Holomorphic functions with nonzero derivative are conformal

Statement

Let $U \subseteq \mathbb{C}$ be an open set, let $f : U \to \mathbb{C}$ be holomorphic, and let $z_0 \in U$. If $f'(z_0) \neq 0$, then $f$ is conformal at $z_0$. In particular, if $f'(z) \neq 0$ for each $z \in U$, then $f$ is conformal on $U$.

Remarks

Write $f'(z_0) = r e^{i\varphi}$ with $r > 0$; then the real derivative of $f$ at $z_0$ is $h \mapsto r e^{i\varphi} h$, a rotation by $\varphi$ followed by a scaling by $r$. The converse also holds: a map that is conformal at each point of an open set $U$ is holomorphic on $U$ with a nonzero derivative. The condition $f'(z_0) \neq 0$ is necessary: $z \mapsto z^2$ doubles the angles between curves through $0$. The exponential function is conformal at each point but not injective, so a conformal map need not be injective.

Explanations

L1 human

Near $z_0$, a complex differentiable function acts almost like $z \mapsto f(z_0) + f'(z_0)(z - z_0)$. Multiplication by the complex number $f'(z_0)$ turns the plane and scales it. Turning and scaling keep all angles, so $f$ keeps the angles between curves through $z_0$.