The least positive argument of complex number z, satisfying the inequality  $|Z-1-2i|\le 1$  is 

$|Z-1-2i|\le 1$   …………… (1)
(1) $\Rightarrow$  Z is a point on or inside the circle with centre $(1,2)$ and radius 1
  ${(x-1)}^2+{(y-2)}^2\le 1\Rightarrow x^2+y^2-2x-4y+4\le 0$
Clearly the circle touches y-axis at (0,2)
$\therefore$  If  $2\alpha$  is angle between  
   $Tan\alpha =\frac{r}{\sqrt{S_{11}}}=\frac{1}{2}$
   
   $2\alpha =2Ta{n}^{-1}\frac{1}{2}=Ta{n}^{-1}\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}=Ta{n}^{-1}\frac{4}{3}$
  Least argument $\theta$  From the figure is  $\theta =\frac{\pi }{2}-2\alpha$
 $\theta =\frac{\pi }{2}-Ta{n}^{-1}\frac{4}{3}={\cot }^{-1}\frac{4}{3}=Ta{n}^{-1}\frac{3}{4}$