1. Given the complex number $z$ satisfies $\frac{z-\mathrm{i}}{z-1}$ is a pure imaginary number (where $\mathrm{i}$ is the imaginary unit), then the maximum value of $|z|$ is
Let $z=x+y \mathrm{i} (x, y \in \mathbf{R})$, then $\frac{z-\mathrm{i}}{z-1}=\frac{x+(y-1) \mathrm{i}}{x-1+y \mathrm{i}}=\frac{[x+(y-1) \mathrm{i}][(x-1)-y \mathrm{i}]}{(x-1)^{2}+y^{2}}$ $=\frac{x(x-1)+y(y-1)+[(x-1)(y-1)-x y] \mathrm{i}}{(x-1)^{2}+y^{2}}$ is a pure imaginary number,
thus $\left\{\begin{array}{l}x(x-1)+y(y-1)=0, \\ (x-1)(y-1)-x y \neq 0\end{array} \Rightarrow\left\{\begin{array}{l}x^{2}-x+y^{2}-y=0, \\ x+y-1 \neq 0\end{array}\right.\right.$ $\Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$ (excluding the points $(1,0),(0,1)$). Therefore, the maximum value of $|z|$ is $\sqrt{2}$, at this point $z=1+\mathrm{i}$.