Question

suppose the equation z^2+4z+20+iz(A+1)=0 where A is constant has complex conjugate root if one of the root of this quadratic is z=B+2i where B is real constant find possible value of A ? please help me I need your help​

Answer 1

If one of the roots is
`w=B+2i`, then the other root is its conjugate
`\bar w=B-2i`. So we can factorize the quadratic to

`z^2 + 4z + 20 + iz (A+1) = (z-(B+2i)) (z-(B-2i))`

Expand the right side and collect all the coefficients.

`z^2 + (4+(A+1)i) z + 20 = z^2 - 2B z + B^2+4`

From the
`z` and constant terms, we have

`\begin{cases}4 + (A+1)i = -2B \\ 20 = B^2 + 4 \end{cases}`

From the second equation we get

`B^2 = 16 \implies B = \pm4`

Then

`4+(A+1)i = \pm8`

`(A+1)i = 4 \text{ or } (A+1)i = -12`

Since
`\frac1i=-i`, we have

`-\frac{A+1}i = 4 \text{ or } -\frac{A+1}i = -12`

`A+1 = -4i \text{ or } A+1 = 12i`

`\boxed{A = -1 - 4i \text{ or } A = -1 + 12i}`