Show that $ζ$ is a Quadratic Integer in $Q[\sqrt{−3}]$

Question

So in the complex plane, there are three cube roots of one. Suppose we let $ζ$ be the cube root of one which has positive imaginary part. How can we show that $ζ$ is a quadratic integer in $Q[\sqrt{−3}]$ by writing it in the form $(a+b\sqrt{−3})/2$, where $a$ and $b$ are rational integers and $a$ and $b$ are either both even or both odd. Additionally what steps would I take to also write it in the form $$m + n(\frac{1 + \sqrt{−3}}{2})$$, where $m$ and $n$ are rational integers.

Answer 1

$$p(x)=x^3-1=(x-1)(x^2+x+1)$$

and since $\;\zeta\in\Bbb C\setminus\Bbb R\;$ and it is a root of $\;p(x)\;$ , it must be a root of the above quadratic.

We also have that

$$\zeta=-\frac12+\frac{\sqrt3}2i=-1+\frac{1+\sqrt{-3}}{2}$$