Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, then $x^{2}+y^{2}-\mathrm{i}(x-y \mathrm{i})=x^{2}+y^{2}-y-x \mathrm{i}=1+\mathrm{i}$, thus $x=-1 \Rightarrow y^{2}-y=0 \Rightarrow y=0$ or 1. Therefore, the imaginary part of $z$ is 0 or 1.
1. If the complex number $z$ satisfies $z \bar{z}-\mathrm{i} \bar{z}=1+\mathrm{i}$, then the imaginary part of $z$ is $\qquad$