Answer 6.
Analysis Let $z=x+y i(x, y \in \mathbf{R})$, then $\bar{z}=x-y i, z-\bar{z}=(x+y i)-(x-y i)=2 y i,|z-\bar{z}|=2|y|$, from $|z-i|=2$, we know $|(x+y i)-i|=2, x^{2}+(y-1)^{2}=4$, so $(y-1)^{2} \leq 4,-1 \leq y \leq 3$, thus $|z-\bar{z}|=2|y| \leq 6$, when and only when $y=3$, i.e., $z=3 i$, the equality holds, so the maximum value of $|z-\bar{z}|$ is 6.
3. Let the complex number $z$ satisfy $|z-i|=2$, then the maximum value of $|z-\bar{z}|$ is $\qquad$ (where $i$ is the imaginary unit, $\bar{z}$ is the conjugate of the complex number $z$)