Exercise 15.4.1

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 (15.36).
\begin{align*}
\varphi\left((1+i)zight) &= \frac{(1+i) \varphi(z) \varphi'(z)}{1 - \varphi^4(z)},\
\varphi\left((1-i)zight) &= \frac{(1-i) \varphi(z) \varphi'(z)}{1 - \varphi^4(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} Using the addition law together with (15.24):
$$\varphi(iz) = i \varphi(z), \qquad \varphi'(iz) = \varphi'(z),$$
 we obtain
\begin{align*}
\varphi((1+i)z) &= \varphi(z+iz)\
&=\frac{\varphi(z) \varphi'(iz) + \varphi'(z) \varphi(iz)}{1 + \varphi^2(z) \varphi^2(iz)}\
&=\frac{\varphi(z) \varphi'(z) + i\varphi'(z) \varphi(z)}{1 - \varphi^4(z)}\
&=\frac{(1+i) \varphi(z) \varphi'(z)}{1 - \varphi^4(z)}.
\end{align*}
Similarly, using $\varphi(-z) = - \varphi(z), \varphi'(-z) = \varphi'(z)$, we have
\begin{align*}
\varphi((1+i)z) &= \varphi(z-iz)\
&=\frac{\varphi(z) \varphi'(iz) - \varphi'(z) \varphi(iz)}{1 + \varphi^2(z) \varphi^2(iz)}\
&=\frac{\varphi(z) \varphi'(z) - i\varphi'(z) \varphi(z)}{1 - \varphi^4(z)}\
&=\frac{(1-i) \varphi(z) \varphi'(z)}{1 - \varphi^4(z)}.
\end{align*}
\end{proof}

Exercise 15.4.1

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Exercise 15.4.1

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