Exercise 15.4.5

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}}

Derive the two formulas for $\varphi\left((2+i)zight)$ stated in Example 15.4.3.

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 formula, together with (15.2) and the duplication formula (15.14), we obtain
\begin{align*}
\varphi\left((2+i)zight)&= \varphi(2z + iz)\
&=\frac{\varphi(2z)\varphi'(iz) + \varphi'(2z)\varphi(iz)}{1+ \varphi^2(2z) \varphi^2(iz)}\
&=\frac{\varphi(2z)\varphi'(z) +i \varphi'(2z)\varphi(z)}{1- \varphi^2(2z) \varphi^2(z)}\
\end{align*}
Starting from the duplication formula
$$\varphi(2z) = \frac{2 \varphi(z) \varphi'(z)}{1+ \varphi^4(z)},$$
we obtain by differentiation,
\begin{align*}
2 \varphi'(2z) &= 2 \frac{\left(\varphi'^2(z) + \varphi(z) \varphi''(z)ight)\left(1+ \varphi^4(z)ight)- 4 \varphi^4(z)\varphi'^2(z)}{(1+ \varphi^4(z))^2}\
\end{align*}
Then, using $\varphi'(z)^2 = 1 - \varphi^4(z), \varphi''(z) = -2 \varphi^3(z)$,
\begin{align*}
\varphi'(2z) &=\frac{(1 -3 \varphi^4(z))(1+ \varphi^4(z))- 4 \varphi^4(z)(1 - \varphi^4(z))}{(1+ \varphi^4(z))^2}.\
&= \frac{1 -6\varphi^4(z) + \varphi^8(z)}{\left(1+\varphi^4(z)ight)^2}.
\end{align*}
Thus
\begin{align*}
\varphi\left((2+i)zight)&=\frac{ \left(\frac{2 \varphi(z) \varphi'(z)}{1+ \varphi^4(z)}ight)\varphi'(z) +i  \left(\frac{1 -6\varphi^4(z) + \varphi^8(z)}{\left(1+\varphi^4(z)ight)^2}ight)\varphi(z)}{1-  \left(\frac{2 \varphi(z) \varphi'(z)}{1+ \varphi^4(z)}ight)^2 \varphi^2(z)}\
&= \varphi(z)\frac{2 \varphi'^2(z)(1+ \varphi^4(z)) + i (1-6\varphi^4(z) + \varphi^8(z))}{1+ \varphi^4(z))^2 - 4 \varphi^4(z) \varphi'^2(z)}\
&=\varphi(z) \frac{2(1-\varphi^8(z)) + i (1-6\varphi^4(z) + \varphi^8(z))}{1 + 2 \varphi^4(z) + \varphi^8(z) - 4 \varphi^4(z)(1 - \varphi^4(z))}\
&=\varphi(z) \frac{(-2+i) \varphi^8(z) - 6i \varphi^4(z) + 2+i}{5 \varphi^8(z) - 2 \varphi^4(z) + 1}.
\end{align*}

We factor these expressions, writing $u = \varphi(x)$:
\begin{align*}
5(5u^8 -2 u^4 + 1) &= 25u^8 - 10 u^4 + 5\
&=(5u^4 -1)^2 + 4\
&=(5u^4 - 1-2i)(5u^4 - 1 + 2i)
\end{align*}
Since $5 = (1+2i)(1-2i)$,
\begin{align*}
 5u^8 - 2 u^4 + 1 &= \left(\frac{5}{1+2i} u^4 - 1ight)\left(\frac{5}{1-2i} u^4 - 1 ight)\
&=((1-2i) u^4 - 1)((1+2i) u^4 - 1).
\end{align*}
The discriminant $\Delta$ of $f(x) = (-2+i)x^2 - 6ix + 2+i$ is given by $\frac{\Delta}{4} = (3i)^2 -(2+i)(-2+i) = -9+ 5 = -4 = (2i)^2$. Thus the roots of $f$ are
\begin{align*}
x_1 &= \frac{3i +2i}{-2+i} = \frac{5i}{-2+i} = (-2-i)i  = 1-2i,\
x_2 &= \frac{3i -2i}{-2+i} = \frac{i}{-2+i} = \frac{i(-2+i)}{(-2+i)(-2-i)} = \frac{1-2i}{5}.
\end{align*}
Therefore $f(x) = (-2+i)( x-1+2i)\left(x+\frac{2i-5}{5}ight)$, thus the substitution $x \leftarrow u^4$ gives
\begin{align*}
(-2+i)u^8 - 6i u^4 + 2+i &= (-2+i)( u^4-1+2i)\left(u^4+\frac{2i-5}{5}ight)\
&=( u^4-1+2i)\left((-2+i)u^4 + \frac{(-2+i)(2i-1)}{5}ight)\
&=( u^4-1+2i)((-2+i)u^4 - i)\
&= -i(u^4 - 1+2i)((-1-2i)u^4+1).
\end{align*}
Therefore $\varphi((2+i)z) = h(\varphi(z))$, where
\begin{align*}
h(u) &= u \frac{(-2+i)u^8 - 6i u^4 + 2+i}{ 5u^8 - 2 u^4 + 1} \
&= -iu \frac{(u^4 - 1+2i)((-1-2i)u^4+1)}{((1-2i) u^4 - 1)((1+2i) u^4 - 1)}\
&= -iu \frac{u^4 - 1 + 2i}{(-1+2i)u^4+1}.
\end{align*}
This gives (15.39):
$$\varphi((2+i)z) = -i\varphi(z) \frac{\varphi^4(z) + (-1+2i)}{(-1+2i) \varphi^4(z) + 1}.$$
\end{proof}

Exercise 15.4.5

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Exercise 15.4.5

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