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

Printable View

- May 10th 2013, 01:19 AMPlato13Show 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, 06:46 AMxxp9Re: Show f is a conformal mapping
deleted.

- May 10th 2013, 10:14 AMxxp9Re: Show f is a conformal mapping
yes I was right. A parametrization is conformal if its first fundamental form is of the form . So just compute its first fundamental form.

Let . The coordinate vectors are:

It is easy to verify that

And - May 10th 2013, 10:43 AMPlato13Re: 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, 11:08 AMxxp9Re: Show f is a conformal mapping
Your theorem is nothing but the statement of "the first fundamental form is of the form "

Note that in my post, , and , given and . - May 10th 2013, 11:12 AMPlato13Re: Show f is a conformal mapping
thanks for you help and patience