Exercise 15.4.6

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}}

Prove the third and fourth lines of (15.42):
\begin{align*}
\varphi((\beta + 1)z) &= -\varphi((\beta-1)z) + \frac{2 \varphi(\beta z)\varphi'(z)}{1 + \varphi^2(\beta z) \varphi^2(z)},\
\varphi((\beta + i )z) &= -\varphi((\beta-i)z) + \frac{2 \varphi(\beta z)\varphi'(z)}{1 - \varphi^2(\beta z) \varphi^2(z)}.
\end{align*}

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} 
By (15.13),
$$\varphi(x+y)+ \varphi(x-y) = \frac{2 \varphi(x) \varphi'(y)}{1 + \varphi^2(x) \varphi^2(y)},$$
and, by the Principle of Analytic Continuation, this remains true for all $x,y$ such that both members are defined.

If we use the substitution $(x,y) \leftarrow (\beta z, z)$, we obtain
$$\varphi((\beta + 1)z) + \varphi((\beta-1)z) = \frac{2 \varphi(\beta z)\varphi'(z)}{1 + \varphi^2(\beta z) \varphi^2(z)},$$
and the substitution $(x,y) \leftarrow (\beta z, iz)$ gives, using $\varphi(iz) = i \varphi(z), \varphi'(iz) = \varphi'(z)$,
\begin{align*}
\varphi((\beta + i)z) + \varphi((\beta-i)z) &= \frac{2 \varphi(\beta z)\varphi'(iz)}{1 + \varphi^2(\beta z) \varphi^2(iz)}\
&=\frac{2 \varphi(\beta z)\varphi'(z)}{1 - \varphi^2(\beta z) \varphi^2(z)}.
\end{align*}
\end{proof}

Exercise 15.4.6

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Exercise 15.4.6

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