Can I approximate a unit vector in $C^d$ with complex algebraic vectors in the unit ball in $C^d$?

Question

My answer: Let $z=(z_1,.....z_d) \in C^d$ be so that $||z||=1$ and let $\epsilon>0$ be given. The algebraic complex numbers are dense in $C$. So, for all $i$, find $a_i$, complex algebraic such that $|a_i|<|z_i|....(1)$ and $|a_i-z_i|<(\epsilon)^{1/2}/2^d....(2)$. let $a=(a_1,.....a_d)$ By (1), $||a||<||z||=1$. So, a is in the unit ball. By (2), $||a-z||<\epsilon$. So, any point on the unit sphere can be approximated by a complex algebraic vector in the unit ball in $C^d$. So, the answer is yes. Is this proof correct???

Answer 1

Let $(z_0, \dots, z_n)\in \mathbb{C}$ with $|z|^2 = |z_0|^2 + \cdots + |z_n|^2 = 1$. Choose algebraic $w_0, w_1, \dots, w_n\in \mathbb{C}$ close to the respective $z_i$, and assume without loss of generality that $w_0\not = 0$ and that $r = 0 < |w_1|^2 + \cdots + |w_n|^2 \leq 1$. Then $w'_0 = (1 - r^2)^{1/2} |w_0|^{-1} w_0$ is also algebraic, and $w' = (w'_0, w_1 \dots, w_n)$ has $|w'|^2 = 1$.

(The only potential snag in making this argument more precise is the case $w_0 = 0$, which you can deal with either by treating that case separately or by induction. What makes this argument work is that $\bar{\mathbb{Q}}\subset \mathbb{C}$ is closed under taking square roots, whereas $\mathbb{Q}$ itself is not.)