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Let $G$ be an open connected set in $\mathbb{C}^n$. Let $u,v\in G$. Now we know that their exists a path $\gamma:[0,1]\longrightarrow G$ Such that $\gamma(0)=u$ and $\gamma(1)=v$. Then it is said that by the Weierstrass approximation theorem, we find a polynomial $P:[0,1]\longrightarrow G$ such that $P(0)=u$ and $P(v)=v$. Then it is said that it is easy to choose a simply connected domain $D\in \mathbb{C}, [0,1]\subset D,$ such that $P(\lambda)\in G$ for every $\lambda\in D$.

My questions are;

1) How do we get such a $P$. I know that $ \gamma$ can be approximated uniformally by polynomials $P_n$. But how do we get a $P$ among them such that $P(0)=u$ and $P(v)=v$. Like the sequence $\frac{1}{n}\longrightarrow 0$. But no member of the sequence is equal to zero.

2)What theorem guarantees the existence of such a simply connected domain $D$?

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Suppose $\|\gamma-P\|_\infty < \epsilon$ and let $l(t) = (1-t)(\gamma(0)-P(0)) + t (\gamma(1)-P(1))$ and let $Q=P+l$. Then $\|\gamma-Q\|_\infty \le \|\gamma-P\|_\infty+ \|l\|_\infty < 2 \epsilon$.

Since $G$ is open, for sufficiently small $\epsilon>0$ we have $Q(t) \in G$ and $Q(t) = \gamma(t)$ for $t \in \{0,1\}$.

Suppose you have a polynomial $P$ such that $P([0,1]) \subset G$. This is smooth and defined everywhere on $\mathbb{C}$. In particular, $P^{-1}(G)$ is open and contains $[0,1]$. In particular, we can find some $\delta>0$ such that $[0,1]+B(0,\delta) \subset P^{-1}(G)$. This is a suitable simply connected domain.

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