If ${\displaystyle \sum _{k=0}^{200}{i}^k+{\displaystyle \prod _{p=1}^{50}{i}^p=x+iy}}$ , (where $i=\sqrt{-1}$ ) then q$\left(x,y\right)$  is:

$\because {\displaystyle \sum _{k=0}^{200}{i}^k}+\underset{p=1}{\overset{50}{II}}{i}^p=x+iy$ 

$\Rightarrow$          $i^0+0+i^{1+2+\mathrm{...}+50}=x+iy$

$\Rightarrow$              $1+i^{1275}=x+iy$

$\Rightarrow$              $1-i=x+iy$

On comparing $x=1$  and $y=-1$

$\therefore \left(x,y\right)=\left(1,-1\right)$ .