I see that this problem involves complex numbers, which have both real and imaginary parts. To add complex numbers, I can add their real parts and their imaginary parts separately. So, if the sum of the numbers is $2i$, that means the real part of the sum is zero, and the imaginary part is 2. This gives me two equations: $a+c+e=0$ and $b+d+f=2$. I already know that $b=3$ and $e=-a-c$, so I can substitute them into the equations. This gives me: $a+c-a-c=0$ and $3+d+f=2$. The first equation simplifies to $0=0$, which is always true, so it does not help me. The second equation simplifies to $d+f=-1$. This is the answer I am looking for. 
There are 3 complex numbers $a+bi$, $c+di$, and $e+fi$. If $b=3$, $e=-a-c$, and the sum of the numbers is $2i$, find $d+f$.