5. The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|=$ $\left|z_{3}\right|=\left|z_{4}\right|=1$, and $z_{1}+z_{2}+z_{3}+z_{4}=0$. Then the quadrilateral formed by the points corresponding to these four complex numbers must be ( ).
(A) Trapezoid
(B) Square
(C) Parallelogram
(D) Rectangle
5. (D).

Since $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$, these points all lie on the unit circle. From $z_{1}+z_{2}+z_{3}+z_{n}=0$, we get
$$
\frac{z_{1}+z_{2}}{2}=-\frac{z_{3}+z_{4}}{2} \text {. }
$$

Let the four points be $A_{1}, A_{2}, A_{3}, A_{1}$, then 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 $M, N, O$ coincide, then chord $A_{1} A_{2}$ and $A_{3} A_{4}$ coincide and are diameters of the circle, which can be understood as a special case.
If $M, N, O$ do not coincide, then $M, O, N$ are collinear and
$$
O M=O N, O M \perp A_{1} A_{2}, O N \perp A_{3} A_{4} \text {, }
$$

then $A_{1} A_{2} / / A_{3} A_{4}$ and $A_{1} A_{8}=\approx A_{3} A_{4}$.
Thus, $A_{1} A_{2} A_{3} A_{4}$ is a parallelogram, and since it is inscribed in the circle, $A_{1} A_{2} A_{3} A_{4}$ is a rectangle.