Exercise 1.1.2

Question

\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}}

In Example 1.1.1, show that $y_2$ and $y_3$ are complex conjugates of each other.

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}
$z_1 = \sqrt[3]{\frac{-1+\sqrt{5}}{2}}$ and $z_2 = \sqrt[3]{\frac{-1-\sqrt{5}}{2}}$ are the real cube roots of $\frac{-1+\sqrt{5}}{2}$ and $\frac{-1-\sqrt{5}}{2}$.

Then
\begin{align*}
y_2 &= \omega z_1 +\omega^2 z_2,\
y_3 &= \omega^2 z_1 +\omega z_2,\
\end{align*}
are complex conjugates.

Indeed, $ \omega^3 =1$, so $\omega^2 = \frac{1}{\omega} = \overline{\omega}$, since $\omega \overline{\omega} = \vert \omega \vert ^2= 1$.

Moreover $\overline{\omega^2} = \overline{\omega}^2 = \omega^4 = \omega$.

Consequently 
\begin{align*}
\overline{y_2} &= \overline{\omega}\, \overline{z_1} + \overline{\omega}^2\,  \overline{z_2}\
&= \omega^2 z_1 + \omega z_2\
&=y_3.
\end{align*}
\end{proof}

Exercise 1.1.2

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Exercise 1.1.2

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