Let z be a complex number with Im(z)>0 and arg ((z-1)/(z+1))=(π)/(4). The minimum value of |z-2| is

$\arg\frac{z-1}{z+1}=\arg(z-1)-\arg(z+1)$ is the angle at $z$ subtended by the segment joining $-1$ and $1$, so $z$ lies on an arc through $\pm1$ on which this inscribed angle is $\frac{\pi}{4}$. The chord has length $2$, so the circumradius is $R=\frac{2}{2\sin(\pi/4)}=\sqrt2$. For $z=i$ the argument is $\frac{\pi}{2}>\frac{\pi}{4}$, so the required arc (angle acute) is the major arc and the centre lies above the axis at $C=(0,1)$: the circle is $x^2+(y-1)^2=2$ with $y>0$. The distance from $P=(2,0)$ to the centre is $\sqrt5$, so the nearest point on the circle is at distance $\sqrt5-\sqrt2$. That nearest point is $C+\sqrt2\,\frac{(2,-1)}{\sqrt5}$ whose ordinate is $1-\sqrt{2/5}>0$, hence it lies on the permitted arc. The minimum is $\sqrt5-\sqrt2\approx0.8219$. A numerical scan of the arc agrees. $\sqrt5-1$ would come from taking radius $1$, and $\sqrt2-1$ from taking centre $(0,0)$.