For the constructible numbers, any quadratic extension is a radical extension, and there must be a first non-real field, which is gotten by adjoining $\sqrt a$ for some real $a<0$; then we can just write $\sqrt a=i\sqrt{|a|}$. Thus, after repeatedly applying the formula for complex square roots, performing the required complex arithmetic, and finally discarding the $i$ component, any constructible $\alpha\in\mathbb R$ can be written in terms of real square roots.
Similarly, for the cubic-constructible numbers, there must be a first non-real field, which is gotten by adjoining $\beta=\omega\sqrt[3]{a+\sqrt b\,}+\omega^2\sqrt[3]{a-\sqrt b\,}$ where $b>0$. This has the form $c+di$ where $c$ and $d$ are real-cubic-constructible:
$$\beta=\frac{-1+i\sqrt3}{2}\sqrt[3]{a+\sqrt b\,}+\frac{-1-i\sqrt3}{2}\sqrt[3]{a-\sqrt b\,}$$
$$=-\frac12\left(\sqrt[3]{a+\sqrt b\,}+\sqrt[3]{a-\sqrt b\,}\right)+\frac{\sqrt3}{2}\left(\sqrt[3]{a+\sqrt b\,}-\sqrt[3]{a-\sqrt b\,}\right)i.$$
Clearly, complex numbers of this form are closed under field operations. They're also closed under cube roots:
$$(x+yi)^3=a+bi$$
$$x^3-3xy^2=a,\quad3x^2y-y^3=b$$
$$3xy^2=x^3-a,\quad(3x^2-y^2)y=b$$
$$\big(9x^3-(x^3-a)\big)y=3bx$$
$$(8x^3+a)^2y^2=9b^2x^2$$
$$(8x^3+a)^2(x^3-a)-27b^2x^3=0.$$
This equation is $9$th degree in $x$, but it's $3$rd degree in $x^3$; furthermore, evaluating at $x^3\to-\infty,\;x^3=-a/8,\;x^3=a,\;x^3\to+\infty$ shows three sign changes, so we can solve for three different values of $x^3$. Taking a real cube root gives $x$, and the middle equation gives $y$ in terms of $x$. So we get three cube roots of $a+bi$. (This applies to the general case $ab\neq0$, but the special cases are easy to handle.)
This shows that the real part of anything expressible with complex square and cube roots is expressible with real solutions of cubics.
This approach probably doesn't generalize to degrees higher than $3$ or $4$.
Best Answer
The identification $(x+\mathrm i\mkern1mu y)^2=a+\mathrm i\mkern1mu b$ leads to the equality $2xy=b$, hence the signs of $x$ and $y$ are the same if $b> 0$, opposite if $b<0$. This shows you cannot combine the signs of the real and imaginary parts of $z$ in an arbitrary way.