4. Given that $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$ (where $a, b, c$ are real numbers), and $c$ is an imaginary number, $\frac{\alpha^{2}}{\beta}$ is a real number. Then the value of $\sum_{k=1}^{5985}\left(\frac{\alpha}{\beta}\right)^{k}$ is ( ).
(A) 1
(B) 2
(C) 0
(D) $\sqrt{3} \mathrm{i}$
4. (C).

From $\frac{\alpha^{2}}{\beta}$ being a real number, we get $\frac{\alpha^{3}}{\alpha \beta} \in \mathbf{R}$.
$$
\because \alpha \beta \in \mathbf{R}, \therefore \alpha^{3} \in \mathbf{R} \text {. }
$$
$\because \alpha, \beta$ are conjugates, $\therefore \beta^{3} \in \mathbf{R}$, and $\alpha^{3}-\beta^{3}=0$.
Also, $\because \alpha \neq \beta$, we get $\alpha^{2}+\alpha \beta+\beta^{2}=0$,
$\therefore \frac{\alpha}{\beta}=-\frac{1}{2} \pm \frac{\sqrt{3}}{2}$ i. Thus,
$$
\sum_{k=1}^{5985}\left(\frac{\alpha}{\beta}\right)^{k}=\sum_{k=1}^{3 \times 1995}\left(\frac{\alpha}{\beta}\right)^{k}=0 .
$$