The key point is that rationalizing (here $\color{#c00}{\textit{real-izing}}$) denominators allows one to lift "existence of inverses of elements $\ne 0\,$" from $\mathbb R$ to $\mathbb C,\,$ i.e. since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\ \Rightarrow\ r^{-1}\in \mathbb R,\:$ so
$$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne{\alpha\alpha}' = r\in \mathbb R\ \ \Rightarrow \underbrace{\frac{1}\alpha\, =\, \frac{1\ \alpha'}{\alpha\:\alpha'}\, =\, \frac{\alpha'}r\in\mathbb C}_{\textstyle{\color{#c00}{\textit{real-ize}}\ the\ denominator} }$$
Thus $ $ field $\mathbb R\, \Rightarrow\, $ field $\mathbb C\ $ by using the norm $\rm\:\alpha\to\alpha\!\ \alpha'\:$ to lift existence of inverses from $\mathbb R$ to $\mathbb C.$
More generally we can construct "rational" ($\in$ Z) multiples $\ne 0$ of algebraic elements $\alpha\ne 0$ (of a domain D algebraic over a subring Z) via the constant term of a minimal polynomial of $\alpha$ over Z (vs. above use of norm = product of conjugates). Namely, since $\rm\:0\ne\alpha\in$ D is algebraic over Z, it is a root of a polynomial $\rm\:0\ne f(x)\in Z[x].\:$ W.l.o.g. $\rm\:f(0)\ne 0\:$ by $\rm\:f(\alpha)\ \alpha^n = 0\ \Rightarrow\ f(\alpha) = 0,\:$ since nonzero elements of a domain are cancellable. Thus we may write $\rm\:f(x) = x\ g(x)-n\:$ for $\rm\ 0 \ne n\in Z.\:$ So $\rm\ f(\alpha) = 0\ \Rightarrow\ \alpha\ g(\alpha) = n\in Z.\:$ So $\rm\:n\:$ is our "rational" ($\in$ Z) multiple $\ne 0\,$ of $\rm\,\alpha.\,$ As above, this enables us to "rationalize" a denominator $\rm\:\alpha\in D\:$ by multiplying by $\rm \alpha' = g(\alpha), $ viz.
$$\rm 0\ne\alpha\in D\ \ \Rightarrow\ \ 0\ne\alpha\alpha' = n\in Z\ \ \Rightarrow\ \ \frac{1}\alpha\ =\ \frac{1\ \alpha'}{\alpha\:\alpha'}\ = \frac{\alpha'}n\in D $$
This is a prototypical instance of the method of simpler multiples.
In particular, if domain D is an algebraic extension of a field F, then D is a field, since, as shown, every element $\ne 0\,$ of D divides an element $\ne 0\,$ of F. But elements $\ne 0\,$ of the field F are units (which remain units in D), and divisors of units are units. Yours is the special $\rm\ D = \mathbb C,\ \ F = \mathbb R.$
Generalizing the above from domains to rings allows one to conclude that integral (or primitive) extensions cannot increase Krull dimension (= max length of prime ideal chains), see here. Recall that a primitive extension is a ring extension $\rm R \subset E$ where every element of E is primitive over R, i.e. every element of E is a root of a polynomial $\rm\:f(x)\in R[x]\:$ that is primitive, i.e. content($\rm f$) = $1,\:$ i.e. the ideal in R generated by the coefficients of $\rm\:f\:$ is $\rm (1) = R.\:$ One easily shows that an element is primitive over R iff it is a root of a polynomial $\rm\:f\in R[x]\:$ have some coefficient being $1$ (or a unit). Thus primitive extensions are generalizations of integral extensions. Like integral elements, primitive elements satisfy the crucial property that they remain primitive modulo a prime P, since they are roots of a polynomial f with some coefficient $= 1,\: $ so f cannot vanish mod P. Due to this, the above-mentioned proof on Krull-dimension still works for primitive extensions. They play a key role in various characterizations of integral extensions, e.g. see papers by David E. Dobbs, e.g. see here.