# 9.3: Tartaglia-Cardano Revisited

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

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.

This page titled 9.3: Tartaglia-Cardano Revisited is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Bob Dumas and John E. McCarthy via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.