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 $z^7-1\;\;$?

  • $\begingroup$ There is at least the following easy observation: if $f(z)\in\mathbf{R}[z]$ has all its roots real then a (reduced) writting of a root of $f(z)$ in terms of radicals should only involve $\sqrt[m]{}$ with integers $m$ such that $\varphi(m)=2^r$, since for $p>2$ prime the group of $p$-th roots of unity cannot be embedded in the splitting field of $f(z)$ which is included in $\mathbf{R}$ by hypothesis. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 17:50
  • $\begingroup$ Of course one has to be careful by the meaning of a "reduced writting". For example $\sqrt[5]{1}$ would not be a reduced writting for me since the 4 complex 5-th roots of unity can be written in terms of $\sqrt[2]$. I'm not sure at this point that I would be able to define properly what I mean by a reduced writting... But definitely an expression like $\sqrt[n]{1}$ (for $n>1$) would not be reduced but its reduced writing would only involves a succession of $\sqrt[p]{}$ with $p$ prime dividing $\varphi(n)$. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 18:00
  • $\begingroup$ I can't understand! What is not clear is the primitive solution of the problem. It also Pythagorean triples. $$u^2+v^2=\sqrt{a^2+b^2}$$ The decision boils down to this primitive idea. $$a=2uv$$ $$b=u^2-v^2$$ $\endgroup$ – individ Oct 8 '15 at 4:48

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.

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    $\begingroup$ Hi Will, indeed I knew about the casus irreducibilis which might one of the key reason (swing factor) for the acceptation by certain mathematicians of complex numbers for solving cubic polynomials. Since as you explained, when an irreducible real cubic polynomial has 3 real roots (so Galois group $\simeq C_3$) then none of the roots can be written in terms of radicals of positive quantities. This means that the cancellation of the imaginary part (which can only takes place in the complex world) is an unavoidable phenomenon when you shoehorn yourself to only allow the use of radicals. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 17:14
  • $\begingroup$ Note also that the proof that the $3$ real roots cannot be written in terms of radicals of positive quantities follows from the key observation that for a prime number $p$ and an arbitrary field $K$, the polynomial $z^p-a$ with $a\in K$ is either irreducible or $a$ is a $p$-th in $K$. Therefore at the last step of the sequence of fields appearing in the tower one must take a cube root. Since the splitting field is normal and real this would mean that the all 3rd roots of unity are real which is a contradiction. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 17:31
  • $\begingroup$ I meant $a$ is a $p$-th power in $K$. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 17:32
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    $\begingroup$ The restriction to degree 3 can be relaxed (and Cardano bypassed): if $f$ is an irreducible polynomial of any odd degree $> 1$ over a field $F$ of characteristic 0 then its splitting field over $F$ contains a root of unity of odd prime order and so cannot be embedded into $\mathbf{R}$. In particular, when $F$ is given as a subfield of $\mathbf{R}$ and the Galois group of $f$ is abelian over $F$ (e.g., cyclic), so the splitting field is generated over $F$ by a single root of $f$, then it is impossible to find any root of $f$ inside a radical tower over $F$ inside $\mathbf{R}$. $\endgroup$ – grp Sep 18 '12 at 20:48
  • $\begingroup$ Hi @grp, that is quite a nice result. So in particuar using your observation we see that $z^n-1$ is positive solvable iff $\varphi(n)$ is a power of $2$ so iff $n=2^rq_1q_2\ldots q_s$ where the $q_i$'s are distinct Fermat's primes. $\endgroup$ – Hugo Chapdelaine Sep 18 '12 at 21:02

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.

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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.

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  • $\begingroup$ "The answer to your question follows directly from the paper mentioned above." above is not constant across platforms and viewing modes. Could you please employ some more invariant term, so I can know what paper you mean? $\endgroup$ – Gerry Myerson Dec 19 '14 at 5:47
  • $\begingroup$ Done. Note that there are some inaccuracies in Weisstein's overview, but the key result is the same: $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals(i.e. composition of algebraic operations and rational powers at rational numbers) iff $\varphi(n)=2^m$. $\endgroup$ – Alexander Dec 19 '14 at 11:35
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    $\begingroup$ Dear Alexander, already in the answer that I had posted on Sep 21, 2012, I was quoting Brian Conrad's note: "radical tower and roots of unity" $\endgroup$ – Hugo Chapdelaine Dec 19 '14 at 17:12
  • $\begingroup$ Dear Hugo, in my answer I mentioned that this article was already quoted "above". By "above" I meant exactly your post. And now I gave a complete answer to your question. Although this observation is based on the result from "radical tower and roots of unity", it required some additional analysis, now I think that my phrase "follows directly" is a bit incorrect. $\endgroup$ – Alexander Dec 19 '14 at 18:53
  • $\begingroup$ Ooh I see! I probably misread your question: did you mean a criterion for every specific a,b,n? It's a different problem... I just found all $n$ such that $z^n=a+ib$ is positive solvable for all $a,b\in R$. $\endgroup$ – Alexander Dec 19 '14 at 19:10

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