Question

Complex numbers help needed. Will give medal and fan. *question attached below*

Answer 1

So I have been working on this question for ages. I found another complex root of the equation which is 2+i, but when it comes down to finding the real root, I end up with 13/5, but I don't know if it is correct. Can I have some assistance, please?

Answer 2

First, you would have to divide: \(z^3+pz^2+qz+13\) by \(2-i\)
Then, using T&E, you would need to find the other 2 possible roots.
Using those roots, you would get 2 equations. Using either substitution or elimination, you can find both \(p\) and \(q\). I am kinda busy, so I can't write the individual steps. I can only tell you that is what you got to do :)

Answer 3

But wouldn't two of the roots be complex numbers (2-i, 2+i) since p and q are real and by the Complex Conjugate Theorem?

Answer 4

Assuming \(p\) and \(q\) are real, then yes. (I didn't solve it, going with your answer)

Answer 5

Well the question said "real constants p and q"

Answer 6

Then yes, the two roots would be complex numbers

Answer 7

It is possible to get complex roots, just that the equation won't hit the x-axis

Answer 8

Wouldn't it be troublesome to divide the equation by (2-i) first since the coefficients p and q are unknown? :S

Answer 9

Yes, very troublesome. You would have to either keep it like:
\[\frac{p}{2-i}\] or something of that nature. Though question you got, not even sure myself how to do it ;)

Answer 10

Thank you!

Answer 11

No problem @iTiaax :)