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} A key observation of these formulas is that it involves only square roots of positive real numbers.

Let $f(z)\in\mathbf{R}[z]$ 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 coefficients of $f(z)$ and a successive use of the operators $\sqrt[m]{}$ (for all $m$) applied to positive quantities. The positivity condition here makes sense since we start here with a polynomial with coefficients in $\mathbf{R}$. 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 $z^7-1\;\;$?