(i) Show that for every $\epsilon>0$ there is a polynomial $p: \mathbb{R} \rightarrow \mathbb{R}$ such that $\left|\frac{1}{x}-p(x)\right| \leqslant \epsilon$ for all $x \in \mathbb{R}$ satisfying $\frac{1}{2} \leqslant|x| \leqslant 2$.

[You may assume standard results provided they are stated clearly.]

(ii) Show that there is no polynomial $p: \mathbb{C} \rightarrow \mathbb{C}$ such that $\left|\frac{1}{z}-p(z)\right|<1$ for all $z \in \mathbb{C}$ satisfying $\frac{1}{2} \leqslant|z| \leqslant 2$.