CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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If both $f$ and $\overset{ˉ}{f}$ are holomorphic on a domain, then $f$ is constant

Statement

Let $U$ be a complex domain. If $f$ and $\overset{ˉ}{f}$ are both holomorphic on $U$ , then $f$ is constant.

Facts & Assumptions

Given: A domain $U$ and a function $f = u + i v$ such that both $f$ and $\overset{ˉ}{f} = u - i v$ are holomorphic on $U$ .

[L1]

A holomorphic map $p + i q$ satisfies $p_{x} = q_{y}$ and $p_{y} = - q_{x}$ , with derivative $p_{x} + i q_{x}$ (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).

[L2]

A holomorphic function with zero derivative on a domain is constant (A holomorphic function with zero derivative on a domain is constant).

Proof

technique · direct

1.1

Applying [L1] to $f$ gives $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ , while applying it to $\overset{ˉ}{f} = u + i (- v)$ gives $u_{x} = - v_{y}$ and $u_{y} = v_{x}$ .

givenL1

2.1

The paired equations imply $u_{x} = u_{y} = v_{x} = v_{y} = 0$ , so [L1] gives $f^{'} = 0$ throughout $U$ .

step 1.1L1algebra

3.1

The domain theorem [L2] now makes $f$ constant.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 50 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 2.2.9 (standard reference, not scraped)

If both ff and fˉfˉ​ are holomorphic on a domain, then ff is constant