It's complicated

Multiply the second equation by $i$ :

$bi+\dfrac{8ai-bi}{a^2+b^2}=0$

And add it to the first one:

$a+bi+\dfrac{a+8b+i(8a-b)}{a^2+b^2}=2$

Factor the numerator and the denominator of the fraction:

$a+bi+\dfrac{(1+8i)(a-bi)}{(a+bi)(a-bi)}=2 \\ a+bi+\dfrac{1+8i}{a+bi}=2$

Multiply both sides by $a+bi$ :

$(a+bi)^2-2(a+bi)+1+8i=0$

Solve the quadratic in $a+bi$ , using the quadratic formula:

$a+bi=1 \pm \sqrt{1-(1+8i)} \\ a+bi=1 \pm 2\sqrt{-2i} \\ a+bi=1\pm 2(-1+i)$

We have two pairs of real solutions, comparing both real and imaginary parts:

$(a,b)=(-1,2)$ , $(a,b)=(3,-2)$

So, the sum is $-1+2+3-2=\boxed{2}$ .

Note: this method only gives us the real solutions, if we want all complex solutions, we can find $a-bi$ with a similar method, and then obtain $a$ and $b$ by adding and subtracting all the four possible combinations of values.