False statementConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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FALSE: real differentiability as a map $R^{2} \to R^{2}$ implies complex differentiability; conjugation is the counterexample

Statement

False claim: if a map $f : C \to C$ , regarded as a map $R^{2} \to R^{2}$ , is real totally differentiable at a point, then it is complex differentiable there.

Facts & Assumptions

Given: The conjugation map $f (z) = \overset{ˉ}{z}$ .

[L1]

Under the real-coordinate identification $Φ (a + b i) = (a, b)$ ( $C$ is the real coordinate plane, with coordinate arithmetic), $x + i y$ corresponds to $(x, y)$ ; conjugation is $\overline{x + i y} = x - i y$ (Real and imaginary parts, complex conjugation, and modulus), so it corresponds to $(x, y) \mapsto (x, - y)$ .

[L2]

Complex differentiability is equivalent to real total differentiability together with the Cauchy–Riemann equations $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ (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).

Refutation

technique · direct counterexample

1.1

By [L1], $f$ is the real-linear map $(x, y) \mapsto (x, - y)$ . Its increment is exactly its linear action, so it is real totally differentiable everywhere with derivative matrix $diag (1, - 1)$ .

L1algebra

1.2

Directly, for nonzero real $t$ the quotient $(\overline{z + t} - \overset{ˉ}{z}) / t$ is $1$ , whereas the quotient for the increment $i t$ is $(\overline{z + i t} - \overset{ˉ}{z}) / (i t) = - 1$ . Thus the complex difference quotient has incompatible directional limits.

givenalgebra

2.1

Its components $u = x$ and $v = - y$ have $u_{x} = 1$ and $v_{y} = - 1$ , so the first Cauchy–Riemann equation fails at every point. By [L2], $f$ is nowhere complex differentiable.

step 1.1L2

3.1

The same map is real totally differentiable everywhere by step 1.1 and complex differentiable nowhere by steps 1.2 and 2.1, so it refutes the claim.

step 1.1step 2.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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 (standard reference, not scraped)

FALSE: real differentiability as a map R2→R2R2→R2 implies complex differentiability; conjugation is the counterexample