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Part CCSIR NET December 2024minimum-modulus-puts-a-zero-in-every-component

Minimum modulus puts a zero in every component

Let P(z) be a non-constant polynomial over C\mathbb{C}. Given R > 0, let S_R = {z∈Cz \in \mathbb{C} : |P(z)| < R}. Which of the following statements are true?

  1. A.S_R is an open subset of C\mathbb{C}.✓ Correct
  2. B.S_R is a bounded subset of C\mathbb{C}.✓ Correct
  3. C.|P(z)| = R for every z on the boundary of S_R.✓ Correct
  4. D.Every connected component of S_R contains a zero of P(z).✓ 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: f′(z) ≠ 0 everywhere implies f is injective

More on this topic

The chapter behind this: Argument principle, Rouché and zero counting — free to read

From Zeros and Mappings › Argument principle, Rouché's theorem, open mapping

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

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