# Conformal Proof

• Nov 7th 2010, 09:40 PM
Haven
Conformal Proof
Here's an interesting question I can't seem to get my head around.

A function f is called a diffeomorphism if $\displaystyle J_f \neq 0$, where $\displaystyle J_f = u_xv_y - u_yv_x$.
Let $\displaystyle \alpha_1(t)$ and $\displaystyle \alpha_2(t)$ denote contours $\displaystyle C_1$ and $\displaystyle C_2$ in a domain D, where$\displaystyle \alpha_1(a)= \alpha_2(b)=z_0$.
Let $\displaystyle \Theta(C_1,C_2) = \arg{\alpha_1'(a)}-\arg{\alpha_2'(b)}$.
A function f is called conformal at a point if $\displaystyle \Theta(C_1,C_2) = \Theta(f(C_1),f(C_2))$, and f is called angle-preserving at a point $\displaystyle |\Theta(C_1,C_2)| = |\Theta(f(C_1),f(C_2))|$.

I have shown a function f is conformal at $\displaystyle z_0$ if and only if $\displaystyle J_f(z_0)>0$ and is angle-preserving at $\displaystyle z_0$.

However I am having difficulties proving that if f is conformal at $\displaystyle z_0$ then f is analytic at $\displaystyle z_0$. I would greatly appreciate any hints, but please don't solve the problem for me.

EDIT: Without using all the theta terms, we wish to show that if the angle between two curves on the z-plane, at $\displaystyle z_0$ is preserved by the mapping $\displaystyle f(z)$ and so is the orientation, the f is analytic at $\displaystyle z_0$