1. Let the geometric sequence $z_{1}, z_{2}, \cdots, z_{n}, \cdots$ be such that $z_{1}=$ $1, z_{2}=a+b \mathrm{i}, z_{3}=b \mathrm{i}(a, b \in \mathbf{R}, ab>0)$. Then the smallest natural number $n$ for which $z_{1} z_{2} \cdots z_{n}<0$ is $\qquad$ .
$Ni, 1.8$.
$$
\begin{array}{l}
\text { Given } z_{2}^{2}=z_{1} z_{3} \Rightarrow(a+b \mathrm{i})^{2}=b \mathrm{i} \\
\Rightarrow a^{2}-b^{2}+2 a b \mathrm{i}=b \mathrm{i} \\
\Rightarrow\left\{\begin{array} { l } 
{ a ^ { 2 } = b ^ { 2 } , } \\
{ 2 a b = b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\frac{1}{2}, \\
b=\frac{1}{2}
\end{array}\right.\right. \\
\Rightarrow \text { common ratio } q=\frac{z_{2}}{z_{1}}=\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+\mathrm{i} \sin \frac{\pi}{4}\right) \\
\Rightarrow z_{1} z_{2} \cdots z_{n}=\left(z_{1}\right)^{n} q^{\frac{n(n-1)}{2}} \\
=\left(\frac{\sqrt{2}}{2}\right)^{\frac{n(n-1)}{2}}\left[\cos \frac{n(n-1)}{8} \pi+i \sin \frac{n(n-1)}{8} \pi\right] \\
<0 \\
\Rightarrow\left\{\begin{array}{l}
\cos \frac{n(n-1)}{8} \pi=-1, \\
\sin \frac{n(n-1)}{8} \pi=0
\end{array}\right. \\
\Rightarrow \frac{n(n-1)}{8}=k \text { (odd number). } \\
\end{array}
$$

The smallest value of $n$ that satisfies the condition is 8.