9. (16 points) Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|<1$, $\left|z_{2}\right|<1, z_{1}+z_{2}$ is a real number, and $z_{1}+z_{2}+z_{1} z_{2}=0$. Find the range of $z_{1} z_{2}$.
Let $z_{1}=a+b \mathrm{i}, z_{2}=c+d \mathrm{i}$.
Given that $a+b \mathrm{i}+c+d \mathrm{i}$ is a real number, we have $b=-d$.
$$
\begin{array}{l}
\text { Also } z_{1}+z_{2}+z_{1} z_{2}=0 \\
\Leftrightarrow a+c+(a+b \mathrm{i})(c-b \mathrm{i})=0 \\
\Leftrightarrow\left\{\begin{array}{l}
a+c+a c+b^{2}=0, \\
(c-a) b=0 .
\end{array}\right.
\end{array}
$$
(1) When $b=0$, we have $z_{1}=a, z_{2}=c$, thus
$$
a+c+a c=0 \Leftrightarrow a=\frac{-c}{1+c} \text {. }
$$

Given $|a|<1,|c|<1$, solving the inequality yields
$$
a \in\left(-\frac{1}{2}, 1\right) \text {. }
$$

Therefore, $z_{1} z_{2}=a c=a \cdot \frac{-a}{1+a}=\frac{-a^{2}}{1+a} \in\left(-\frac{1}{2}, 0\right]$.
(2) When $b \neq 0$, we get $a=c$.

Thus, $2 a+a^{2}+b^{2}=0$.
Also, $\left|z_{1}\right|=a^{2}+b^{2}<1$, then
$$
z_{1} z_{2}=a^{2}+b^{2} \in[0,1) \text {. }
$$

In summary, $z_{1} z_{2} \in\left(-\frac{1}{2}, 1\right)$.