Problem 749

Noticing $z_{2}=\overline{z_{1}}$, we have $\frac{z_{1}}{z_{2}^{2}}=\frac{\overline{z_{1}}}{z_{2}^{2}}=\frac{\overline{z_{1}}}{\bar{z}_{2}^{2}} \Rightarrow z_{1} \cdot \bar{z}_{2}^{2}=z_{2}^{2} \cdot \overline{z_{1}} \Rightarrow z_{1}^{3}=z_{2}^{3}$, thus $\left\{\begin{array}{l}z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}=0, \\ \left(z_{1}-z_{2}\right)\left(\overline{z_{1}}-\overline{z_{2}}\right)=48\end{array} \Rightarrow\left\{\begin{array}{l}z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}=0, \\ -z_{1}^{2}+2 z_{1} z_{2}-z_{2}^{2}=48\end{array} \Rightarrow\left|z_{1}\right|^{2}=z_{1} z_{2}=16\right.\right.$. Therefore, $\left|z_{1}\right|=4$.