Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 1A / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The fixed-point equation is

\frac{a z + b}{z + 1} = z ⟺ z^{2} + (1 - a) z - b = 0.

A unique solution at

z = i

means that this polynomial is

(z - i)^{2} = z^{2} - 2 i z - 1.

Comparing coefficients gives

a = 1 + 2 i, b = 1 .

This is the repeated-root case of prescribed fixed points of a Möbius transformation.

ii

Solution

3 i

and

1 + i

, the fixed-point polynomial must be

(z - 3 i) (z - 1 - i) = z^{2} - (1 + 4 i) z + (3 - 3 i) .

Comparing this with

z^{2} + (1 - a) z - b

gives

a = 2 + 4 i, b = - 3 + 3 i .

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