Finding the inverse of $phi_{lambda}^2$ where $phi_{lambda}$ is the Mobius transform on $mathbb{D}$
$begingroup$
Let $mathbb{D}$ denote the open unit disk. Fix $lambda in mathbb{D}$. Define the Mobius transform $phi_{lambda}:mathbb{D}rightarrowmathbb{D}$ by
$$phi_{lambda}(z) = frac{z-lambda}{1-overline{lambda}z}$$
If possible, I'm trying to find the inverse of $phi_{lambda}^2$; that is, I want to find an analytic function $psi:mathbb{D}rightarrowmathbb{D}$ such that
$$phi_{lambda}^2Big(psi(z)Big) = psiBig(phi_{lambda}^2(z)Big) = z$$
I attempted to find $psi$ analytically by trying to solve the equation
$$phi_{lambda}^2Big(psi(z)Big) = z$$
but I know that "taking the square root on both sides" is not always a valid move in $mathbb{C}$.
Any help will be appreciated.
complex-analysis inverse-function mobius-transformation
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
add a comment |
$begingroup$
Let $mathbb{D}$ denote the open unit disk. Fix $lambda in mathbb{D}$. Define the Mobius transform $phi_{lambda}:mathbb{D}rightarrowmathbb{D}$ by
$$phi_{lambda}(z) = frac{z-lambda}{1-overline{lambda}z}$$
If possible, I'm trying to find the inverse of $phi_{lambda}^2$; that is, I want to find an analytic function $psi:mathbb{D}rightarrowmathbb{D}$ such that
$$phi_{lambda}^2Big(psi(z)Big) = psiBig(phi_{lambda}^2(z)Big) = z$$
I attempted to find $psi$ analytically by trying to solve the equation
$$phi_{lambda}^2Big(psi(z)Big) = z$$
but I know that "taking the square root on both sides" is not always a valid move in $mathbb{C}$.
Any help will be appreciated.
complex-analysis inverse-function mobius-transformation
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48
add a comment |
$begingroup$
Let $mathbb{D}$ denote the open unit disk. Fix $lambda in mathbb{D}$. Define the Mobius transform $phi_{lambda}:mathbb{D}rightarrowmathbb{D}$ by
$$phi_{lambda}(z) = frac{z-lambda}{1-overline{lambda}z}$$
If possible, I'm trying to find the inverse of $phi_{lambda}^2$; that is, I want to find an analytic function $psi:mathbb{D}rightarrowmathbb{D}$ such that
$$phi_{lambda}^2Big(psi(z)Big) = psiBig(phi_{lambda}^2(z)Big) = z$$
I attempted to find $psi$ analytically by trying to solve the equation
$$phi_{lambda}^2Big(psi(z)Big) = z$$
but I know that "taking the square root on both sides" is not always a valid move in $mathbb{C}$.
Any help will be appreciated.
complex-analysis inverse-function mobius-transformation
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
Let $mathbb{D}$ denote the open unit disk. Fix $lambda in mathbb{D}$. Define the Mobius transform $phi_{lambda}:mathbb{D}rightarrowmathbb{D}$ by
$$phi_{lambda}(z) = frac{z-lambda}{1-overline{lambda}z}$$
If possible, I'm trying to find the inverse of $phi_{lambda}^2$; that is, I want to find an analytic function $psi:mathbb{D}rightarrowmathbb{D}$ such that
$$phi_{lambda}^2Big(psi(z)Big) = psiBig(phi_{lambda}^2(z)Big) = z$$
I attempted to find $psi$ analytically by trying to solve the equation
$$phi_{lambda}^2Big(psi(z)Big) = z$$
but I know that "taking the square root on both sides" is not always a valid move in $mathbb{C}$.
Any help will be appreciated.
complex-analysis inverse-function mobius-transformation
complex-analysis inverse-function mobius-transformation
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 22 at 14:26
Benedict VoltaireBenedict Voltaire
1,308928
1,308928
$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48
add a comment |
$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48
$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48
$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48
add a comment |
1 Answer
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$begingroup$
The Moebius transforms $$ f(z) = frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices
$$ A = begin{pmatrix} a & b \ c & d end{pmatrix},.$$
Applied for your case, the Moebius transform $phi_lambda$ is associated to the matrix
$$A _ lambda =begin{pmatrix} 1 & -lambda \
-barlambda & 1 end{pmatrix}. $$
You are searching for the inverse of $phi_lambda circ phi_lambda$ which is the Moebius transform associated with the matrix
$$ [(A_lambda)^2 ]^{-1} = [(A_lambda)^{-1}]^2 =frac{1}{(1-|lambda|^2)^2} begin{pmatrix} 1 & lambda \
barlambda & 1 end{pmatrix}^2 =frac{1}{(1-|lambda|^2)^2}begin{pmatrix} 1+ |lambda|^2 &2 lambda\ 2 barlambda& 1+ |lambda|^2 end{pmatrix} .$$
In other words, the inverse transform is given by
$$psi(z) = frac{(1+ |lambda|^2) z + 2 lambda}{2barlambda z + (1+ |lambda|^2)},.$$
answered Jan 22 at 16:10
FabianFabian
19.8k3774
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
$begingroup$
The Moebius transforms $$ f(z) = frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices
$$ A = begin{pmatrix} a & b \ c & d end{pmatrix},.$$
Applied for your case, the Moebius transform $phi_lambda$ is associated to the matrix
$$A _ lambda =begin{pmatrix} 1 & -lambda \
-barlambda & 1 end{pmatrix}. $$
You are searching for the inverse of $phi_lambda circ phi_lambda$ which is the Moebius transform associated with the matrix
$$ [(A_lambda)^2 ]^{-1} = [(A_lambda)^{-1}]^2 =frac{1}{(1-|lambda|^2)^2} begin{pmatrix} 1 & lambda \
barlambda & 1 end{pmatrix}^2 =frac{1}{(1-|lambda|^2)^2}begin{pmatrix} 1+ |lambda|^2 &2 lambda\ 2 barlambda& 1+ |lambda|^2 end{pmatrix} .$$
In other words, the inverse transform is given by
$$psi(z) = frac{(1+ |lambda|^2) z + 2 lambda}{2barlambda z + (1+ |lambda|^2)},.$$
answered Jan 22 at 16:10
FabianFabian
19.8k3774
$endgroup$
add a comment |
$begingroup$
The Moebius transforms $$ f(z) = frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices
$$ A = begin{pmatrix} a & b \ c & d end{pmatrix},.$$
Applied for your case, the Moebius transform $phi_lambda$ is associated to the matrix
$$A _ lambda =begin{pmatrix} 1 & -lambda \
-barlambda & 1 end{pmatrix}. $$
You are searching for the inverse of $phi_lambda circ phi_lambda$ which is the Moebius transform associated with the matrix
$$ [(A_lambda)^2 ]^{-1} = [(A_lambda)^{-1}]^2 =frac{1}{(1-|lambda|^2)^2} begin{pmatrix} 1 & lambda \
barlambda & 1 end{pmatrix}^2 =frac{1}{(1-|lambda|^2)^2}begin{pmatrix} 1+ |lambda|^2 &2 lambda\ 2 barlambda& 1+ |lambda|^2 end{pmatrix} .$$
In other words, the inverse transform is given by
$$psi(z) = frac{(1+ |lambda|^2) z + 2 lambda}{2barlambda z + (1+ |lambda|^2)},.$$
answered Jan 22 at 16:10
FabianFabian
19.8k3774
$endgroup$
add a comment |
$begingroup$
The Moebius transforms $$ f(z) = frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices
$$ A = begin{pmatrix} a & b \ c & d end{pmatrix},.$$
Applied for your case, the Moebius transform $phi_lambda$ is associated to the matrix
$$A _ lambda =begin{pmatrix} 1 & -lambda \
-barlambda & 1 end{pmatrix}. $$
You are searching for the inverse of $phi_lambda circ phi_lambda$ which is the Moebius transform associated with the matrix
$$ [(A_lambda)^2 ]^{-1} = [(A_lambda)^{-1}]^2 =frac{1}{(1-|lambda|^2)^2} begin{pmatrix} 1 & lambda \
barlambda & 1 end{pmatrix}^2 =frac{1}{(1-|lambda|^2)^2}begin{pmatrix} 1+ |lambda|^2 &2 lambda\ 2 barlambda& 1+ |lambda|^2 end{pmatrix} .$$
In other words, the inverse transform is given by
$$psi(z) = frac{(1+ |lambda|^2) z + 2 lambda}{2barlambda z + (1+ |lambda|^2)},.$$
answered Jan 22 at 16:10
FabianFabian
19.8k3774
$endgroup$
The Moebius transforms $$ f(z) = frac{a z + b}{c z + d}$$ under composition are homorphic as a group to the matrices
$$ A = begin{pmatrix} a & b \ c & d end{pmatrix},.$$
Applied for your case, the Moebius transform $phi_lambda$ is associated to the matrix
$$A _ lambda =begin{pmatrix} 1 & -lambda \
-barlambda & 1 end{pmatrix}. $$
You are searching for the inverse of $phi_lambda circ phi_lambda$ which is the Moebius transform associated with the matrix
$$ [(A_lambda)^2 ]^{-1} = [(A_lambda)^{-1}]^2 =frac{1}{(1-|lambda|^2)^2} begin{pmatrix} 1 & lambda \
barlambda & 1 end{pmatrix}^2 =frac{1}{(1-|lambda|^2)^2}begin{pmatrix} 1+ |lambda|^2 &2 lambda\ 2 barlambda& 1+ |lambda|^2 end{pmatrix} .$$
In other words, the inverse transform is given by
$$psi(z) = frac{(1+ |lambda|^2) z + 2 lambda}{2barlambda z + (1+ |lambda|^2)},.$$
answered Jan 22 at 16:10
FabianFabian
19.8k3774
answered Jan 22 at 16:10
FabianFabian
19.8k3774
answered Jan 22 at 16:10
FabianFabian
19.8k3774
answered Jan 22 at 16:10
FabianFabian
19.8k3774
19.8k3774
add a comment |
add a comment |
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$begingroup$
I am afraid such an inverse might do not exist. Note that $f(z)=z^2:mathbb{D}tomathbb{D}$ is onto but not one-to-one. Hence, $phi_{lambda}^2:mathbb{D}tomathbb{D}$ is also onto but not one-to-one.
$endgroup$
– hypernova
Jan 22 at 14:48