&d_2end{bmatrix}.begin{bmatrix}a_1&b_1\c_1&d_1end{bmatrix}right)\&=begin{vmatrix}a_2&b_2\c_2&d_2end{vmatrix}.begin{vmatrix}a_1&b_1\c_1&d_1end{vmatrix}\&neq0.end{align}






    share|cite|improve this answer














    Note thatbegin{align}Sbigl(T(z)bigr)&=frac{a_2frac{a_1z+b_1}{c_1z+d_1}+b_2}{c_2frac{a_1z+b_1}{c_1z+d_1}+d_2}\&=frac{a_2(a_1z+b_1)+b_2(c_1z+d_1)}{c_2(a_1z+b_1)+d_2(c_1z+d_1)}\&=frac{(a_2a_1+b_2c_1)z+a_2b_1+b_2d_1}{(c_2a_1+d_2c_1)z+c_2b_1+d_2b_1}end{align}



    So,begin{align}begin{vmatrix}a_3&b_3\c_3&d_3end{vmatrix}&=begin{vmatrix}a_2a_1+b_2c_1&a_2b_1+b_2d_1\c_2a_1+d_2c_1&c_2b_1+d_2b_1end{vmatrix}\&=detleft(begin{bmatrix}a_2&b_2\c_2&d_2end{bmatrix}.begin{bmatrix}a_1&b_1\c_1&d_1end{bmatrix}right)\&=begin{vmatrix}a_2&b_2\c_2&d_2end{vmatrix}.begin{vmatrix}a_1&b_1\c_1&d_1end{vmatrix}\&neq0.end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 20 '18 at 22:05

























    answered Nov 20 '18 at 21:33









    José Carlos Santos

    151k22123224




    151k22123224












    • In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
      – 72D
      Nov 20 '18 at 21:37








    • 1




      I have expanded and corrected my answer. Is it clear now?
      – José Carlos Santos
      Nov 20 '18 at 22:06


















    • In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
      – 72D
      Nov 20 '18 at 21:37








    • 1




      I have expanded and corrected my answer. Is it clear now?
      – José Carlos Santos
      Nov 20 '18 at 22:06
















    In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
    – 72D
    Nov 20 '18 at 21:37






    In your answer (for example) how $a_3 =a_1a_2+b_1c_2$? I calculate $S(T(z))$ by direct inserting..
    – 72D
    Nov 20 '18 at 21:37






    1




    1




    I have expanded and corrected my answer. Is it clear now?
    – José Carlos Santos
    Nov 20 '18 at 22:06




    I have expanded and corrected my answer. Is it clear now?
    – José Carlos Santos
    Nov 20 '18 at 22:06











    0














    There is a natural identification of these transformations with that of $2times 2$ matrices with non-zero determinant:



    $$frac{az+b}{cz+d}longleftrightarrow begin{pmatrix}a&b\
    c&d
    end{pmatrix}.$$



    You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.






    share|cite|improve this answer


























      0














      There is a natural identification of these transformations with that of $2times 2$ matrices with non-zero determinant:



      $$frac{az+b}{cz+d}longleftrightarrow begin{pmatrix}a&b\
      c&d
      end{pmatrix}.$$



      You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.






      share|cite|improve this answer
























        0












        0








        0






        There is a natural identification of these transformations with that of $2times 2$ matrices with non-zero determinant:



        $$frac{az+b}{cz+d}longleftrightarrow begin{pmatrix}a&b\
        c&d
        end{pmatrix}.$$



        You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.






        share|cite|improve this answer












        There is a natural identification of these transformations with that of $2times 2$ matrices with non-zero determinant:



        $$frac{az+b}{cz+d}longleftrightarrow begin{pmatrix}a&b\
        c&d
        end{pmatrix}.$$



        You can, relatively easily, prove that composition of two of these transformations coincides with matrix multiplication, which makes the result easier to see.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 21:41









        Chickenmancer

        3,319723




        3,319723






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3006920%2fshow-that-ad-bc-ne-0-for-a-composition-of-two-mobius-transformations%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

            'app-layout' is not a known element: how to share Component with different Modules

            Image
            0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
            Read more

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
            Read more

            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
            Read more