Limit of sequence in complex plane Suppose $Z_{n+1}=\frac{\alpha}{Z_n}+\frac{\beta}{Z_{n-1}}$. where $\alpha$ and $\beta$ and $Z_n$ and $Z_{n-1}$ are all complex numbers. 

It is easy to show that $\frac{Z_n}{Z_{n+1}}=\frac{\alpha.\frac{Z_{n-2}}{Z_{n-1}}+\beta}{\alpha.\frac{Z_{n-1}}{Z_{n}}+\beta}$ for all $n$.

From this equation, is it possible to talk about limit of the sequence $\{Z_n\}$?
In general: One can always talk about limits of complex sequences since $\mathbb{C}$ is a metric space where $d(z_1,z_2) = |z_1 - z_2|$.

Regarding your sequence: Let $Z_0 = Z_1 = 1$, $\alpha \neq 1$ and $\beta = 0$. Then
$$Z_{n+1} = \frac{\alpha}{Z_n} ~ , ~ n &gt; 1$$
This means that $(Z_1,Z_2,Z_3,\ldots) = (1, \alpha, 1, \alpha, \ldots)$. Hence, the sequence is not convergent (for this particular choice of $\alpha, \beta$.
In general: One can always talk about limits of complex sequences since $\mathbb{C}$ is a metric space where $d(z_1,z_2) = |z_1 - z_2|$.

Regarding your sequence: Let $Z_0 = Z_1 = 1$, $\alpha \neq 1$ and $\beta = 0$. Then
$$Z_{n+1} = \frac{\alpha}{Z_n} ~ , ~ n &gt; 1$$
This means that $(Z_1,Z_2,Z_3,\ldots) = (1, \alpha, 1, \alpha, \ldots)$. Hence, the sequence is not convergent (for this particular choice of $\alpha, \beta$.