5 Complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy
$$
\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1 \text {, and } z_{1}+z_{2}+z_{3}+z_{4}=0 \text {. }
$$

Then the quadrilateral formed by the points corresponding to these four complex numbers is
A. Rhombus
B. Square
C. Trapezoid
D. Rectangle
5 D. Since $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$, the corresponding four points $A_{1}, A_{2}, A_{3},$ and $A_{4}$ are all on the unit circle. From $z_{1}+z_{2}+z_{3}+z_{4}=0$ we get
$$
\frac{z_{1}+z_{2}}{2}=-\frac{z_{3}+z_{4}}{2}.
$$

This means the midpoint $M$ of chord $A_{1} A_{2}$ and the midpoint $N$ of chord $A_{3} A_{4}$ are symmetric with respect to the origin. If $M$ and $N$ do not coincide, then $M, O, N$ are collinear,
$$
O M \perp A_{1} A_{2}, \quad O N \perp A_{3} A_{4},
$$

so $A_{1} A_{2} \parallel A_{3} A_{4}$, and similarly $A_{1} A_{3} \parallel A_{2} A_{4}$. Therefore, $A_{1} A_{2} A_{3} A_{4}$ is a parallelogram. Since it is inscribed in a circle, it is a rectangle. If $M$ and $N$ coincide, then they coincide with each other, and in this case, the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is clearly a rectangle.