This could not be more imaginary

Let $S_1=a_1+a_2+a_3$ , $S_2=a_1a_2+a_1a_3+a_2a_3$ and $S_3=a_1a_2a_3$ . Using Newton's sums we obtain:

$f(1)=S_1$

$f(2)=S_1^2-2S_2$

$f(3)=S_1^3-3S_1S_2+3S_3$

With the known values, the following system is formed;

$S_1=0$

$S_1^2-2S_2=i$

$S_1^3-3S_1S_2+3S_3=i^i$

Solving that we get: $S_1=0$ , $S_2=-\dfrac{i}{2}$ and $S_3=\dfrac{i^i}{3}$ .

By Newton's sums we know that $f(n)=\dfrac{i}{2}f(n-2)+\dfrac{i^i}{3}f(n-3)$ :

$f(4)=\dfrac{i}{2}f(2)+\dfrac{i^i}{3}f(1)=\dfrac{i}{2}(i)+\dfrac{i^i}{3}(0)=-\dfrac{1}{2}$

$f(5)=\dfrac{i}{2}f(3)+\dfrac{i^i}{3}f(2)=\dfrac{i}{2}(i^i)+\dfrac{i^i}{3}(i)=\dfrac{5i(i^i)}{6}$

$f(6)=\dfrac{i}{2}f(4)+\dfrac{i^i}{3}f(3)=\dfrac{i}{2}\left(-\dfrac{1}{2}\right)+\dfrac{i^i}{3}(i^i)=-\dfrac{4i^i+3i}{12}$

$f(7)=\dfrac{i}{2}f(5)+\dfrac{i^i}{3}f(4)=\dfrac{i}{2}\left(\dfrac{5i(i^i)}{6}\right)+\dfrac{i^i}{3}\left(-\dfrac{1}{2}\right)=-\dfrac{7i^i}{12}$

Hence, $-(f(4)+f(7))=\dfrac{1}{2}+\dfrac{7i^i}{12}=\dfrac{6+7i^i}{12}=\dfrac{6+7e^{-\frac{\pi}{2}}}{12}$ .

Comparing we get $a=6$ , $b=7$ , $c=-2$ , $d=12$ and $a+b+c+d=\boxed{23}$ .

Yes, I used Newton's Sums method too. It is tedious but systematic. I am attempting to make it simpler here.

Let $S_1 = a_1+a_2+a_3\quad S_2 = a_1a_2+a_2a_3+a_3a_1\quad S_3 = a_1a_2a_3$

Then by Newton's Sums we have:

$\begin {cases} f(1) = S_1 = 0 \\ f(2) = S_1f(1) - 2S_2 = 0 - 2S_2 = i \quad \quad \Rightarrow S_2 = -\frac {1}{2} i \\ f(3) = S_1f(2) - S_2f(1) + 3S_3 = 0 - 0 + 3S_3 = i^i \quad \quad \Rightarrow S_3 = \frac {1}{3} i^i \\ f(4) = S_1f(3) - S_2f(2) + S_3f(1) = 0 - (-\frac {i}{2} )(i) + (\frac {i^i}{3})(0) = -\frac {1}{2} \\ f(5) = \frac {i}{2} f(3) + \frac {i^i}{3} f(2) = \frac {i}{2} i^i + \frac {i^i}{3} i = \frac {1}{2} i^{i+1} + \frac {1}{3}i^{i+1} = \frac {5}{6}i^{i+1} \\ f(7) = \frac {i}{2} f(5) + \frac {i^i}{3} f(4) = \frac {i}{2} (\frac {5}{6}i^{i+1} ) + \frac {i^i}{3} (-\frac {1}{2}) = \frac {1}{12} (5i^{i+2} - 2i^i) \end {cases}$

Therefore,

$-(f(4)+f(7)) = \dfrac {1}{2} - \dfrac {5i^{i+2} - 2i^i} {12} = \dfrac {6 - (-5i^i-2i^i)} {12}$

$\quad \quad \quad \quad \quad \quad \quad = \dfrac {6+7i^i} {12} = \dfrac {6+7 \left( e^{-\frac{\pi}{2}} \right)} {12}$

$\Rightarrow a = 6$ , $b=7$ , $c=-2$ and $d = 12\quad \Rightarrow a+b+c+d = \boxed {23}$

For simplicity lets represent $a_1 = x, ~ a_2 = y ~ \& ~ a_3 = z$ .

Hence $x+y+z = 0 \\ x^2+y^2+z^2 = i \\ x^3 + y^3 +z^3 = 3xyz = i^i = e^{\frac{-\pi}{2}}$

$(x+y+z)^2 = 0 \\ \implies x^2+y^2+z^2 + 2(xy+yz+xz) = 0 \\ \therefore xy+yz+xz = \dfrac{-i}{2} \dots \dots (i)$

Now $(xy+yz+xz)^2 = \dfrac{-1}{4} \\ \implies x^2y^2+y^2z^2+z^2x^2 + 2xyz(x+y+z) = \dfrac{-1}{4} \\ \implies x^2y^2+y^2z^2+z^2x^2 = \dfrac{-1}{4} \dots \dots (ii)$

Now $(x^2 + y^2+z^2)^2 = -1 \\ \implies x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + z^2x^2) = -1 \\ \implies x^4 + y^4 + z^4 +2\times (\dfrac{-1}{4}) = -1 \\ \therefore x^4 + y^4 + z^4 = f(4) = \dfrac{-1}{2}$

Now $(x^3+y^3+z^3)(x^4 + y^4 + z^4) = x^7 + y^7 + z^7 + x^3y^3(x+y) + y^3z^3(y+z)+x^3z^3(x+z) \\ = x^7 + y^7 + z^7 - (x^3y^3z + y^3z^3x + x^3z^3y) \\ = x^7 + y^7 + z^7 - xyz(x^2y^2+y^2z^2+z^2x^2) \\ \implies \dfrac{-1}{2} \times e^{\frac{-\pi}{2}} = x^7+y^7+z^7 - e^{\frac{-\pi}{2}} \times (\dfrac{-1}{4}) \\ \therefore x^7+y^7+z^7 = f(7) = \dfrac{-7e^{\frac{-\pi}{2}}}{12}$

So $-(f(4)+f(7)) = \dfrac{6+7e^{\frac{-\pi}{2}}}{12} \\ \therefore ~a = 6,~b=7,~c=-2 ~ \& ~ d=12$

Hence correct answer is $23$ .

Same thing but much less calculation than Newton's Sums

$\quad \quad Let\quad z\quad =\quad { i }^{ i }\\ \Rightarrow \quad \ln { (z) } =\quad i\ln { (i) } \\ \Rightarrow \quad \ln { (z) } =\quad i\ln { ({ e }^{ i\frac { \pi }{ 2 } }) } \\ \Rightarrow \quad \ln { (z) } =\quad -\frac { \pi }{ 2 } \\ \Rightarrow \quad \quad \quad z\quad =\quad { e }^{ -\frac { \pi }{ 2 } }$

There is no statement that a,b,c,d are integers, nor positive.