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Part BCSIR NET December 2024harmonic-modulus-squared-forces-constant

Harmonic modulus squared forces constant

Let 𝔻 = {z = x + iy ∈C\in \mathbb{C} : |z| < 1} be the open unit disc and f : 𝔻 →Ca\to \mathbb{C} a holomorphic function such that f(0) = 0. Let ψ(z)=\psi(z) = |f(z)|2^{2}, and ∂2ψ/∂x2+∂2ψ/∂y2≡0\partial^{2}\psi/\partial{}x^{2} + \partial^{2}\psi/\partial{}y^{2} \equiv 0. Which of the following statements is FALSE?

  1. A.f can be extended to C\mathbb{C} as an entire function.
  2. B.f must have infinitely many zeros in 𝔻.
  3. C.f is not a polynomial.✓ Correct
  4. D.exp(f) cannot take every complex value.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: A harmonic function on any domain has a harmonic conjugate

More on this topic

The chapter behind this: Cauchy–Riemann — what they do and do not give you — free to read

From Analytic Functions › Cauchy–Riemann equations, harmonic functions

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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