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

Is $\{(|a_{0}|,|a_{1}|,\ldots,|a_{n}|)\mid (a_{i})\in A\}$ a  semi algebraic set in $\mathbb{R}^{n+1}$?

Of course this is true for $n=2$
 A: 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. 
