Skip to content
Part BCSIR NET June 2024one-identity-decides-all-four

One identity decides all four

For a complex number a such that 0 < |a| < 1, which of the following statements is true?

  1. A.If |z| < 1, then |1 − āz| < |z − a|
  2. B.If |z − a| = |1 − āz|, then |z| = 1✓ Correct
  3. C.If |z| = 1, then |z − a| < |1 − āz|
  4. D.If |1 − āz| < |z − a|, then |z| < 1

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 execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

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

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

ShareWhatsAppTelegram