3. In the complex plane, the points corresponding to the complex numbers $z_{1}, z_{2}, z_{3}$ are $Z_{1}, Z_{2}, Z_{3}$, respectively. If $\left|z_{1}\right|=\left|z_{2}\right|=\sqrt{2}, \overrightarrow{O Z_{1}} \cdot \overrightarrow{O Z_{2}}=0,\left|z_{1}+z_{2}-z_{3}\right|=2$, then the range of $\left|z_{3}\right|$ is $\qquad$.
Answer $[0,4]$.
From $\overrightarrow{O Z_{1}} \cdot \overrightarrow{O Z_{2}}=0$, we know that $\overrightarrow{O Z_{1}} \perp \overrightarrow{O Z_{2}},\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=4$, which means $\left|z_{1}+z_{2}\right|=2$. By the property of the modulus of complex numbers, we have
$$
2=\left|z_{1}+z_{2}-z_{3}\right| \geq|| z_{1}+z_{2}|-| z_{3}||=|2-| z_{3}||
$$

Thus, $-2 \leq 2-\left|z_{3}\right| \leq 2,0 \leq\left|z_{3}\right| \leq 4$, hence the answer is $[0,4]$.