jmc

Since $i$ is not a real number, $f(i)=i^2=-1$. Since $1$ is a real number, $f(1)=1+2=3$. Also, $-1$ is a real number, so $f(-1)=-1+2=1$. Finally, $-i$ is not a real number, so $f(-i)=(-i{)}^2=-1$. Therefore, $f(i)+f(1)+f(-1)+f(-i)=-1+3+1+(-1)=\boxed{2}$.