(A)
5 【Analysis and Solution】The points in $M$ lie on the curve
$M:\left\{\begin{array}{l}x=\frac{t}{1+t} \\ y=\frac{1+t}{t}\end{array} \quad(t \in \mathbf{R}, t \neq 0,-1)\right.$: The points in $N$ lie on the curve
$N:\left\{\begin{array}{l}x=\sqrt{2\left(1-t^{2}\right)} \\ y=\sqrt{2 t}\end{array} \quad(t \in \mathbf{R},|t| \leqslant 1)\right.$, The ordinary equations of lines $M$ and $N$ are $M: x y=1 \quad(x \neq 0,1)$,
$N: x^{2}+y^{2}=2 \quad(0 \leqslant x \leqslant \sqrt{2})$. Therefore, the x-coordinates of the intersection points of curves $M$ and $N$ satisfy $x^{2}+\frac{1}{x^{2}}=2$, i.e., $x= \pm 1$. It is evident that $M \cap N=\Phi$.
5. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{1+t}{t}\right., t \in \mathbf{R}, t \neq-1, t \neq 0\right\}, N=\{z \mid z=\sqrt{2}[\cos (\arcsin t)+i \cos$ $(\arccos t)], t \in \mathbf{R},|t| \leqslant 1\}$, then the number of elements in $M \cap N$ is
( A ) 0
(B) 1
(C) 2
(D) 4