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The Jacobian determinant of a holomorphic map is $∣ f^{'} ∣^{2}$ and is positive exactly where $f^{'} \neq 0$

Statement

Let $f = u + i v$ be holomorphic on an open set $U \subseteq C$ . At every $z \in U$ ,

$\det J f (z) = ∣ f^{'} (z) ∣^{2} \geq 0.$

The determinant is positive exactly where $f^{'} (z) \neq 0$ , and it is zero exactly where $f^{'} (z) = 0$ ; it is never negative.

Facts & Assumptions

Given: A holomorphic map $f = u + i v$ and a point $z$ in its open domain.

[L1]

If $f^{'} (z) = a + i b$ , then the real derivative matrix is $(\begin{array}{cc} a & - b \\ b & a \end{array})$ (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).

[F1]

The Jacobian matrix is the matrix of the first partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L2]

Complex modulus satisfies $∣ a + i b ∣^{2} = a^{2} + b^{2}$ , and $∣ w ∣ = 0$ if and only if $w = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

By [L1] and [F1], $\det J f (z) = a^{2} + b^{2} = ∣ f^{'} (z) ∣^{2}$ .

givenL1F1L2algebra

2.1

The sum $a^{2} + b^{2}$ is nonnegative and, by [L2], is zero exactly when $f^{'} (z) = 0$ ; otherwise it is positive.

step 1.1L2algebra∎

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Sources

J. Lebl, Guide to Cultivating Complex Analysis, §2.2.4 (standard reference, not scraped)

The Jacobian determinant of a holomorphic map is ∣f′∣2∣f′∣2 and is positive exactly where f′≠0f′=0