4. The complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3,\left|z_{1}+z_{2}\right|=4$, then $\frac{z_{1}}{z_{2}}=$ $\qquad$
Given that $z_{1} \overline{z_{1}}=4 \Rightarrow \overline{z_{1}}=\frac{4}{z_{1}}, z_{2} \overline{z_{2}}=9 \Rightarrow \overline{z_{2}}=\frac{9}{z_{2}}$,
let $\frac{z_{1}}{z_{2}}=z$, then $\left(z_{1}+z_{2}\right)\left(\overline{z_{1}}+\overline{z_{2}}\right)=z_{1} \overline{z_{1}}+z_{2} \overline{z_{2}}+z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=13+z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=16$
$\Rightarrow z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=3 \Rightarrow \frac{9 z_{1}}{z_{2}}+\frac{4 z_{2}}{z_{1}}=3 \Rightarrow 9 z^{2}-3 z+4=0 \Rightarrow z=\frac{1 \pm \sqrt{15} \mathrm{i}}{6}$,
so $\frac{z_{1}}{z_{2}}=\frac{1}{6} \pm \frac{\sqrt{15}}{6} \mathrm{i}$.