1. Given the complex number $z$ satisfies
$$
|z-1|=|z-\mathrm{i}| \text {. }
$$

If $z-\frac{z-6}{z-1}$ is a positive real number, then $z=$ $\qquad$
$-、 1.2+2$ i.
From the problem, we know that the real part and the imaginary part of $z$ are equal.
Let $z=x+x \mathrm{i}(x \in \mathbf{R})$. Then
$$
\begin{aligned}
z- & \frac{z-6}{z-1}=z-1+\frac{5}{z-1} \\
& =x+x \mathrm{i}-1+\frac{5}{x+x \mathrm{i}-1} \\
& =(x-1)+x \mathrm{i}+\frac{5((x-1)-x \mathrm{i})}{(x-1)^{2}+x^{2}} \\
& =(x-1)+\frac{5(x-1)}{(x-1)^{2}+x^{2}}+\left(x-\frac{5 x}{(x-1)^{2}+x^{2}}\right) \mathrm{i} .
\end{aligned}
$$

From the fact that $z-\frac{z-6}{z-1}$ is a positive real number, we know
$$
(x-1)+\frac{5(x-1)}{(x-1)^{2}+x^{2}}>0 \text {, }
$$

and $x-\frac{5 x}{(x-1)^{2}+x^{2}}=0$.
Solving this, we get $x=2$.
Therefore, $z=2+2 \mathrm{i}$.