For $\left|z - 1\right| = 1$, find tan $[arg(\dfrac{(z -1)}{(2 - 2 \frac{i}{z})})].$

Let ;$z=x+iy$

Hence,& ;$|z-1|=1$

$(x-1)^{2}+y^{2}=1$

Hence by using parametric equation of circle.

$x-1=cos\theta$

$x=1+cos\theta$ and 

$y=sin\theta$

Then, $z=(1+cos\theta)+isin\theta$

Let  $\theta=\frac{\pi}{2}$

Then, $z=1+i$ ...(i)

Hence, $z-1=i$ and 

$2-2i\dfrac{1}{z}$

$=2(1-\dfrac{i}{1+i})$

$=2(\dfrac{1+i-i}{1+i})$

$=\dfrac{2}{1+i}$

$=1-i$...(ii)

Hence, $\dfrac{z-1}{2-\dfrac{2i}{z}}$

$=\dfrac{i}{1-i}$

$=\dfrac{i(1+i)}{2}$

$=\dfrac{i-1}{2}$

$=\dfrac{1}{\sqrt{2}}(\dfrac{-1+i}{\sqrt{2}})$

$=\dfrac{1}{\sqrt{2}}.e^{i\frac{3\pi}{4}}$

Hence, $tan(\frac{3\pi}{4})$

$=-1$.