CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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$z \mapsto Re z$ is real differentiable but nowhere complex differentiable

Statement refuted

A real-linear map from $C$ to itself is complex differentiable.

Facts & Assumptions

Given: $f (z) = Re z$ .

[L1]

Complex differentiability is equivalent to real total differentiability together with $u_{x} = v_{y}$ 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).

Counterexample

technique · direct

1.1

In real coordinates, $f (x + i y) = x + i 0$ , so it is the real-linear map $(x, y) \mapsto (x, 0)$ and is real totally differentiable everywhere with that same linear map as derivative.

algebra

2.1

Its components have $u_{x} = 1$ and $v_{y} = 0$ . The first Cauchy–Riemann equation therefore fails everywhere, so [L1] makes $f$ nowhere complex differentiable.

step 1.1L1

3.1

Equivalently, at any $z$ , the difference quotient is $1$ along nonzero real increments and $0$ along nonzero imaginary increments. These incompatible limits independently confirm step 2.1 and refute the claim.

step 2.1givenalgebra∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Howell and Mathews, Complex Analysis, §3.1 (standard reference, not scraped)

z↦Re⁡zz↦Rez is real differentiable but nowhere complex differentiable