TITLE: What's the value of $n+\frac{n}{n+\frac{n}{n+\frac{n}{\ddots}}}$ for $n\in\mathbb{C}$?
QUESTION [13 upvotes]: Write $$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$ so that $\phi_n=n+\frac{n}{\phi_n},$ which gives $\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$ We know $\phi_1=\phi$, the Golden Ratio, so let's take $\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$. (Is that justified?)
Wolfram Alpha states that, with $(2)$, $$\lim\limits_{n\to -\infty}\phi_n=-1.$$ Why? Can I infer that this is true for $(1)$ and, if so, why?
I wonder what happens in $(1)$ for $n\in\mathbb{C}\backslash\mathbb{Z}$ too. I got something horrendous looking in $(2)$ for $n=i$.

Clarification: I'm trying to find $\phi_n$ in terms of $n$. See the comments below.

REPLY [5 votes]: For any given $n \ne 0, -4$, let $( a_{n,m} )_{m\in\mathbb{Z}_{+}}$ be the sequence
defined by
$$a_{n,m} = \begin{cases}n,& m = 1\\\displaystyle n + \frac{n}{a_{n,m-1}},&m > 1\end{cases}$$
Let $\displaystyle\;\mu_n = \frac{n+\sqrt{n(n+4)}}{2}\;$ and
$\displaystyle\;\nu_n = \frac{n-\sqrt{n(n+4)}}{2}\;$. It is easy to verify the following expression
$$a_{n,m} = \frac{\mu_n^{m+1} - \nu_n^{m+1}}{\mu_n^{m}-\nu_n^{m}}$$
provided a closed-form solution for $a_{n,m}$.
On those portion of complex plane where $|\mu_n|$ differs from $|\nu_n|$,
if is clear one of $\mu_n$ or $\nu_n$ will completely dominate the other one
for large $m$. As a result, we can make the following partial summary about $\phi_n$.
$$\phi_n = \lim_{m\to\infty} a_{n,m} = 
\begin{cases}
\mu_n, & |\mu_n| > |\nu_n|\\
\\
???& |\mu_n| = |\nu_n|\quad\leftarrow
\begin{array}{c}
\small\verb/This includes the special/\\
\small\verb/special case when /n = 0, -4.
\end{array}
\\
\nu_n, & |\mu_n| < |\nu_n|
\end{cases}
$$
In particular,
$$
\begin{array}{lcl}
n \in (0,\infty)   & \implies & |\mu_n| > |\nu_n| \implies \phi_n = \mu_n\\
n \in (-\infty,-4) & \implies & |\mu_n| < |\nu_n| \implies \phi_n = \nu_n
\end{array}
$$
and hence
$$\lim\limits_{n\to-\infty}\phi_n = \lim\limits_{n\to-\infty}\nu_n
= \lim\limits_{k\to\infty}\frac{-k+\sqrt{k(k-4)}}{2}
= \lim\limits_{k\to\infty}\frac{-2k}{k+\sqrt{k(k-4)}} = -1
$$
Update
Let us switch to the case $|\mu_n| = |\nu_n|$ but $n \ne 0, -4$. In particular, this cover
the range where $n \in (-4,0)$. Since $\mu_n\nu_n = -n$ and $\mu_n \ne \nu_n$, we can find
a $\theta_n \in (0,\pi)$ such that
$$\big\{\; \mu_n, \nu_n \;\big\} = \big\{\; \sqrt{-n}e^{i\theta_n}, \sqrt{-n}e^{-i\theta_n}\;\big\}$$
In terms of $\theta_n$, we have
$$a_{n,m} = \sqrt{-n}\frac{\sin((m+1)\theta_n)}{\sin(m\theta_n)}$$
There are two sub-cases. 

If $\displaystyle\;\frac{\theta_n}{\pi} \in \mathbb{Q}$, then $a_{n,m}$ is periodic in $m$. Notice when $\theta \in (0,\pi)$, $\displaystyle\;\frac{\sin((m+1)\theta)}{\sin\theta}$ is never an constant sequence. So $\phi_n$ diverges.
If $\displaystyle\;\frac{\theta_n}{\pi} \notin \mathbb{Q}$, then $a_{n,m}$ form a dense subset of the line $\big\{\; \sqrt{-n} t : t \in \mathbb{R} \;\big\} \subset \mathbb{C}$.
Once again, $\phi_n$ diverges.

Finally, it leaves us the cases $n = 0$ and $n = -4$. 

For $n = 0$, it is easy because start at $m = 2$, we encounter an undefined expression
like $a_{0,2} = 0 + \frac{0}{0}$. So all $a_{0,m}, m \ge 2$ and hence $\phi_0$ are undefined.
For $n = -4$, we have $\mu_{-4} = \nu_{-4} = -2$. Notice
$$\lim_{\mu\to\nu}\frac{\mu^{m+1}-\nu^{m+1}}{\mu^{m}-\nu^{m}} = \mu\frac{m+1}{m}$$
We will suspect $\displaystyle\;a_{-4,m} = -2\frac{m+1}{m}$. By direct substitution,
one can check that this is indeed the case. As a result,
$$\phi_{-4} = \lim\limits_{m\to\infty} a_{-4,m} = -2\lim_{m\to\infty}\frac{m+1}{m} = -2 = \nu_n$$

Combine all these, we obtain:
$$\phi_n = \lim_{m\to\infty} a_{n,m} = 
\begin{cases}
\mu_n, & |\mu_n| > |\nu_n|\\
\nu_n, & |\mu_n| < |\nu_n|\quad\text{ or }\quad n = -4\\
\text{undefined},& n = 0\\
\text{diverges},& |\mu_n| = |\nu_n|
\end{cases}
$$
Update2
The final question is what are the set that $|\mu_n| = |\nu_n|$. 
It turns out when $n \ne 0$,
$$\begin{align}
|\mu_n| = |\nu_n|
&\iff  \left|1 + \sqrt{1+\frac{4}{n}}\right| = \left|1 - \sqrt{1+\frac{4}{n}}\right|
\iff \Re\left(\sqrt{1+\frac{4}{n}}\right) = 0\\
&\iff
n \in [-4,0)
\end{align}$$
In fact, if we define $\lambda(z)$ by
$$\mathbb{C}\setminus (-4,0] \ni z\quad\mapsto\quad \lambda(z) = \frac{z}{2}\left(1 + \sqrt{1 + \frac{4}{z}}\right) \in \mathbb{C},$$
$\lambda(z)$ will be a single-valued function over $\mathbb{C} \setminus (-4,0]$ and analytic over its interior $\mathbb{C} \setminus [-4,0]$.
Furthermore, $\lambda(n)$ coincides with $\mu_n$ and $\nu_n$ on $(0,\infty)$ and $(\infty,-4]$ respectively! What this means is the apparent switching of value of $\phi_n$ 
between $\mu_n$ and $\nu_n$ is really an artifact of how we label them.
At the end, we have a much simpler description for $\phi_n$.
$$\phi_n = \begin{cases} 
\frac{n}{2}\left(1 + \sqrt{1 + \frac{4}{n}}\right),& n \notin (-4,0]\\
\\
\text{ undefined/diverges },& n \in (-4,0]
\end{cases}$$