Let $a+bi\in\mathbf{C}$ be a complex number with $a,b\in\mathbf{R}$. Then
it is easy to find exact solutions of $z^2=a+ib$. For example let $z=u+iv$. Then
$$
u^2+v^2=z\overline{z}=\sqrt{a^2+b^2}
$$
and
$$
u^2-v^2+2uvi=z^2=a+ib.
$$
From this we deduce that
\begin{align}
u=\pm \sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}\;\;\;\mbox{and}\;\;\; v=\pm \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}\;\;\; (\star)
\end{align}
So here the sign combinations which are allowed are $(+,+)$ and $(-,-)$ if $b\geq 0$
and $(-,+)$ and $(+,-)$ if $b<0$. A key observation of these formulas is that it involves only square roots of **positive real numbers**.

Let $f(z)\in\mathbf{C}[z]$ and let $u_i+iv_i$ be the roots of $f(z)$ with $u_i,v_i\in\mathbf{R}$. We will say that the equation $f(z)$ is **positive solvable** if it is possible to write the $u_i$'s and $v_i$'s as "algebraic" expressions over the rationals involving only the real and imaginary parts of the coefficients of $f(z)$ and successive applications of the operators $\sqrt[m]{}$ (for all $m$) applied to **positive quantities**. So this stimulates the following question:

**Q**: Is there some algebraic criterion plus some positivity condition which allows one to determine when is $z^n=a+ib$ positive solvable?

For example $z^3-1$, and $z^{2^r}-(a+ib)$ (use induction on $r$ and apply inductively the formulas $(\star)$). Also if $p=2^r+1$ is prime (a Fermat's prime) then the splitting field of $z^p-1$ can be constructed by taking a succession of quadratic extensions and again by the formulas $(\star)$ we see that $z^p-1$ is positive solvable. More generally, we see that $z^m-1$ is positive solvable if $m=2^rq_1q_2\ldots q_r$ where the $q_i$'s are distinct Fermat's primes. So what about $z^7-1\;\;$?