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Part CCSIR NET June 2025a-bounded-singularity-is-removable-so-the-puncture-is-not-essential

A bounded singularity is removable so the puncture is not essential

Let 𝔻^× = {z∈C:0<z \in \mathbb{C} : 0 < |z| < 1} be the punctured unit disk and f be a bijective holomorphic map of 𝔻^× onto itself. Which of the following statements are true?

  1. A.limz→0lim_{z\to0} f(z) does not exist.
  2. B.limz→0lim_{z\to0} f(z) exists and has absolute value ≤ 1.✓ Correct
  3. C.limz→0lim_{z\to0} f(z) = 0.✓ Correct
  4. D.There exists θ∈R\theta \in \mathbb{R} such that f(z) = eiθze^{i\theta}z for all z ∈ 𝔻^×.✓ Correct

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.

See pricing

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: ℂ and the unit disc are biholomorphic (both are simply connected)

More on this topic

The chapter behind this: Conformal maps and Möbius transformations — free to read

From Zeros and Mappings › Conformal maps, Möbius transformations, Schwarz lemma

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