9.3: Tartaglia-Cardano Revisited

Let us consider again Example 9.9. We wanted to find the cube roots of \[\zeta_{\pm}=\frac{-1 \pm \sqrt{-3}}{2} .\] If we take the \(+\) sign, we get \[\zeta_{+}=\operatorname{Cis}(2 \pi / 3),\] and if we take the - sign, we get \[\zeta_{-}=\operatorname{Cis}(4 \pi / 3) .\] So \(\zeta_{+}\)has 3 roots, namely \[\left\{\operatorname{Cis}\left(\frac{2 \pi}{9}+\frac{2 k \pi}{3}\right): k=0,1,2\right\} \text {, }\] and \(\zeta_{-}\)has 3 roots, namely \[\left\{\operatorname{Cis}\left(\frac{4 \pi}{9}+\frac{2 k \pi}{3}\right): k=0,1,2\right\} \text {, }\] Knowing \(w\), we want to find \(x\), which for Example \(9.9\) is given by \(w+1 / w\). For any number \(w\) that can be written as \(\operatorname{Cis}(\theta)\) (i.e. any complex number of modulus 1), we have \[\begin{aligned} w+\frac{1}{w} &=\cos \theta+i \sin \theta+\cos (-\theta)+i \sin (-\theta) \\ &=2 \cos \theta . \end{aligned}\] Therefore the roots of the polynomial given in (9.10) are \[\left\{2 \cos \frac{2 \pi}{9}, 2 \cos \frac{8 \pi}{9}, 2 \cos \frac{14 \pi}{9}, 2 \cos \frac{4 \pi}{9}, 2 \cos \frac{10 \pi}{9}, 2 \cos \frac{16 \pi}{9}\right\} .\] Are these 6 different roots? Theorem \(4.10\) says that \(p\) can have at most 3 different roots. As \(\cos (\theta)=\cos (2 \pi-\theta)\), we see our set (9.27) may be written as \[\left\{2 \cos \frac{2 \pi}{9}, 2 \cos \frac{4 \pi}{9}, 2 \cos \frac{8 \pi}{9}\right\} .\] It turns out that the Tartaglia-Cardano formula (9.7) does give all three roots of the cubic, and moreover it does not matter whether one chooses the \(+\) or \(-\) sign, as long as one calculates all 3 cube roots of (9.6) for some choice of sign. We shall use \(\mathbb{C}[z]\) to denote the set of polynomials in \(z\) with coefficients from \(\mathbb{C}\).

THEOREM 9.29. Consider the polynomial \[p(z)=z^{3}+a z+b\] in \(\mathbb{C}[z]\), and assume \(a \neq 0\). Let \(c=-a / 3\), and let \(\zeta\) be \[\zeta=\frac{-b+\sqrt{b^{2}-4 c^{3}}}{2} .\] Let \(w_{1}, w_{2}, w_{3}\) be the three distinct cube roots of \(\zeta\). For each \(w_{i}\), define \(z_{i} b y\) \[z_{i}=w_{i}+\frac{c}{w_{i}} .\] Then \[p(z)=\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right) .\] REMARK. It will follow from the proof that it doesn’t matter which square root of \(b^{2}-4 c^{3}\) one chooses in (9.31).

Proof. If \(p\) is given by (9.32), then \[p(z)=z^{3}-\left(z_{1}+z_{2}+z_{3}\right) z^{2}+\left(z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right) z-\left(z_{1} z_{2} z_{3}\right) .\] We must show that the coefficients in (9.33) match those in (9.30). By Proposition 9.26, we can assume \[w_{1}=\omega w_{3}, \quad w_{2}=\omega^{2} w_{3}\] where \(\omega=-\frac{1}{2}+i \frac{\sqrt{3}}{2}\) is a primitive cube root of unity. In the following calculations, we use the facts that \(\omega^{2}=1 / \omega\) and \(1+\omega+\omega^{2}=0\). (Why are these true?) Notice that \(w_{3} \neq 0\), as that would force \(c=0\).

The coefficient of \(z^{2}\) in (9.33) is \[\begin{aligned} -\left(z_{1}+z_{2}+z_{3}\right) &=-w_{3}\left(\omega+\omega^{2}+1\right)-\frac{1}{w_{3}}\left(\omega^{2}+\omega+1\right) \\ &=0 . \end{aligned}\] The coefficient of \(z\) is \[\begin{aligned} z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=&\left(\omega w_{3}+c \omega^{2} \frac{1}{w_{3}}\right)\left(\omega^{2} w_{3}+c \omega \frac{1}{w_{3}}\right) \\ &+\left(\omega^{2} w_{3}+c \omega \frac{1}{w_{3}}\right)\left(w_{3}+c \frac{1}{w_{3}}\right) \\ &+\left(w_{3}+c \frac{1}{w_{3}}\right)\left(\omega w_{3}+c \omega^{2} \frac{1}{w_{3}}\right) \\ =& w_{3}^{2}\left(1+\omega^{2}+\omega\right)+3 c\left(\omega+\omega^{2}\right)+\frac{c^{2}}{w_{3}^{2}}\left(1+\omega+\omega^{2}\right) \\ =&-3 c \\ =& a . \end{aligned}\] The constant term in (9.33) is \[\begin{aligned} -z_{1} z_{2} z_{3} &=-\left(\omega w_{3}+c \omega^{2} \frac{1}{w_{3}}\right)\left(\omega^{2} w_{3}+c \omega \frac{1}{w_{3}}\right)\left(w_{3}+\frac{1}{w_{3}}\right) \\ &=-w_{3}^{3}-c w_{3}\left(1+\omega^{2}+\omega\right)-\frac{c^{2}}{\frac{1}{w_{3}}}\left(\omega+1+\omega^{2}\right)-\\ &=-\zeta-\frac{c^{3}}{\zeta} \\ &=-\frac{-b+\sqrt{b^{2}-4 c^{3}}}{2}-\frac{2 c^{3}}{-b+\sqrt{b^{2}-4 c^{3}}} \\ &=\frac{-b^{2}+2 b \sqrt{b^{2}-4 c^{3}}-\left(b^{2}-4 c^{3}\right)-4 c^{3}}{2\left(-b+\sqrt{b^{2}-4 c^{3}}\right)} \\ &=\frac{b\left(-b+\sqrt{b^{2}-4 c^{3}}\right)}{-b+\sqrt{b^{2}-4 c^{3}}} \\ &=b . \end{aligned}\] Therefore all the coefficients of (9.30) and (9.32) match, so they are the same polynomial.

The Tartaglia-Cardano formula therefore gives all three roots to a reduced cubic polynomial \(p\) with complex coefficients (repeated roots can occur). If the coefficients \(a\) and \(b\) are real, we know from the Intermediate Value Theorem that at least one of the three roots of \(p\) will be real (See Exercise 8.31). As Example 9.9 shows, however, it may still be necessary to take the cube root of a complex \(\zeta\) to obtain the real roots of a real cubic. This realization was what led to the acceptance of complex numbers as useful objects rather than a bizarre fantasy.