Exercise 15.2.8

Question

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

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

ewcommand{\ssi} {si et seulement si }

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

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

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

Let $n$ be even, and let $P_n(u)$ be the polynomial from Theorem 15.2.5. Complete the proof of Corollary 15.2.6 by showing that the polar distances of the $n$-division points of the lemniscate are roots of $uP_n(u^4)(1-u^2)$.

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} 
The polar distances of the $n$-division points are
$$u_m = \varphi\left( m \frac{2 \varpi}{n}ight),\qquad m=0,1,\ldots,n-1.$$
If $n$ is even, then 
$$\varphi(nx) = \varphi(x)\frac{P_{n}\left(\varphi^4(x)ight)}{Q_{n}\left(\varphi^4(x)ight)} \varphi'(x).$$
With $x = \frac{2 \varpi}{n}$, we obtain
$$0 = \varphi(m \cdot 2 \varpi) = \varphi\left(n\cdot  m \frac{2 \varpi}{n}ight) = \varphi\left( m \frac{2 \varpi}{n}ight)\frac{P_n\left(\varphi^4\left( m \frac{2 \varpi}{n}ight)ight)}{Q_n\left(\varphi^4\left( m \frac{2 \varpi}{n}ight)ight)}\varphi'\left(m \frac{2 \varpi}{n}ight),$$
where, by Exercise 9, the denominator $Q_n\left(\varphi^4\left( m \frac{2 \varpi}{n}ight)ight)$ is non vanishing.

Since $\varphi'\left(m \frac{2 \varpi}{n}ight) = \pm\sqrt{1 - \varphi^4\left(m \frac{2 \varpi}{n}ight)}$, we obtain
$$0 = u_m P_n(u_m^4) \sqrt{1 - u_m^4}.$$
$\sqrt{1 - u_m^4} = \sqrt{1 - u_m^2} \sqrt{1+u_m^2}$, where $1+ u_m^2 
e 0$, thus $\sqrt{1 -u_m^4} = 0 \iff 1 - u_m^2 = 0$. Therefore $u_m= \varphi\left( m \frac{2 \varpi}{n}ight)$ is a root of 
$$u P_n(u^4)(1-u^2).$$
\end{proof}

Exercise 15.2.8

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Exercise 15.2.8

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