Solving z^n=a+ib using only radicals of positive real numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:33:43Z http://mathoverflow.net/feeds/question/107479 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107479/solving-znaib-using-only-radicals-of-positive-real-numbers Solving z^n=a+ib using only radicals of positive real numbers Hugo Chapdelaine 2012-09-18T15:34:29Z 2012-09-22T14:35:42Z <p>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 $(-,-)$. A key observation of these formulas is that it involves only square roots of <strong>positive real numbers</strong>. </p> <p>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 <strong>positive solvable</strong> 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 <strong>positive quantities</strong>. So this stimulates the following question: </p> <p><strong>Q</strong>: Is there some algebraic criterion plus some positivity condition which allows one to determine when is $z^n=a+ib$ positive solvable?</p> <p>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 $z^7-1\;\;$? </p> http://mathoverflow.net/questions/107479/solving-znaib-using-only-radicals-of-positive-real-numbers/107486#107486 Answer by Will Jagy for Solving z^n=a+ib using only radicals of positive real numbers Will Jagy 2012-09-18T16:49:56Z 2012-09-18T16:49:56Z <p>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 <a href="http://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow">CASUS IRREDUCIBILIS</a>, 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.</p> <p>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. </p> http://mathoverflow.net/questions/107479/solving-znaib-using-only-radicals-of-positive-real-numbers/107783#107783 Answer by Hugo Chapdelaine for Solving z^n=a+ib using only radicals of positive real numbers Hugo Chapdelaine 2012-09-21T16:40:58Z 2012-09-22T14:35:42Z <p>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 <strong>positive solvability</strong>.</p> <p>So here is a useful result:</p> <p><strong>Theorem</strong> 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 odd there exists a root of unity of odd order inside $M$. </p> <p>A proof of this result may be found for example in Brian Conrad's note: <a href="http://math.stanford.edu/~conrad/210BPage/handouts/radreal.pdf" rel="nofollow">radical tower and roots of unity</a>.</p> <p>We thus have the following corollary</p> <p><strong>Corollary</strong> 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 odd. Then $K$ cannot be embedded in a radical extension contained in $\mathbf{R}$. In particular, $f(x)$ is not positive solvable.</p> <p>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. Since $[K_n:\mathbf{Q}]=\varphi(n)/2$ we find that if $\varphi(n)/2$ is not a power of $2$ then $f_n(x)$ is not positive solvable.</p>