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Counterexample

Is this true?

A bounded holomorphic function on an unbounded domain is constant

No — it is false.

The counterexample

f(z) = eᶻ on {Re z < 0}

Why it fails

|eᶻ| < 1 there, but f is not constant. Liouville needs the whole plane.

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Hypothesis dropped

A theorem applied without checking one of its preconditions.

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Go deeper· Explained in full in Liouville, Morera and the maximum principle· Asked in CSIR NET June 2025· From Cauchy Theory › Liouville, Morera, maximum modulus principle

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