5. Given 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) Rectangle
(B) Square
(C) Parallelogram
(D) Trapezoid
5. (1).

Let the complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ correspond to points $A_{1}, A_{2}, A_{3}, A_{4}$, respectively. From $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$, we know that $A_{1} A_{2} A_{3} A_{4}$ is a quadrilateral inscribed in the unit circle.

From $z_{1}+z_{2}+z_{3}+z_{4}=0$, we get $\frac{z_{1}+z_{2}}{2}=-\frac{z_{3}+z_{4}}{2}$. Thus, the midpoints $M, N$ of sides $A_{1} A_{2}$ and $A_{3} A_{4}$ are symmetric with respect to the origin.

Clearly, points $M, O, N$ do not coincide (otherwise, $A_{1} A_{2}$ and $A_{3} A_{4}$ would coincide), hence, points $M, O, N$ are collinear, and
$|O M|=|O N|, O M \perp A_{1} A_{2}, O N \perp A_{3} A_{4}$.
Thus, $A_{1} A_{2} \parallel A_{3} A_{4}$, and $\left|A_{1} A_{2}\right|=\left|A_{3} A_{4}\right|$. Therefore, $A_{1} A_{2} A_{3} A_{4}$ is a parallelogram,
and since $A_{1} A_{2} A_{3} A_{4}$ is inscribed in the circle, $A_{1} A_{2} A_{3} A_{4}$ is a rectangle.