Periods of Negative-regular Continued Fractions for rational numbers and idempotents in the Modular group

If $x<0$ then there exists the shortest developments of $x$ and $1/x$ into finite negative-regular continued fractions:
\begin{equation*} x=\frac{-1}{a_1}\,\underset{+}{}\frac{-1}{a_2}\,\underset{+\cdots+}{}\frac{-1}{a_k}\,,\quad \frac{1}{x}=\frac{-1}{c_1}\,\underset{+}{}\frac{-1}{c_2}\,\underset{+\cdots+}{}\frac{-1}{c_m}\,. \end{equation*}
These expansions are obtained by a generalized Euclidean algorithm. Then
\begin{equation}\label{minperiod2011} x\oplus s(x^{-1})\overset{def}{=}\{a_1,a_2,\ldots,a_{k-1},a_k+c_m,c_{m-1},\ldots,c_1\}\mapsto x \end{equation}
is the minimal period for $x$. We call this $b_k=a_k+c_m$ as the marked element of the minimal period $\{b_1,\ldots,b_n\}$.
Theorem. Let $P=\{b_1,\ldots, b_n\}$ be the minimal period of a rational number $x\neq -1$ with marked element $b_k$. Then the period $P^*$, in which $b_k$ is replaced with $b_k-2$, corresponds to a Möbius transform of order $2$. Any Möbius transform of order $2$ in the modular group is obtained this way.
Example 1:
\begin{equation*} \begin{aligned} &\{1,3,2,\overset{\blacktriangledown}{6},2\}\mapsto -\frac{7}{4}\\ &\{1,3,2,\overset{\blacktriangledown}{4},2\}\rightarrow-\frac{17\pm i}{10}\\ &\begin{matrix} &1&3&2&4&2&\\ \frac{1}{0}&\frac{0}{1}&\frac{-1}{1}&\frac{-3}{2}&\frac{-5}{3}&\frac{-17}{10}&\frac{-\mathbf{29}}{\mathbf{17}} \end{matrix} \end{aligned}\quad \begin{aligned} \frac{7}{4}&=1+\frac{1}{1}\,\underset{+}{}\,\frac{1}{2+1}\,;\\ \frac{\mathbf{29}}{\mathbf{17}}&=1+\frac{1}{1}\,\underset{+}{}\,\frac{1}{2}\,\underset{+}{}\,\frac{1}{2}\,\underset{+}{}\,\frac{1}{1}\,\underset{+}{}\,\frac{1}{1}\,. \end{aligned} \end{equation*}
Example 2:
\begin{equation*} \begin{aligned} &\{1,1,\overset{\blacktriangledown}{4},1,1\}\mapsto 1\\ &\{1,1,\overset{\blacktriangledown}{2},1,1\}\rightarrow \pm i\\ &\begin{matrix} &1&1&2&1&1&\\ \frac{1}{0}&\frac{0}{1}&\frac{-1}{1}&\frac{-1}{0}&\frac{-1}{-1}&\frac{0}{-1}&\frac{\mathbf{1}}{\mathbf{0}} \end{matrix} \end{aligned}\quad \begin{aligned} 1&=0+\frac{1}{0+1}\,;\\ \frac{\mathbf{1}}{\mathbf{0}}&=0+\frac{1}{0}\,\underset{+}{}\,\frac{1}{0}\,\underset{+}{}\,\frac{1}{0}\,. \end{aligned} \end{equation*}

This is a joint result with Mikhail Yu. Tyaglov.