Exercise 1.3.7

Question

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\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

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ewcommand{e}{\,\mathrm{Re}\,}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Example 1.3.2 expressed the root $y=4$ of $y^3-15y-4$ in terms of Cardan's formulas. Find the other two roots, and eplain how Cardan's formulas give these roots.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} Since
\begin{align*}
y^3-15y-4 &= (y-4)(y^2+4y+1)\
&=(y-4)[(y+2)^2-3]\
&=(y-4)(y+2-\sqrt{3})(y+2+\sqrt{3}),
\end{align*}
the three roots of $y^3-15y-4$ are $4,-2+\sqrt{3},-2-\sqrt{3}$ and are all real.

The Cardan's formulas, with $q/2 = -2,p/3=-5 $, give 
$$-\frac{q}{2}+ \sqrt{\left (\frac{q}{2}ight)^2+\left (\frac{p}{3}ight)^3} = 2 + \sqrt{4 - 125} = 2 + 11i.$$

Using the Bombelli's note :
$$(2+i)^3 = 2+11i,$$
so we can take for $z_1$ a cube root of $2+11i$ given by 
$$z_1 = 2+i.$$
$z_2$ is the cube root of $-\frac{q}{2}- \sqrt{\left (\frac{q}{2}ight)^2+\left (\frac{p}{3}ight)^3}= 2 - i \sqrt{11}$ verifying $z_1 z_2 = -p/3 = 5$.

By Ex. 1.3.6,  $$z_2 = \overline{z_1} = 2 -i.$$
The roots of $y^3-15y-4$, using Cardan's formulas, are
\begin{align*}
y_1 &= z_1+ z_2,\
y_2 &= \omega z_1 + \omega^2 z_2 = 2\, \mathrm{Re}(\omega z_1),\
y_3 &= \omega^2 z_1 + \omega z_2 = 2\, \mathrm{Re}(\omega^2 z_1),\
\end{align*}
so
\begin{align*}
y_1 &= 2+i + 2 -i = 4,\
y_2 &=\mathrm{Re}[(-1+i\sqrt{3})(2+i)] = -2 - \sqrt{3},\
y_3 &= \mathrm{Re}[(-1-i\sqrt{3})(2+i)] = -2 + \sqrt{3}.\
\end{align*}
\end{proof}

Exercise 1.3.7

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Exercise 1.3.7

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