Question

Periodic Points and Multipliers 1.13. We defined the multiplier of ϕ at a fixed point α to be the derivative λα​(ϕ)=ϕ′(α). This definition obviously must be modified if α=[infinity]. In order to be compatible with Proposition 1.9, we must set λ[infinity]​(ϕ)=(ϕf)′(f⁻¹([infinity])) for all f∈PGL₂(C). Show that taking f(z)=1/z leads to the definition λ[infinity]​(ϕ)=limz→0​z⁻²ϕ′(z⁻¹)/ϕ(z⁻¹)²​. Prove that λ[infinity]​(ϕ) is finite and may be computed without taking a limit by showing that the fraction on the righthand size is a rational function in C(z) with no pole at z=0, so it may be evaluated at z=0.

Answer 1

To define λ at infinity for φ, we consider the transformation f(z)=1/z and prove that the limit expression for the multiplier is a rational function without a pole at z=0. This allows us to compute λ without taking a limit by direct evaluation at z=0.

Step-by-step explanation:

The question asks for the definition of the multiplier λ at infinity for a transformation φ and how to prove it is finite without taking a limit. By selecting the function f(z)=1/z, one can find the derivative at the inverse of infinity, which translates to evaluating the limit as z approaches 0. To show that λ∞​(φ) is finite and can be calculated without taking a limit, we must prove that the fraction z⁻²φ'(z⁻¹)/φ(z⁻¹)² is a rational function with no pole at z=0, therefore it can be evaluated directly at z=0.