It seems like if $U$ is an open subset of the complex plane, $\mathbb{C}$, then a function $$f: U \rightarrow \mathbb{C}$$ is conformal if and only if it is holomorphic and its derivative is everywhere non-zero on $U$.
Can anyone explain why conformal map means holomorphic and vice versa?
Thanks in advance
First, conformal at $z_0$ means differentiable at $z_0$, with non zero Jacobian and angles-preserving. With Cauchy-riemann formula, you show that $f'(z_0)\neq0$ implies $J(f)_{z_0} \neq 0$. Then, you take $\gamma_1, \gamma_2$ two curves with origin at $z_0$, and compute the angle $\hat{(f(\gamma_1),f(\gamma_2))}$, by the formula $\hat{(\tau_1,\tau_2)}=Arg \frac{\tau_1'(0)}{\tau_2'(0)}$, and find that f is conformal.
For the converse, let $f$ be a conformal map. We need to prove $d_{z_0}f$ to be $\mathbb{C}$-linear.
$\textbf{Lemma}$: it's enough to prove that $d_{z_0}f$ is a $\mathbb{R}$-linear, non constant, conformal map.
The two first facts are obvious, the third comes from the fact that $f$ is conformal, as well.
For this lemma, you take a $\mathbb{R}$-linear, non constant, conformal map $L$, normalize such that $L(1)=1$, but L is conformal gives $L(i) \in i\mathbb{R}$, and then $L(i)=i$, so $L=id$ up to homothetic transformation, then $L$ is $\mathbb{C}$-linear, and you are done.