Let $f:U\to \mathbb{C}$ be a $C^1$ map, for some open set $U\subseteq \mathbb{C}$. Then $f$ is (locally) conformal iff 1;; it is (locally) biholomorphic.

We have shown previously that $f$ is 1;;locally biholomorphic iff $f^{\prime }(z_0)\ne 0$ for all $z_0\in U$, see 4. Theorem 2.2.1.