117 Let the geometric sequence $\left\{z_{n}\right\}$ be such that $z_{1}=1, z_{2}=a+b i, z_{3}=b+a i(a, b \in \mathbf{R}, a>0)$, then the smallest $n$ that satisfies $z_{1}+z_{2}+\cdots+z_{n}=0$ is
A. 6
B. 12
C. 18
D. 24
117 B. According to the problem,
$$
(a+b i)^{2}=b+a i \quad(a, b \in \mathbf{R}),
$$

which means
$$
\left\{\begin{array}{l}
a^{2}-b^{2}=b, \\
2 a b=a .
\end{array}\right.
$$

Solving these equations, we get
$$
\left\{\begin{array}{l}
a=\frac{\sqrt{3}}{2}, \\
b=\frac{1}{2},
\end{array}\right.
$$

Therefore,
$$
q=z_{2}=\frac{\sqrt{3}}{2}+\frac{1}{2} i=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6} .
$$

From $S_{n}=\frac{1-q^{n}}{1-q}=0$, we have
$$
\cos \frac{n \pi}{6}+i \sin \frac{n \pi}{6}=1
$$

Thus,
$$
\frac{n \pi}{6}=2 k \pi,(n=12 k, n \in \mathbf{N}) .
$$
The smallest value of $n$ is 12.