3. Let the complex number $z$ satisfy $|z-\mathrm{i}|=2$. Then the maximum value of $|z-\bar{z}|$ is $\qquad$ (i is the imaginary unit, $\bar{z}$ is the conjugate of the complex number $z$).
3. 6 .

Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { By }|z-\mathrm{i}|=2 \Rightarrow|(x+y \mathrm{i})-\mathrm{i}|=2 \\
\Rightarrow x^{2}+(y-1)^{2}=4 \\
\Rightarrow(y-1)^{2} \leqslant 4 \\
\Rightarrow-1 \leqslant y \leqslant 3 .
\end{array}
$$

Therefore, $|z-\bar{z}|=2|y| \leqslant 6$, when and only when $y=3$, i.e., $z=3 \mathrm{i}$, $|z-\bar{z}|$ reaches its maximum value of 6.