5. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{l+t}{t}\right., t \in R, t \neq-1, t \neq 0\right\}$,
$$
N=\{z|z=\sqrt{2}[\cos (\arcsin t)+i \cos (\arccos t)], t \in R,| t \mid \leqslant 1\} \text {, }
$$
then the number of elements in $M \cap N$ is
(A) 0 ;
(B) 1 ;
(C) 2 ;
(D) 4 .
5. (A)
The points in $M$ lie on the curve
$$
M:\left\{\begin{array}{l}
x=\frac{t}{1+t} \\
y=\frac{1+t}{t}
\end{array} \quad(t \in \mathbb{R}, t \neq 0,-1)\right.
$$

The points in $N$ lie on the curve
$$
N:\left\{\begin{array}{l}
x=\sqrt{2\left(1-x^{2}\right)} \\
y=\sqrt{2} t
\end{array} \quad(t \in \mathbb{R},|t| \leqslant 1)\right.
$$

The ordinary equations of the curves $M$ and $N$ are
$$
\begin{array}{ll}
M: & x y=1 \quad(x \neq 0,1), \\
N: & x^{2}+y^{2}=2 \quad(0 \leqslant x \leqslant \sqrt{2}) .
\end{array}
$$

Thus, the x-coordinates of the intersection points of curves $M$ and $N$ satisfy $x^{2}+\frac{1}{x^{2}}=2$, i.e., $x= \pm 1$. It is evident that $M \cap N=\Phi$.