104 The solutions $z_{1}, z_{2}$ of the equation $4 z_{1}^{2}-2 z_{1} \cdot z_{2}+z_{2}^{2}=0$ correspond to points $A$ and $B$, respectively. If the origin $O$, $A$, and $B$ form a triangle, then $\triangle A O B$ is
A. an isosceles triangle
B. a right triangle
C. an acute triangle
D. an obtuse triangle
104 B.
$$
z_{2}=(1 \pm \sqrt{3} i) z_{1}=2\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right) z_{1} .
$$

So $O B=2 O A$ and $\angle A O B=\frac{\pi}{3}$. Therefore, $\triangle A O B$ is a right triangle.