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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 $n=2$

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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.

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  • $\begingroup$ Thank you very much for your answer.May you elaborate on your véry elegant answer in particular on the logical part? $\endgroup$
    – Ali Taghavi
    Commented Jun 8, 2016 at 19:52
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    $\begingroup$ Let me know if you're asking about something else, but essentially I was converting a formula of the form $\forall_{z, w} A(z, w) \Rightarrow B(z, w)$ into $\neg \exists_{z, w} \neg (A(z, w) \Rightarrow B(z, w))$ (i.e., using $\forall = \neg \exists \neg$), and then converting $\neg (A \Rightarrow B)$ into $\neg (\neg A \vee B)$, and that into $A \wedge \neg B$. The reason for those manipulations is that Tarski-Seidenberg refers to taking images, directly tied to existential quantification since (for $\pi: (x, y) \mapsto y$) the image $\pi(A)$ is exactly $\{y: \exists_x (x, y) \in A\}$. $\endgroup$
    – Todd Trimble
    Commented Jun 8, 2016 at 20:19
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    $\begingroup$ Quantification is all we have to worry about, as semi-algebraic sets are closed under the other logical operations $\neg, \wedge, \vee$, practically by definition. By the way, I first learned of all this by reading the book Tame Topology and O-minimal Structures by van den Dries, which I recommend as it is written for mathematicians with varying backgrounds (not including a background in logic particularly, besides what everyone knows). $\endgroup$
    – Todd Trimble
    Commented Jun 8, 2016 at 20:22

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