Mixing
Inverse function of a conformal mapping
$begingroup$
I'm trying to prove that if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.
I proved the existence of $g$ using the Inverse Function Thereom. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,vin C^1(D)$.
Tha Jacobian is
$begin{equation*}
J =begin{vmatrix}
u_x & u_y\
v_x & v_y\
end{vmatrix}
=u_xv_y-u_yv_x
end{equation*}$
and using the Cauchy-Riemann equations, we have that
begin{equation*}
J=(u_x)^2+(v_x)^2=|f'(z)|^2
end{equation*}
Since $f'(z_0)neq 0$, $Jneq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.
How can I prove that $g$ is analytic in $w_0$?
That's the only thing that is missing, since I also proved that
begin{equation*}
g'(w)=frac{1}{f'(z_0)}
end{equation*}
and since $f'(z_0)neq 0$, then $g'(w_0)neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$
complex-analysis conformal-geometry
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.
I proved the existence of $g$ using the Inverse Function Thereom. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,vin C^1(D)$.
Tha Jacobian is
$begin{equation*}
J =begin{vmatrix}
u_x & u_y\
v_x & v_y\
end{vmatrix}
=u_xv_y-u_yv_x
end{equation*}$
and using the Cauchy-Riemann equations, we have that
begin{equation*}
J=(u_x)^2+(v_x)^2=|f'(z)|^2
end{equation*}
Since $f'(z_0)neq 0$, $Jneq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.
How can I prove that $g$ is analytic in $w_0$?
That's the only thing that is missing, since I also proved that
begin{equation*}
g'(w)=frac{1}{f'(z_0)}
end{equation*}
and since $f'(z_0)neq 0$, then $g'(w_0)neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$
complex-analysis conformal-geometry
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.
I proved the existence of $g$ using the Inverse Function Thereom. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,vin C^1(D)$.
Tha Jacobian is
$begin{equation*}
J =begin{vmatrix}
u_x & u_y\
v_x & v_y\
end{vmatrix}
=u_xv_y-u_yv_x
end{equation*}$
and using the Cauchy-Riemann equations, we have that
begin{equation*}
J=(u_x)^2+(v_x)^2=|f'(z)|^2
end{equation*}
Since $f'(z_0)neq 0$, $Jneq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.
How can I prove that $g$ is analytic in $w_0$?
That's the only thing that is missing, since I also proved that
begin{equation*}
g'(w)=frac{1}{f'(z_0)}
end{equation*}
and since $f'(z_0)neq 0$, then $g'(w_0)neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$
complex-analysis conformal-geometry
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
$endgroup$
I'm trying to prove that if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.
I proved the existence of $g$ using the Inverse Function Thereom. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,vin C^1(D)$.
Tha Jacobian is
$begin{equation*}
J =begin{vmatrix}
u_x & u_y\
v_x & v_y\
end{vmatrix}
=u_xv_y-u_yv_x
end{equation*}$
and using the Cauchy-Riemann equations, we have that
begin{equation*}
J=(u_x)^2+(v_x)^2=|f'(z)|^2
end{equation*}
Since $f'(z_0)neq 0$, $Jneq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.
How can I prove that $g$ is analytic in $w_0$?
That's the only thing that is missing, since I also proved that
begin{equation*}
g'(w)=frac{1}{f'(z_0)}
end{equation*}
and since $f'(z_0)neq 0$, then $g'(w_0)neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$
complex-analysis conformal-geometry
complex-analysis conformal-geometry
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
asked Nov 15 '15 at 22:59
Fawcett512Fawcett512
30019
30019
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This would be basically an application of the Lagrange inversion theorem, which states the following:
Suppose $w$ is defined as a function of $z$ by an equation of the form
$$f(z)=w,$$
where $f$ is analytic at a point $z_0$ and $f'(z_0)neq 0$. Then it is possible to invert (or solve) the equation for $z$:
$$z=g(w),$$
on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.
Since you have (or have proved) all the hypothesis, you can just apply the theorem.
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
$endgroup$
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
This would be basically an application of the Lagrange inversion theorem, which states the following:
Suppose $w$ is defined as a function of $z$ by an equation of the form
$$f(z)=w,$$
where $f$ is analytic at a point $z_0$ and $f'(z_0)neq 0$. Then it is possible to invert (or solve) the equation for $z$:
$$z=g(w),$$
on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.
Since you have (or have proved) all the hypothesis, you can just apply the theorem.
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
$endgroup$
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
add a comment |
$begingroup$
This would be basically an application of the Lagrange inversion theorem, which states the following:
Suppose $w$ is defined as a function of $z$ by an equation of the form
$$f(z)=w,$$
where $f$ is analytic at a point $z_0$ and $f'(z_0)neq 0$. Then it is possible to invert (or solve) the equation for $z$:
$$z=g(w),$$
on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.
Since you have (or have proved) all the hypothesis, you can just apply the theorem.
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
$endgroup$
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
add a comment |
$begingroup$
This would be basically an application of the Lagrange inversion theorem, which states the following:
Suppose $w$ is defined as a function of $z$ by an equation of the form
$$f(z)=w,$$
where $f$ is analytic at a point $z_0$ and $f'(z_0)neq 0$. Then it is possible to invert (or solve) the equation for $z$:
$$z=g(w),$$
on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.
Since you have (or have proved) all the hypothesis, you can just apply the theorem.
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
$endgroup$
This would be basically an application of the Lagrange inversion theorem, which states the following:
Suppose $w$ is defined as a function of $z$ by an equation of the form
$$f(z)=w,$$
where $f$ is analytic at a point $z_0$ and $f'(z_0)neq 0$. Then it is possible to invert (or solve) the equation for $z$:
$$z=g(w),$$
on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.
Since you have (or have proved) all the hypothesis, you can just apply the theorem.
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
answered Oct 5 '16 at 16:12
EduEdu
1,4051619
1,4051619
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
add a comment |
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
$begingroup$
Thanks man. Better late than never I guess. I can't even remember what happened with this particular problem, whether I solved it or not. I'll think about it again...
$endgroup$
– Fawcett512
Oct 7 '16 at 4:52
add a comment |
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