Asymptotic development of the sequence $z_{n+1} = \alpha z_n + \frac{\beta}{z_n}$ Let $\alpha$ and $\beta$ be two positive real numbers with $\alpha + \beta = 1$. Let $z_0\in \mathbb C^*$ and define the sequence $$z_{n+1} = \alpha z_n + \frac{\beta}{z_n}$$
One can prove that if $\Re(z_0)=0$, then the sequence either is not well-defined, either does not converge in $\mathbb C$. However, if $\Re(z_0)\not =0$, the sequence converges to $1$ if the real part is positive, $-1$ if it is negative. In order to prove this, one can consider the biholomorphism $\phi : z \mapsto \frac{z-1}{z+1}$, which maps $\{z\in \mathbb C \, | \,\Re(z) > 0\}$ to the unit open disc. 
Knowing this result, I would like to find an asymptotic development of the sequence $(z_n)$ in the case $\Re(z_0) > 0$. Using the asymptotic expansion of $\frac{1}{1-(1-z_n)}$, I found out the result $$z_{n+1}-1 = (\alpha - \beta)(z_n - 1) + o(z_n-1)$$
We may assume that $\alpha \not = \beta$ first and that $(z_n)$ is not stationnary to $1$, which allows us to write 
$$\frac{z_{n+1}-1}{z_n-1} \sim \alpha - \beta$$
Is there any theorem of some kind I could use here in order to "take the product" in this equivalence and obtain an estimation of $z_n-1$ ? The fact that the left-hand side is complex numbers is bothering me, as I prefer not having to justify the existence of a logarithm. 
What would be your take on this exercise ? How could I progress further ?
I thank you very much in advance.
 A: I assume $\alpha \neq \beta$.
(if $\alpha = \beta = 1/2$ then letting $w_n = \frac {z_n-1}{z_n+1}$, you should get $w_{n+1} = w_n^2$, which is easy to study)
From $\frac {z_{n+1}-1} {z_n -1} \sim \alpha - \beta$, you probably want to show next that $z_n = 1 + k (\alpha - \beta)^n + o((\alpha-\beta)^n)$ for some complex number $k$.
This is equivalent to showing that $(z_n-1)(\alpha-\beta)^{-n}$ converges (to $k$).
Let $w_n = (z_n-1)(\alpha-\beta)^{-n}$. You have shown so far that $w_{n+1} \sim w_n$. However, a more precise calculation leads to 
$w_{n+1} = w_n + \beta(w_n^2/z_n)(\alpha-\beta)^{n-1} $
Since we already know that $w_{n+1}/w_n \to 1$, we get that $w_n = o((1+\epsilon)^n)$ for any $\epsilon>0$. We also know that $z_n \to 1$, so taking all those together we obtain $w_{n+1} - w_n = o((|\alpha-\beta|+\epsilon')^n)$ for any $\epsilon' >0$, which is enough to prove the convergence of $w_n$.
In fact, it is possible to continue that process indefinitely and show that for any integer $m$, there are complex coefficients $k_1=k, k_2,\ldots k_m$ such that $z_n = 1 + \sum_{i=1}^m k_i (\alpha-\beta)^{in} + o((\alpha-\beta)^{mn})$.
The coefficients $k_i$ for $i>1$ can be expressed algebraically in terms of $k_1,\alpha,\beta$ only. For example, $k_2 = k_1^2 \beta/(\alpha- \beta)(\alpha-\beta-1)$. Those expressions can be obtained by looking for solutions to the recurrence in the ring of formal power series $\Bbb C[[(\alpha-\beta)^n]]$. Also, I am not making any claim of convergence of the infinite series for any $n$ and any $z_0$.
The relationship between $z_0 \in \{ z \in \Bbb C \mid \Re z > 0 \}$ and $k \in \Bbb C$ is pretty complicated and fractal-esque. $k$ is holomorphic in $z_0$, and I see no easier way to compute $k$ other than $k = \lim_{n \to \infty} (z_n-1)(\alpha-\beta)^{-n}$
