Column – I		Column – II
A)	$t\in R$  such that there is at least one z satisfying  $\left|z\right|=3,|z-\{t(1+i)-i\}|\le 3$ and  $|z+2t-(t+1)i|>3$	p)	6
B)	solve for  $x:\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$	q)	0
C)	The non zero  integer $n$  for which ${\left(\frac{1+i}{1-i}\right)}^n$  is real	r)	3
D)	The greatest and least absolute value of $z+1,$  where $|z+4|\le 3$  are	s)	4
		t)	8

A) To satisfy all at a time z should lie on the circle $\left|z\right|=3$ .
Inside the circle  $|z-\{(1+i)-i\}|=3$ and outside the circle

$|z+2t-(t+1)i|=3$
For this, 
$\sqrt{{(t-0)}^2+{(t-1-0)}^2}\le 3+3$ and  $\sqrt{4t^2+{(4+1)}^2}>3+3$
$\Rightarrow 2t^2-2t-35\le 0$ and  $5t^2+2t-35>0$
Using sign scheme we have.
Hence,  $1-(1-\frac{\sqrt{71}}{2},\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\frac{1+\sqrt{71}}{2})$
Hence, 3, 4 lies in above interval,
B) We have to solve for  $x,y$
$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$
$\Rightarrow \frac{(1+i)(3-i)x-2i(3-i)+(2-3i)(3+i)y+i(3+1)}{9-i^2}=i$
$\Rightarrow (4+2i)x-6i-2+(9-7i)y+3i-1=10i$
$\Rightarrow (4x+9y-3)+i(2x-7y-3)=0+10i$
Comparing the real and imaginary parts 
$4x+9y-3=0\mathrm{.........}\left(i\right)$
$2x-7y-13=0\mathrm{..........}\left(ii\right)$
$\left(i\right)-2\times \left(ii\right)$ gives, 
$9y+14y-3+26=0$
$\Rightarrow 23y=-23\Rightarrow y=-1$
Putting $y=-1$  in (i), we get  $x=3$
C)  ${\left(\frac{1+i}{1-i}\right)}^n=\left(\frac{1+i^2+2i}{1+i}\right)=i^n$
Hence  $n=0,4,8,6$
D) Greatest and least absolute values of $z+1$  are 1 and 6.
  $\because |z+4|\le 3$