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







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    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






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share|cite|improve this question













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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
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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}






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  • 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





















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    2 Answers
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    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














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    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






























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