# Conformal maps

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

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

$$\arg u_2 - \arg u_1 = \arg v_2 - \arg v_1\quad\mod\;2\pi$$

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