If one of the roots is [tex]w=B+2i[/tex], then the other root is its conjugate [tex]\bar w=B-2i[/tex]. So we can factorize the quadratic to
[tex]z^2 + 4z + 20 + iz (A+1) = (z-(B+2i)) (z-(B-2i))[/tex]
Expand the right side and collect all the coefficients.
[tex]z^2 + (4+(A+1)i) z + 20 = z^2 - 2B z + B^2+4[/tex]
From the [tex]z[/tex] and constant terms, we have
[tex]\begin{cases}4 + (A+1)i = -2B \\ 20 = B^2 + 4 \end{cases}[/tex]
From the second equation we get
[tex]B^2 = 16 \implies B = \pm4[/tex]
Then
[tex]4+(A+1)i = \pm8[/tex]
[tex](A+1)i = 4 \text{ or } (A+1)i = -12[/tex]
Since [tex]\frac1i=-i[/tex], we have
[tex]-\dfrac{A+1}i = 4 \text{ or } -\dfrac{A+1}i = -12[/tex]
[tex]A+1 = -4i \text{ or } A+1 = 12i[/tex]
[tex]\boxed{A = -1 - 4i \text{ or } A = -1 + 12i}[/tex]
