$12 \cdot 35$ If two sets are given as
$$
\begin{array}{l}
M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{1+t}{t}\right., \quad t \in R, \quad t \neq-1, \quad t \neq 0\right\}, \\
N=\{z|z=\sqrt{2}[\cos (\arcsin t)+i \cos (\arccos t)], t \in R,| t \mid \leqslant 1\},
\end{array}
$$

then the number of elements in $M \cap N$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 4 .
(China High School Mathematics League, 1989)
[Solution] In the Cartesian coordinate system, the elements of sets $M$ and $N$ are points on the curves
$M:\left\{\begin{array}{l}x=\frac{t}{1+t}, \\ y=\frac{1+t}{t} ;\end{array}\right.$
$(t \in R, \quad t \neq 0, \quad t \neq-1)$,
$N:\left\{\begin{array}{l}x=\sqrt{2\left(1-t^{2}\right)}, \\ y=\sqrt{2} t .\end{array}\right.$
$(t \in R, \quad|t| \leqslant 1)$

The parametric equations of $M$ and $N$ can be converted to the following ordinary equations:
$$
\begin{array}{l}
M: x y=1(x \neq 1, x \neq 0), \\
N: x^{2}+y^{2}=2 \quad(0 \leqslant x \leqslant \sqrt{2}) .
\end{array}
$$

Thus, the x-coordinates of the intersection points of curves $M$ and $N$ should satisfy the equation $x^{2}+\frac{1}{x^{2}}=2$, which yields
$$
x= \pm 1 .
$$

However, $\pm 1 \notin(0,1) \cup(1, \sqrt{2})$, meaning that $M \cap N=\varnothing$.
Therefore, the answer is $(A)$.