$12 \cdot 37$ Let $i=\sqrt{-1}$, the complex sequence $z_{1}=0, n \geqslant 1$ when, $\approx_{n+1}=$ $z_{n}^{2}+i$. Then the modulus of $z_{111}$ in the complex plane is
(A) 1 .
(B) $\sqrt{2}$.
(C) $\sqrt{3}$.
(D) $\sqrt{110}$.
(E) $\sqrt{2^{55}}$.
(43rd American High School Mathematics Examination, 1992)
［Solution］Suppose $z_{n+1}=z_{n}^{2}+i$, then we have
$$
\begin{array}{lll}
z_{1}=0, & z_{2}=i, & z_{3}=-1+i, \\
z_{4}=-i . & z_{5}=-1+i, & z_{6}=-i,
\end{array}
$$

Therefore, when $2 k+1 \geqslant 3$, $z_{2 k+1}=-1+i$.
Thus, $\quad\left|z_{111}\right|=|-1+i|=\sqrt{2}$.
Hence, the answer is $(B)$.