TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with $\partial_{\overset{ˉ}{z}} f = 0$ , or with the Cauchy–Riemann equations

Statement

Let $U \subseteq C$ be open, let $a \in U$ , and write $f = u + i v : U \to C$ . The following are equivalent:

$f$ is complex differentiable at $a$ .
Under $C ≅ R^{2}$ , the map $f$ is real totally differentiable at $a$ and $D f (a)$ is multiplication by a complex number.
The map $f$ is real totally differentiable at $a$ and $\partial_{\overset{ˉ}{z}} f (a) = 0$ .
The map $f$ is real totally differentiable at $a$ and satisfies the Cauchy–Riemann equations

$u_{x} (a) = v_{y} (a), u_{y} (a) = - v_{x} (a) .$

When these conditions hold,

$f^{'} (a) = u_{x} (a) + i v_{x} (a) = v_{y} (a) - i u_{y} (a) = \partial_{z} f (a) .$

Facts & Assumptions

Given: An open set $U \subseteq C$ , a point $a \in U$ , and a map $f = u + i v : U \to C$ .

[F1]

Complex differentiability at $a$ is existence of the limit $(f (a + h) - f (a)) / h$ as $h \to 0$ through nonzero increments with $a + h \in U$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[F2]

Real total differentiability at $a$ means that for some real-linear $L$ , $f (a + h) = f (a) + L h + r (h)$ with $∥ r (h) ∥_{2} / ∥ h ∥_{2} \to 0$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, A linear map $L : R^{m} \to R^{n}$ in Euclidean coordinates).

[L1]

If a map is totally differentiable at $a$ , then its directional derivatives exist and equal $D f (a) v$ ; its partial derivatives are the columns of its Jacobian matrix (A total derivative computes every directional derivative, and its matrix is the Jacobian, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L2]

Every real-linear map between Euclidean spaces has a unique matrix and is bounded by a constant times the Euclidean norm (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

[L3]

Under $Φ (a + b i) = (a, b)$ , complex multiplication satisfies $(a + b i) (x + i y) = (a x - b y) + i (b x + a y)$ ( $C$ is the real coordinate plane, with coordinate arithmetic).

[F3]

For a real-differentiable $f$ , $D f (h) = (\partial_{z} f) h + (\partial_{\overset{ˉ}{z}} f) \overset{ˉ}{h}$ , with $\partial_{\overset{ˉ}{z}} f = \frac{1}{2} (u_{x} - v_{y}) + \frac{i}{2} (v_{x} + u_{y})$ (The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

[L4]

For complex numbers, $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z ∣ = 0$ if and only if $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

Assume condition 1 and write $L = f^{'} (a)$ . For $h \neq 0$ put $r (h) = f (a + h) - f (a) - L h$ ; then $∣ r (h) ∣ / ∣ h ∣ = ∣ (f (a + h) - f (a)) / h - L ∣ \to 0$ .

F1L4givenalgebra

1.2

Conversely assume condition 2, so $f (a + h) - f (a) = L h + r (h)$ with $∣ r (h) ∣ / ∣ h ∣ \to 0$ . For nonzero $h$ , division by $h$ gives $(f (a + h) - f (a)) / h = L + r (h) / h$ , and $∣ r (h) / h ∣ = ∣ r (h) ∣ / ∣ h ∣ \to 0$ ; hence condition 1 holds with $f^{'} (a) = L$ .

F1F2L4givenalgebra

1.3

Write the matrix of $D f (a)$ as $(\begin{array}{cc} u_{x} & u_{y} \\ v_{x} & v_{y} \end{array})$ by [L1]. By [L3], it is multiplication by $α + i β$ exactly when it is $(\begin{array}{cc} α & - β \\ β & α \end{array})$ .

L1L2L3

1.4

By [F3], $\partial_{\overset{ˉ}{z}} f (a) = 0$ exactly when both $u_{x} - v_{y} = 0$ and $v_{x} + u_{y} = 0$ . Hence condition 3 is equivalent to condition 4.

F3algebra

2.1

The map $h \mapsto L h$ is real-linear by the coordinate formula [L3], so step 1.1 is the remainder condition [F2]. Thus condition 2 holds.

step 1.1F2L2L3

2.2

Therefore condition 2 is equivalent to condition 4: equality with the multiplication matrix is exactly $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ . In that case $α = u_{x} = v_{y}$ , $β = v_{x} = - u_{y}$ , so the multiplier is $u_{x} + i v_{x} = v_{y} - i u_{y}$ .

step 1.3algebra

3.1

Under the equivalent conditions, [F3] and the Cauchy–Riemann equations give $\partial_{z} f (a) = u_{x} + i v_{x}$ , while steps 1.2 and 2.2 identify the same number with $f^{'} (a)$ . Thus all four conditions are equivalent and the displayed derivative formulas hold.

step 1.2step 2.2step 1.4F3∎

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Propositions 2.1.4 and 2.2.6 (standard reference, not scraped)
J. Orloff, MIT 18.04 Topic 2, §§2.7–2.8 (standard reference, not scraped)

Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with ∂zˉf=0∂zˉ​f=0, or with the Cauchy–Riemann equations