(4)

    Given $R=\left\{\right(a,b):a+5b=42,a,b\in N\}$

    So, $a=42-5b$

    $\therefore (a,b)=\left\{\right(37,1),(32,2),(27,3),(22,4),(17,5),(12,6),(7,7),(2,8\left)\right\}$

    $\therefore \left|R\right|=8\Rightarrow m=8$

     Now, $\sum _{n=1}^m(1-i^{n!})=x+iy$

    Clearly for $n=4,i^{n!}=i^{4!}=({{\left(i\right)}^4)}^6={\left(1\right)}^6=1$

    $\therefore$   $i^{n!}=1$ for $n\ge 4$

    $\therefore$  $\sum _{n=1}^8(1-i^{n!})=(1-i)+(1-i^{2!})+(1-i^{3!})$

    $=1-i+1+1+1+1=5-i \Rightarrow x=5\text{ and }y=-1$.

    So, $m+x+y=8+5-1=12$

(1)

   Let $z=x+iy, x,y\in R$

   Consider $\frac{z-2i}{z+2i}=\frac{x+iy-2i}{x+iy+2i}=\frac{x+(y-2)i}{x+(y+2)i}\times \frac{x-(y+2)i}{x-(y+2)i}$

   $=\frac{x^2+x(y+2)i+x(y-2)i+(y^2-4)}{x^2+{(y+2)}^2}$

   $\Rightarrow \frac{x^2+y^2-4}{x^2+{(y+2)}^2}=0 \Rightarrow x^2+y^2=4$

   $\Rightarrow |z{|}^2=4 \Rightarrow |z|=2$

   Now, $|z-(6+8i\left)\right|\le \left|z\right|+|-6-8i|\le 2+10=12$

   $\therefore |z-(6+8_i){|}_{\text{max}}=12$

(1)

   Given equation, $z^{1985}+z^{100}+1=0$                ...(i)

   and $z^3+2z^2+2z+1=0$                                  ...(ii)

   From (ii), we have $(z+1)(z^2+z+1)=0$

   $\Rightarrow z=-1,z=\omega ,{\omega }^2$

   Clearly, $z=-1$ does not satisfy equation (i).

   $\therefore$ If $z=\omega ,$ then ${\left(\omega \right)}^{1985}+{\left(\omega \right)}^{100}+1={\omega }^2+\omega +1=0$

   Also, $z={\omega }^2$, then ${\left({\omega }^2\right)}^{1985}+{\left({\omega }^2\right)}^{100}+1={\omega }^2+\omega +1=0$

   $\therefore$  Number of roots = 2

(1)

(S1) : $\frac{z–i}{z+i}=\frac{{\displaystyle \bar{z}}–i}{{\displaystyle \bar{z}}+i}$          [$\because$  Purely real]

$\Rightarrow z\bar{z}–i\bar{z}–iz=z\bar{z}+zi+i\bar{z}–1$

$\Rightarrow z+\bar{z}=0$

$\Rightarrow 2x=0$           [$\because$  Purely real]

$\Rightarrow x=0$

Also, |z| = 1

$\therefore z=i$           ($\because z\ne –i is given$)

(S1) is incorrect.

(S2) : $\frac{z–1}{z+1}=\frac{{\displaystyle \bar{z}–1}}{{\displaystyle \bar{z}+1}}=0$          [$\because$  Purely imaginary]

$\Rightarrow z\bar{z}–\bar{z}–z–1+z\bar{z}–z+\bar{z}–1=0 \Rightarrow z\bar{z}=1 \Rightarrow \left|z\right|=1$,

which represents a circle with radius 1 and centre (0, 0).

(S2) is correct.