Is there a sequence of polynomials $P_n$ such that
$\displaystyle\lim_{n \to \infty} P_n(z) = \begin{cases} 1, & \text{Im } z > 0\\ 0, & z \text{ is real,} \\ -1, & \text{Im } z < 0 \end{cases}$
I have no clue where to start from. Please provide some hints. Thanks in advance.
What we need of Runge's theorem is that if $K \subset\mathbb{C}$ is compact such that $\mathbb{C}\setminus K$ is connected, and $f$ holomorphic on $K$ (that means that there is an open neighbourhood $U$ of $K$ and $f\colon U\to\mathbb{C}$ is holomorphic), then there is a sequence of polynomials converging uniformly on $K$ to $f$.
Here, a single compact $K$ is not sufficient, we need a sequence $K_n$ of compact sets exhausting the plane.
Letting $A_n = \left\{ z \in\mathbb{C} : \lvert z\rvert \leqslant n \land \operatorname{Im} z \geqslant \frac{1}{n} \right\}$, $B_n = \left\{ z \in\mathbb{C} : \lvert z\rvert \leqslant n \land \operatorname{Im} z \leqslant -\frac{1}{n} \right\}$ and $C_n = \left\{ z\in\mathbb{C} : \lvert \operatorname{Re} z\rvert \leqslant n \land \operatorname{Im} z = 0\right\}$, we obtain an increasing sequence of compact sets $K_n = A_n\cup B_n \cup C_n$ with $\bigcup K_n = \mathbb{C}$. For all $n \geqslant 1$, the complement of $K_n$ is connected, so by Runge's theorem, there is a polynomial $P_n$ with $\lvert P_n(z)-1\rvert < \frac{1}{n}$ on $A_n$, $\lvert P_n(z)+1\rvert <\frac{1}{n}$ on $B_n$ and $\lvert P_n(z)\rvert < \frac{1}{n}$ on $C_n$.
