Exercise 15.3.3

Question

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

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

ewcommand{\ssi} {si et seulement si }

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

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

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

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

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

Prove the formula for $\varphi\left(z \pm \frac{\varpi}{2} iight)$ stated in the proof of Theorem 15.3.2.

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} 
By (15.24) and (15.23), 
$$\varphi\left(\frac{\varpi}{2}iight) = i \varphi\left(\frac{\varpi}{2}ight) = i, \qquad \varphi'\left(\frac{\varpi}{2}iight) = \varphi'\left(\frac{\varpi}{2}ight) = 0,$$
and
$$\varphi\left(-\frac{\varpi}{2}iight) = i\varphi\left(-\frac{\varpi}{2}ight) = -i,\qquad  \varphi'\left(-\frac{\varpi}{2}iight) = \varphi'\left(-\frac{\varpi}{2}ight) = 0.$$
Using the addition law (Proposition 15.3.1(b)), we see that
\begin{align*}
\varphi\left( z + \frac{\varpi}{2}i ight) &= 
\frac{\varphi(z) \varphi'\left( \frac{\varpi}{2}iight) + \varphi\left(\frac{\varpi}{2}iight) \varphi'(z)}{1 + \varphi^2(z) \varphi^2\left(\frac{\varpi}{2}iight)}\
&= \frac{i \varphi'(z)}{1 - \varphi^2(z)},
\end{align*}
and similarly, 
\begin{align*}
\varphi\left( z -\frac{\varpi}{2}i ight) &= 
\frac{\varphi(z) \varphi'\left( -\frac{\varpi}{2}iight) + \varphi\left(-\frac{\varpi}{2}iight) \varphi'(z)}{1 + \varphi^2(z) \varphi^2\left(-\frac{\varpi}{2}iight)}\
&= \frac{-i \varphi'(z)}{1 - \varphi^2(z)}.
\end{align*}
We have proved
$$\varphi\left( z \pm \frac{\varpi}{2}i ight) = \pm  \frac{i \varphi'(z)}{1 - \varphi^2(z)}.$$
\end{proof}

Exercise 15.3.3

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Exercise 15.3.3

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