3. 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 4 complex numbers must be ().
A. Rectangle
B. Square
C. Parallelogram, but not a rectangle
D. Trapezoid
3. A

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 quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is a cyclic quadrilateral inscribed in the unit circle. Given $z_{1}+z_{2}+z_{3}+z_{4}=0$, we have $\frac{z_{1}+z_{2}}{2}=-\frac{z_{3}+z_{4}}{2}$. Therefore, the midpoint $M$ of side $A_{1} A_{2}$ and the midpoint $N$ of side $A_{3} A_{4}$ are symmetric with respect to the origin, and points $M, O, N$ are collinear, with $|O M|=|O N|,\left|A_{1} A_{2}\right|=\left|A_{3} A_{4}\right|$. This implies that quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is a parallelogram. Since $\square A_{1} A_{2} A_{3} A_{4}$ is inscribed in a circle, $\square A_{1} A_{2} A_{3} A_{4}$ is a rectangle. Choose $\mathrm{A}$.