If $|z-1|=1$, then 

Since arg $(z-1-i)/z$ is the angle subtended by the chord joining the points O and $1+i$at the circumference of the circle $|z-1|=1$, so it is equal to $-\pi /4$. The line joining the points $z=0$ and $z=2+0i$ is the diameter.

$\arg \frac{z-2}{z}=\pm \frac{\pi }{2}$

$\Rightarrow \frac{z-2}{z-0}$is purely imaginary

We have,

arg$\left(\frac{2-z}{0-z}\right)=\frac{\pi }{2}$

$\frac{z-2}{z}=i\tan \theta$