---
tag: 0Q7M6K6VD7
kind: statement
role: proposition
name: Holomorphic functions with nonzero derivative are conformal
aliases: [holomorphic-maps-are-conformal]
msc: [30C35]
uses: [0DG0Z72CQD, 0D1C9Y1EXA, 0VHNKP1N3K]
sources:
  - url: https://en.wikipedia.org/wiki/Conformal_map
    locator: In two dimensions
---

## Statement

Let $U \subseteq \mathbb{C}$ be an open set, let $f : U \to \mathbb{C}$ be [[holomorphic|0DG0Z72CQD]], and let $z_0 \in U$.
If $f'(z_0) \neq 0$, then $f$ is [[conformal|0VHNKP1N3K]] 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|0D1C9Y1EXA]] 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 $z_0$.