3. Given complex numbers $z_{1}, z_{2}$ satisfy $\left(z_{1}-i\right)\left(z_{2}+i\right)=1$, if $\left|z_{1}\right|=\sqrt{2}$, then the range of $\left|z_{2}\right|$ is $\qquad$
Answer: $[2-\sqrt{2}, 2+\sqrt{2}]$
Let $z_{2}=x+y i$ then $\sqrt{2}=\left|z_{1}\right|=\left|\frac{z_{2} i}{z_{2}+i}\right|=\frac{|-y+x i|}{|x+(y+1) i|}$
So $x^{2}+(y+2)^{2}=2$