TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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A map into $C^{n}$ is holomorphic exactly when each of its components is

Statement

Let $m, n \geq 1$ , let $U \subseteq C^{m}$ be open, let $a \in U$ and let $F : U \to C^{n}$ with components $F_{j} = π_{j} \circ F$ for $j < n$ . Then $F$ is holomorphic at $a$ (Holomorphic maps $C^{m} \to C^{n}$ and the complex Jacobian matrix) if and only if every $F_{j}$ is holomorphic at $a$ (Holomorphic functions on an open subset of $C^{m}$ ), and in that case

$D F (a) h = (D F_{0} (a) h, \dots, D F_{n - 1} (a) h), (J_{C} F (a))_{j k} = \partial_{z_{k}} F_{j} (a) .$

For $n = 1$ this is the scalar definition read back.

Facts & Assumptions

Given: An open $U \subseteq C^{m}$ , $a \in U$ and $F : U \to C^{n}$ with components $F_{j}$ ; the spaces are read through Complex $m$ -space and its real coordinate dictionary.

[L1]

$F$ is holomorphic at $a$ when there is a $C$ -linear $L : C^{m} \to C^{n}$ with $F (a + h) = F (a) + L (h) + r (h)$ and $∥ r (h) ∥ / ∥ h ∥ \to 0$ ; $L$ is unique and its matrix in the standard bases is $J_{C} F (a)$ , whose $(j, k)$ entry is the $j$ th coordinate of $L (e_{k})$ (Holomorphic maps $C^{m} \to C^{n}$ and the complex Jacobian matrix, Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[L2]

The scalar case is the same condition with $∣ r (h) ∣ / ∥ h ∥ \to 0$ and $L$ a $C$ -linear functional (Holomorphic functions on an open subset of $C^{m}$ ).

[L4]

Under the interleaved real-coordinate identification fixed in the Given, $∣ w_{j} ∣^{2} = (Re w_{j})^{2} + (Im w_{j})^{2}$ , so the Euclidean norm is $∥ w ∥ = (\sum_{j < n} ∣ w_{j} ∣^{2})^{1 / 2}$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L5]

Every $h$ satisfies $h = \sum_{k < m} h_{k} e_{k}$ in the standard basis (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L6]

Finite sums in the additive commutative monoid of $C$ may be regrouped termwise, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L7]

If a scalar function is complex differentiable at $a$ , its complex-linear real differential has the form $D g (a) h = \sum_{k < m} (\partial_{z_{k}} g (a)) h_{k}$ (A real-linear functional on $C^{m}$ is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in $C^{m}$ ).

[L8]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

By [L4] and [L6], every $w \in C^{n}$ satisfies $∣ w_{j} ∣ \leq ∥ w ∥$ for each $j < n$ and $∥ w ∥ \leq \sum_{j < n} ∣ w_{j} ∣$ , the first because $∣ w_{j} ∣^{2}$ is one term of a sum of nonnegative terms and the second because the square of the right-hand side dominates that sum.

L4L6L8

2.1

Suppose $F$ is holomorphic at $a$ with $L = D F (a)$ as in [L1]. For each $j < n$ the map $π_{j} \circ L$ is $C$ -linear, being a coordinate of a $C$ -linear map, and $F_{j} (a + h) = F_{j} (a) + (π_{j} \circ L) (h) + r_{j} (h)$ with $∣ r_{j} (h) ∣ \leq ∥ r (h) ∥$ by step 1.1; so $∣ r_{j} (h) ∣ / ∥ h ∥ \to 0$ and [L2] makes $F_{j}$ holomorphic at $a$ with $D F_{j} (a) = π_{j} \circ L$ .

givenstep 1.1L1L2

2.2

Conversely, suppose every $F_{j}$ is holomorphic at $a$ with $L_{j} = D F_{j} (a)$ , and set $L (h) = (L_{0} (h), \dots, L_{n - 1} (h))$ . Then $L$ is $C$ -linear because each coordinate is and the operations on $C^{n}$ are coordinatewise, and the remainder of $F$ has $∥ r (h) ∥ \leq \sum_{j < n} ∣ r_{j} (h) ∣$ by step 1.1; each summand is $o (∥ h ∥)$ and there are finitely many, so [L6] makes the sum $o (∥ h ∥)$ and [L1] makes $F$ holomorphic at $a$ with $D F (a) = L$ .

givenstep 1.1L1L2L6

3.1

In either direction $D F (a) h = (D F_{0} (a) h, \dots, D F_{n - 1} (a) h)$ by steps 2.1 and 2.2 and the uniqueness in [L1]. Evaluating at $h = e_{k}$ and reading the $j$ th coordinate, [L1], [L5] and [L7] give $(J_{C} F (a))_{j k} = D F_{j} (a) e_{k} = \partial_{z_{k}} F_{j} (a)$ . For $n = 1$ the two conditions of [L1] and [L2] coincide.

step 2.1step 2.2L1L2L5L7∎

Depends on

Used by

Cited to discharge well-definedness by Holomorphic maps ℂᵐ → ℂⁿ and the complex Jacobian matrix.

Dependency tree · two levels

93 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, Tasty Bits of Several Complex Variables, §1.3 (standard reference, not scraped)

A map into CnCn is holomorphic exactly when each of its components is