If $\sum _{k=0}^{100}{i}^k=x+iy$, then the values of $x and y$ are

$\sum _{k=0}^{100}{i}^k=x+iy,$ $\Rightarrow 1+i+i^2$$+\mathrm{......}+i^{100}=x+iy$

Given series is G.P.

$\Rightarrow \frac{1.(1-i^{101})}{1-i}=x+iy$ $\Rightarrow \frac{1-i}{1-i}=x+iy$

$\Rightarrow 1+0i=x+iy$ 

Equating real and imaginary parts, we get the required result.