Hi there. I have to prove that if $\displaystyle f:A \rightarrow B$ it's a bijective and analytic function with analyitic inverse, thenfis conformal.

I think I should prove the angle preserving using the analyticity, but I'm not sure how.

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- Nov 19th 2011, 11:35 AMUlyssesProof, bijective aplication and conformal mapping
Hi there. I have to prove that if $\displaystyle f:A \rightarrow B$ it's a bijective and analytic function with analyitic inverse, then

*f*is conformal.

I think I should prove the angle preserving using the analyticity, but I'm not sure how. - Nov 19th 2011, 11:44 AMxxp9Re: Proof, bijective aplication and conformal mapping
Since it's bijective there is no critical point. Otherwise if p is a cirtical point, f(p) can be expanded as f(z)=az^k + ... thus f is a k-folder covering on a small disk of D\{p}. To show that f preserve angles, note that its differential df ( the induced linear map at the tangent space) is a linear map:

df(v) = f' * v. Where the right side is the complex product. And since p is not critical f' is no zero, f'=r e^{i \theta} is a linear map composed by a rotation of \theta followed by a dilation of ratio r, while both the rotation and the dilation preserve angles.

Translate the above idea into math expressions and you're done.