Let $f:U\to \mathbb{C}$ be a $C^1$ mapping of an open set $U\subseteq \mathbb{C}$. Then the condition that it is locally conformal is the geometric condition that $df(z_0):T_{z_0}U\to T_{f(z_0)}f(U)$ is a Conformal linear map.

$T_{z_0}U$ is defined to be the set of equivalence classes of all curves through $z_0\in U$, with the relation being they have the same derivative at $z_0$.

We can further develop intuition by observing that the above condition means that the intersection angle of two curves in $U$ must be the same as the intersection angle of the curves in $f(U)$ (post composition).