Nonlinear recurrence relation over $\mathbb{C}$ A sequence in $\mathbb{C}$ is given: $(a_n)_{n=1}^{\infty}$ s.t.
$a_1 = i$ and $a_{n+1} = \frac{3}{2 + a_n}$.
Assume $\lim(a_n)$ exists.
Using the fact that $\lim(a_n) = \lim(a_{n+1})$ and solving the resulting quadratic gives that $\lim(a_n)\in \{-3,1\}$.
I was trying to figure out which one the limit is. I tried finding a relation on the norm or the argument (using the initial condition, since clearly that is what the limit is determined by) and I also tried looking at individual terms and by this, the real part seems to stay positive and hover around one while the imaginary part we know goes to zero in any case (i.e., I am thinking the limit is 1). I wanted to ask for verification that I did this last part correctly, since $a_2 = \frac{6-3i}{5}$ if we can show that:
$Sgn[Re(a_m)] = Sgn[Re(a_{m+1})]$ $\forall m\geq 2$
we are done and the limit is 1. Suppose that $a_m = x+yi$ ($m\geq 2$) and $x > 0$. Then
$a_{m + 1}= \frac{3}{2 + x + yi}\frac{(2+x - yi)}{(2+x-yi)} = \frac{3(2+x) - 3yi}{(2+x)^2 + y^2}$
meaning that
$Sgn[Re(a_{m+1})] = Sgn\bigg[\frac{6 + 3x}{(2+x)^2 + y^2}\bigg] = 1$ since $x > 0$ and we are done. So the limit is 1. Does that look about right?
 A: A specific approach to convergence is to examine how $a_{n}-1$ changes in each iteration.
\begin{align}
a_{n+1} - 1 
&= \frac{3}{2+a_n} - 1 \\
&= \frac{1-a_n}{3-(1-a_n)}
\end{align}
Then if $|a_n-1|$ is small relative to $3$ it is apparent that $|a_{n+1}-1|$ decreases by  a factor of nearly $1/3$ on each iteration and so $a_{n+1} \to 1$.  In fact, as long as $3-|1-a_n|$ is greater than some bound greater than $1$, convergence is guaranteed.   For example if $|a_{n-1}-1|< 15/8$ we have,
\begin{align}
|3-(1-a_n)| &\geqslant 3 - |1-a_n| \\
&\geqslant \frac{9}{8} 
\end{align}
whence $|a_{n+1} -1| \leqslant \frac{8}{9} | a_n - 1 |$.
Starting with $a_1 = i$, $|a_1-1| = \sqrt 2 < \frac{15}{8}$ so the condition is met and the sequence converges to $1$.

Another more general approach is to recognise that recurrence is a particular case of a continued fraction.  Consider the expression,
\begin{align}
f = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{p_3}{\ddots}}}
\end{align}
If we define functions of $w \in \mathbb C$ by,
\begin{align}
t_0(w) = q_0 + w, \quad t_n(w) = \frac{p_n}{q_n +w}
\end{align}
then when we combine $t_0$ through $t_n$ we obtain,
\begin{align}
c_n(w) = t_0\circ t_1 \circ \cdots \circ t_n(w) = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{\ddots}{q_{n-1}+\dfrac{p_n}{q_n+w}}}}
\end{align}
and we interpret $f$ to be the limit of this as $n \to \infty$ when $w=0$ (if it exists).  The number $c_n(0)$ is called the $n$th convergent.
A useful recurrence relation exists for $c_n(w)$, $n \geqslant 0$,
\begin{align}
c_n(w) = \frac{P_n+P_{n-1}w}{Q_n +Q_{n-1} w} \tag{1}\label{eq1}
\end{align}
where we use initial conditions, $P_0 = q_0, Q_0 = 1, P_{-1} = 1, Q_{-1}=0$ and the $P_n, Q_n$ do not depend on $w$ and can be obtained from the recurrence,
\begin{align}
\begin{array}{l}
P_{n+1} = q_{n+1}P_n + p_{n+1} P_{n-1} \\
Q_{n+1} = q_{n+1}Q_n + p_{n+1} Q_{n-1} .
\end{array}\tag{2}\label{eq2}
\end{align}
This is proved by induction.  We observe that the given initial conditions imply equation \eqref{eq1} holds when $n=0$.  Then, if \eqref{eq1} holds for some $n \geqslant 0$, from the definition,
\begin{align}
c_{n+1}(w) &= c_n(p_{n+1}/(q_{n+1}+w)) \\
&= \frac{P_n+P_{n-1}(p_{n+1}/(q_{n+1}+w))}{Q_n+Q_{n-1}(p_{n+1}/(q_{n+1}+w))} \\
&=\frac{P_{n}q_{n+1}+P_n w + P_{n-1}p_{n+1}}{Q_{n}q_{n+1}+Q_n w + Q_{n-1}p_{n+1}} \\
&=\frac{P_{n+1}+P_n w}{Q_{n+1}+Q_n w}.
\end{align}
Thus \eqref{eq1} also holds for $n+1$ with the new coefficients $P_{n+1}, Q_{n+1}$ given by \eqref{eq2}.
Applied to the case in hand, we take $q_0 = 0, p_n = 3, q_n=2$ and $w = i$, so that $c_0(i) = i, c_1(i) = 3/(2+i)$ and in general $c_n(i) = a_{n+1}$.
We can solve the two recurrence relations for $P_n, Q_n$ using standard techniques with the initial conditions provided above, obtaining
\begin{align}
P_{n} &= \frac{3}{4} (3^{n} - (-1)^n) \\
Q_{n} &= \frac{1}{4} (3^{n+1} +(-1)^n )
\end{align}
In this form, eliminating the factor $4$, we obtain, for general $w$,
\begin{align}
c_{n-1}(w) &= \frac{3(3^{n-1}-(-1)^{n-1})+3w(3^{n-2}-(-1)^{n-2})}{3^n+(-1)^{n-1}+w(3^{n-1}+(-1)^{n-2})} \\
&=\frac{(3+w)(-3)^{n-1}-3+3w}{(3+w)(-3)^{n-1}+1-w} \\
&=1+\frac{12(1-w)}{(3+w)(-3)^{n}-3(1-w)}.
\end{align}
From which it is clear the recurrence will converge to $1$ regardless of the starting value $w$ provided not $-3$.  Setting $w=i$ this becomes after multiplying numerator and denominator by $(1+i)/2$
\begin{align}
a_n &= c_{n-1}(i) \\
&=1+\frac{12}{(1+2i)(-3)^{n}-3}.
\end{align}
A: We have that $-3$ and $1$ are fixed point of $f(z)=\frac{3}{2+z}$. I know that there's a theorem from dynamic systems theory (although I don't know the precise hypothesis) that says that a fixed point $z_0$ of $f$ is an attractor or a repeller depending on whether $|f'(z_0)|$ is less than or greater than $1$.
In this case it would show that the sequence converges to $1$.
May be some expert on the field could confirm if the theorem apply here?
