Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 2F (Topics in Analysis)

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

Yes. Take exterior poles ${\zeta }_n=1+1/n$. By part (i), choose a polynomial $p_n$ withFor any fixed $z\in \Gamma$, $z\ne 1$, and therefore $(z-{\zeta }_n{)}^{-1}\to (z-1{)}^{-1}$. The triangle inequality provesThis is pointwise convergence. It is not uniform convergence on $\Gamma$: every polynomial is bounded on the closed diamond, whereas $(z-1{)}^{-1}$ is unbounded as $z\to 1$ along $\Gamma$.