# Show f is a conformal mapping

• May 10th 2013, 02:19 AM
Plato13
Show f is a conformal mapping
My working: A basis for the tangent space (at any point) is (1,0), (0,1) so I need only show it for this . df(1,0)=df/dx and df(0,1)=df/dy Is this the right approach? Thanks
• May 10th 2013, 07:46 AM
xxp9
Re: Show f is a conformal mapping
deleted.
• May 10th 2013, 11:14 AM
xxp9
Re: Show f is a conformal mapping
yes I was right. A parametrization is conformal if its first fundamental form is of the form $g(du^2+dv^2)$. So just compute its first fundamental form.
Let $A=1+x^2+y^2$. The coordinate vectors are:
$\frac{\partial}{\partial x}=\frac{(2A-4x^2, -4xy, -4x)}{A^2}$
$\frac{\partial}{\partial y}=\frac{(-4xy, 2A-4y^2, -4y)}{A^2}$
It is easy to verify that $\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial x} \rangle=\langle \frac{\partial}{\partial y}, \frac{\partial}{\partial y} \rangle$
And $\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \rangle=0$
• May 10th 2013, 11:43 AM
Plato13
Re: Show f is a conformal mapping
I was hoping to solve by using that theorem that a map is conformal if at each point p there is a basis e,e' such that the df(e).df(e')=lambda x e.e' where lamda is the scalar factor. I've just realised I ought to use polar co-ordinates x=rcost, y=rsint. That will simplify things right?
• May 10th 2013, 12:08 PM
xxp9
Re: Show f is a conformal mapping
Your theorem is nothing but the statement of "the first fundamental form is of the form $\lambda(dx^2+dy^2)$"
Note that in my post, $\frac{\partial}{\partial x} = df(e_1)$, and $\frac{\partial}{\partial y} = df(e_2)$, given $e_1=(1,0)$ and $e_2=(0,1)$.
• May 10th 2013, 12:12 PM
Plato13
Re: Show f is a conformal mapping
thanks for you help and patience