Put $$A=\{(a\_{0},a\_{1},\ldots,a\_{n}) \in \mathbb{C}^{n+1}\mid p(z)=a\_{0}+a\_{1}z+\ldots a\_{n}z^{n} \;\;\text{is a one-to one function on the unit disc} \{z\in \mathbb{C} \mid |z|\leq 1\}$$



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> Is $\{(|a\_{0}|,|a\_{1}|,\ldots,|a\_{n}|)\mid (a\_{i})\in A\}$ a semi algebraic set in $\mathbb{R}^{n+1}$?
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Of course this is true for $n=2$



Is this a semi algebraic set?
It's fairly immediate from the Tarski-Seidenberg theorem that it's semi-algebraic. More exactly, if you can define your set in the language of real closed fields using first-order logic, then the Tarski-Seidenberg theorem guarantees that the set is semi-algebraic. 


As a first step, the set $A$ is semi-algebraic, as it is defined by a formula $\forall\_{z, w} [z \bar{z} \leq 1 \wedge w \bar{w} \leq 1 \wedge a\_0 + a\_1 z + \ldots + a\_n z^n = a\_0 + a\_1 w +\ldots + a\_n w^n] \Rightarrow z = w$. Abbreviating this as $\forall\_{z, w} [|z| \leq 1 \wedge |w| \leq 1 \wedge p\_a(z) = p\_a(w)] \Rightarrow z = w$, this is equivalent to 


$$\neg \exists\_{z, w} (|z| \leq 1 \wedge |w| \leq 1 \wedge p\_a(z) = p\_a(w)) \wedge \neg (z = w)$$ 


where the subset $\{(z, w, a) \in \mathbb{C}^{n+3}: |z| \leq 1 \wedge |w| \leq 1 \wedge p\_a(z) = p\_a(w) \wedge z \neq w\}$ (before quantification) is clearly semi-algebraic. Existentially quantifying over $z, w$ corresponds to taking the image of that subset under the projection $(z, w, a) \mapsto a$; this image is semi-algebraic by Tarski-Seidenberg, and then the complement of that is also semi-algebraic. 


Finally, your set is 


$$\{(r\_0, \ldots, r\_n): \exists\_{a \in \mathbb{C}^{n+1}} a \in A \wedge r\_0^2 = a\_0 \bar{a\_0} \wedge \ldots \wedge r\_n^2 = a\_n \bar{a\_n} \wedge r\_0 \geq 0 \wedge \ldots \wedge r\_n \geq 0\}$$ 


where, just as before (taking an image of a semi-algebraic set, defined in abbreviated form by $a \in A \wedge r^2 = a\bar{a} \wedge r \geq 0$, under the projection $(a, r) \mapsto r$), we easily conclude this is semi-algebraic. 


