Solving $z^n=a+bi$ using only radicals of positive real numbers

Question

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 $(\star)$) are positive solvable. 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 $z^7-1\;\;$?

Answer 1

As soon as you get to general $$ z^3 = a + b i $$ this may be impossible. In solving a cubic $z^3 + p z + q = 0$ with, say, rational coefficients, one may use Cardano's formula. If there is only one real (irrational) root and two complex conjugate roots, then this works in the sense of being able to separately calculate real and imaginary parts using real square and cube roots. However, if there are three real irrational roots, a situation called CASUS IRREDUCIBILIS, then Cardano's formula is just a sum of cube roots of complex numbers $a+bi,$ with no way to separate real and imaginary parts. The terms are summed in a way that guarantees real answers, but that is not satisfying.

Now, if you begin with general $ z^3 = a + b i, $ say with $a,b \in \mathbb Q,$ and carefully write out equations for the real and imaginary parts of this $z,$ you get cubics with three real roots, where you cannot separate parts for those...it is all pretty circular, and hopeless.

Answer 2

This is not an answer to my question but it is just a summary of the relevant results which have been mentioned so far about positive solvability.

So here is a useful result:

Theorem Let $K/F$ be a finite Galois extension and let $M/F$ be a radical extension. Assume that you have an embedding $\iota:K\hookrightarrow M$. Then, if $[K:F]$ is divisible by an odd prime $p$, there exists a non-trivial root of unity of odd order inside $M$.

A proof of this result may be found for example in Brian Conrad's note: radical tower and roots of unity.

We thus have the following corollary

Corollary Let $f(x)\in\mathbf{Q}[x]$ be a polynomial which has all its roots in $\mathbf{R}$. Let $K$ be the (abstract) splitting field of $f(x)$ and assume that $[K:\mathbf{Q}]$ is divisible by an odd prime $p$. Then $K$ cannot be embedded in a radical extension contained in $\mathbf{R}$. In particular, $f(x)$ is not positive solvable.

As an example to illustrate this corollary, let $\zeta_n=e^{2\pi i/n}$. Let $f_n(x)$ be the minimal polynomial of $\zeta_n+\zeta_n^{-1}=2\cos(2\pi/n)$ over $\mathbf{Q}$ and let $K_n$ be its splitting field. We know that $[K_n:\mathbf{Q}]=\varphi(n)/2$. In particular, if $\varphi(n)/2$ is not a power of $2$ then $f_n(x)$ is not positive solvable.

Answer 3

Let's find all $n$ such that $\forall a,b \in R \Rightarrow z^n =a+ib$ is positive solvable. The answer follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable $\forall a,b\in R$ iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f_x(\sin x)$, where function $f_x$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.