What Are Conformal Maps in Complex Analysis?

[Grufey]

Hello!, I was studing the conformal maps in complex analysis, I don't understand this definition:

Definition: A map f:A\rightarrow\mathbb{C} is called conformal at z_0 if there exist a \theta\in[0,2\pi] and r>0 such that for any curve \gamma(t) which is differentiable at t=0, for which \gamma(t) \in A and \gamma(0)=z_0, and which satisfisfies \gamma\prime(0)\neq0 the curve \sigma(t)=f(\gamma(t)) is differentiable at t=0 and, setting u=\sigma\prime(0) and v=\gamma\prime(0), we have \left|u\right|=r\left|v\right| and \arg u =\arg v + \theta (\mod 2\theta)

I only know about the conformal maps, that the angle between the curves after the transform is equal to the before of the transformation. But I cannot find the relation, with the definition. I think that I don't undertand the definition.

Thanks

 

[jostpuur]

\left|u\right|=r\left|v\right|

This says that the tangent vector of the curve is scaled by some number that does not depend on the direction of the tangent vector.

\arg u =\arg v + \theta (\mod 2\theta)

This says that f rotates the tangent vector by some constant angle.

Suppose you had two curves \gamma_1 and \gamma_2, so that \gamma_1(0)=\gamma_2(0), their tangent vectors u_1:=\gamma'_1(0) and u_2:=\gamma'_2(0), and tangent vectors of the image paths v_1:=(f\circ\gamma_1)'(0) and v_2:=(f\circ\gamma_2)'(0).

The result

<br /> \arg u_2 - \arg u_1 = \arg v_2 - \arg v_1\quad\mod\;2\pi<br />

comes quite easily, and that is what the angle preserving means.