---
tag: 0QEBN8DGVE
kind: statement
role: corollary
name: Holomorphic functions are infinitely differentiable
msc: [30A99]
uses: [0DG0Z72CQD, 0826SGV3R0, 0YMP521ZR5]
sources:
  - url: https://en.wikipedia.org/wiki/Holomorphic_function
    locator: Properties
  - url: https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula
    locator: Consequences
---

## Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] on $U$.
Then the complex derivative $f'$ is holomorphic on $U$.
Thus, for each $n \ge 1$, the $n$-th complex derivative $f^{(n)}$ exists and is holomorphic on $U$.
As a function of two real variables $(x, y)$ with $z = x + iy$, $f$ has continuous partial derivatives of all orders on $U$.

## Remarks

Proof: near each point, $f$ is the sum of a power series by [[holomorphic functions are analytic|0826SGV3R0]]. A [[power series can be differentiated term by term|0YMP521ZR5]], and the result is again a power series.
There is no real analogue: $x \mapsto x |x|$ is differentiable on $\mathbb{R}$, but its derivative $2|x|$ is not differentiable at $0$.
The [[formula for the derivatives|0F0H5P5DED]] gives each $f^{(n)}$ as a circle integral.

## Explanations

### L1 human

Once a complex function has one derivative on an open set, it has a second one, a third one, and so on forever.
So in complex analysis there are no functions that are differentiable only once.