Determinate the minimal polynomial of $\sqrt{a}$ where a is an algebraic complex number over $\mathbb{Q}$

Question

I have the following exercise as homework, but is harder than i thought. There is a well know answer to this question?.

Suppose that $a$ is an algebraic complex number over $\mathbb{Q}$, prove that $\sqrt{a}$ is algebraic, and determine its minimal polynomial.

Answer 1

If $a$ is a zero of $f(x) = c_n x^n+\ldots+c_1x+c_0$ over $\Bbb Q$, then $\sqrt a$ is a zero of $f(x^2)$, since $f((\sqrt a)^2) = f(a)=0$.