∑ k = 0 40 i k = x + i y ⇒ x 100 + x 99 y + x 242 y 2 + x 97 y 3 =

Given:

$\sum _{k=0}^{40}{i}^k=x+iy$

$\Rightarrow i^0+\left[\left(i+i^2+i^3+i^4\right)+\left(i^5+i^6+i^7+i^8\right)+...+\left(i^{37}+i^{38}+i^{39}+i^{40}\right)\right]=x+iy$

$\Rightarrow i^0+0=x+iy$

$\Rightarrow x+iy=1$

So, $x=1, y=0$

Therefore,

$x^{100}+x^{99}y+x^{242}{y}^2+x^{97}{y}^3=1$