Solution
= Solution
The powers of the <imaginary unit> have period four. Hence
$$
\sum_{a=0}^{200}i^a
=50(1+i-1-i)+i^{200}=1,
$$
while
$$
\prod_{b=1}^{50}i^b=i^{1+2+\cdots+50}=i^{1275}=i^3=-i.
$$
Thus $x+iy=1-i$, so
$$
\boxed{xy=-1}.
$$
= Solution
The powers of the <imaginary unit> have period four. Hence
$$
\sum_{a=0}^{200}i^a
=50(1+i-1-i)+i^{200}=1,
$$
while
$$
\prod_{b=1}^{50}i^b=i^{1+2+\cdots+50}=i^{1275}=i^3=-i.
$$
Thus $x+iy=1-i$, so
$$
\boxed{xy=-1}.
$$