Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 2F / i

Solution

Let

K = {z : ∣ Re z ∣ + ∣ Im z ∣ \leq 1}

u

normal to a supporting side of this convex set, with

Re (\overline{u} ζ) > h := max_{z \in K} Re (\overline{u} z) .

Such a choice exists explicitly: take

u = (s + i t) / 2

, where

s, t \in {- 1, 1}

are the signs of the real and imaginary parts of

ζ

; then

h = 1/ 2

and the strict inequality is exactly the hypothesis. Set

c = - R u

. Since

∣ z ∣ \leq 1

K

∣ z - c ∣^{2} \leq R^{2} + 2 R h + 1, ∣ ζ - c ∣^{2} = R^{2} + 2 R Re (\overline{u} ζ) + ∣ ζ ∣^{2} .

For sufficiently large

R

, the former is strictly smaller than the latter uniformly on

K

. Consequently

q = max_{K} ∣ z - c ∣/∣ ζ - c ∣ < 1

The geometric series now supplies actual approximating polynomials:

p_{N} (z) = - \frac{1}{ζ - c} \sum_{j = 0}^{N} (\frac{z - c}{ζ - c})^{j}, sup_{z \in Γ} p_{N} (z) - \frac{1}{z - ζ} \leq \frac{q ^{N + 1}}{∣ ζ - c ∣ ( 1 - q )} .

Choose

N

so the right-hand side is below

ϵ

. This proves the required uniform approximation without assuming a general approximation theorem.

z \in Γ sup ∣ p_{N} (z) - (z - ζ)^{- 1} ∣ < ϵ for sufficiently large N .

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