Algebraic numbers are countable

Q2. Basic Topology (Rudin)

A complex number z is said to be algebraic if there are integers $a_0 , a_1 , \ldots , a_n$ , not all zero, such that

 $a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0$ .

 Prove that the set of all algebraic numbers are countable. Hint: For every positive integer N there are only finitely many equations with $n +  | a_0 | + | a_1 | + \ldots +  | a_n | = N$

Solution: 

Set A is countable if we can establish one-one correspondence between A and set of positive integers \(\mathbb{N}\)

By Fundamental theorem of Algebra  $a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0$  has exactly n complex roots.