Mixing
Show that $ad-bc ne 0$ for a composition of two Mobius transformations.
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
asked Nov 20 '18 at 21:20
72D
572116
add a comment |
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
asked Nov 20 '18 at 21:20
72D
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
add a comment |
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
asked Nov 20 '18 at 21:20
72D
572116
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
asked Nov 20 '18 at 21:20
72D
572116
asked Nov 20 '18 at 21:20
72D
572116
edited Nov 20 '18 at 21:25
asked Nov 20 '18 at 21:20
72D
572116
asked Nov 20 '18 at 21:20
72D
572116
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
add a comment |
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
add a comment |
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}
answered Nov 20 '18 at 21:33
José Carlos Santos
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
add a comment |
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.
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
add a comment |
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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}
answered Nov 20 '18 at 21:33
José Carlos Santos
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
add a comment |
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}
answered Nov 20 '18 at 21:33
José Carlos Santos
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
add a comment |
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}
answered Nov 20 '18 at 21:33
José Carlos Santos
151k22123224
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}
answered Nov 20 '18 at 21:33
José Carlos Santos
151k22123224
edited Nov 20 '18 at 22:05
answered Nov 20 '18 at 21:33
José Carlos Santos
151k22123224
answered Nov 20 '18 at 21:33
José Carlos Santos
151k22123224
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
add a comment |
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
add a comment |
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.
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
add a comment |
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.
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
add a comment |
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.
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
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.
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
answered Nov 20 '18 at 21:41
Chickenmancer
3,319723
3,319723
add a comment |
add a comment |
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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