= Solution
Let $K=\{z:|\operatorname{Re}z|+|\operatorname{Im}z|\leq1\}$. Choose a unit complex number $u$ normal to a supporting side of this <convex set>, with
$$
\operatorname{Re}(\overline u\zeta)>h:=\max_{z\in K}\operatorname{Re}(\overline uz).
$$
Such a choice exists explicitly: take $u=(s+it)/\sqrt2$, where $s,t\in\{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt2$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\leq1$ on $K$,
$$
|z-c|^2\leq R^2+2Rh+1,\qquad |\zeta-c|^2=R^2+2R\operatorname{Re}(\overline u\zeta)+|\zeta|^2.
$$
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>:
$$
p_N(z)=-\frac1{\zeta-c}\sum_{j=0}^N\left(\frac{z-c}{\zeta-c}\right)^j,\qquad
\sup_{z\in\Gamma}\left|p_N(z)-\frac1{z-\zeta}\right|\leq\frac{q^{N+1}}{|\zeta-c|(1-q)}.
$$
Choose $N$ so the right-hand side is below $\epsilon$. This proves the required <uniform approximation> without assuming a general approximation theorem.
$$
\boxed{\sup_{z\in\Gamma}|p_N(z)-(z-\zeta)^{-1}|<\epsilon\quad\text{for sufficiently large }N.}
$$
Back to article page
