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Part BCSIR NET December 2025a-simple-pole-forces-a-nonzero-residue-take-the-contrapositive-on-the-singleton-claim

A simple pole forces a nonzero residue take the contrapositive on the singleton claim

Let S be a finite subset of C\mathbb{C} containing 0. Let f : C§\mathbb{C}\S →C\to \mathbb{C} be a holomorphic function which has a simple pole at 0. For R > 0, let γR\gamma_R denote the path γR(t)=Re(2π\gamma_R(t) = Re^(2\piit) for t ∈ [0,1]. Which of the following statements is necessarily true?

  1. A.The function f(1/z) has a removable singularity at 0.
  2. B.The function f(1/z) has an essential singularity at 0.
  3. C.If ∫γR\int_{\gamma{}R} f(z)dz = 0 for some R > 0, then S is not a singleton set.✓
  4. D.If S is not a singleton set, then ∫γR\int_{\gamma{}R} f(z)dz = 0 for some R > 0 such that S ∩ {z∈Cz \in \mathbb{C} : |z| = R} = ∅.

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.

See pricing

50 are analysed free — try those first.

Related counterexample: ∮γ f = 0 implies f is holomorphic inside γ

More on this topic

The chapter behind this: Residue theorem — the five standard patterns — free to read

From Singularities and Residues › Residue theorem and standard contour integrals

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