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Part BCSIR NET December 2024one-interior-point-fixes-which-side-maps-in

One interior point fixes which side maps in

Consider the half planes R₊ = {z = x + iy ∈C:x>0\in \mathbb{C} : x > 0} and R₋ = {z = x + iy ∈C:x<0\in \mathbb{C} : x < 0}, and the fractional linear transformations T1(z)=(z−1)/(z+1)T_{1}(z) = (z - 1)/(z + 1) and T2(z)=(z+1)/(z−1)T_{2}(z) = (z + 1)/(z - 1). Let disc 𝔻 = {z∈Cz \in \mathbb{C} : |z| < 1}. Which of the following statements is true?

  1. A.T1T_{1} and T2T_{2} conformally map R₊ and R₋ respectively, onto the disc 𝔻✓ Correct
  2. B.T1T_{1} and T2T_{2} conformally map R₋ and R₊ respectively, onto the disc 𝔻
  3. C.T1T_{1} and T2T_{2} conformally map the disc 𝔻 onto, respectively R₊ and R₋
  4. D.T1T_{1} and T2T_{2} conformally map the disc 𝔻 onto, respectively R₋ and 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, 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

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