Solution
= Solution
Over the <complex numbers>, the <polynomial factorization> $z^2+1=(z-i)(z+i)$ gives the <partial fraction decomposition>
$$
\frac{z+1}{z^2+1}=\frac{A}{z-i}+\frac{B}{z+i}.
$$
Multiplication by $(z-i)(z+i)$ and comparison of <polynomial> coefficients give $A+B=1$ and $i(A-B)=1$. Hence
$$
\boxed{\frac{z+1}{z^2+1}=\frac{1-i}{2(z-i)}+\frac{1+i}{2(z+i)}},\qquad z\ne\pm i.
$$