I am aware that for any polynomial with real coefficients, the imaginary roots (if there are any) must come in complex pairs. However, I always believed that you could still have an imaginary number - without a conjugate - as a root. However, I came upon the question:
Which of the following is a polynomial with roots $0$, $4$, and $i$.
a. $x^3 - 4x^2 + x - 4$
b. $x^3 - ( 4+i )x^2 + 4ix$
c. $x^4 - 4x^3 + x^2 - 4x$
d. $x^2 - 4x$
3. $x^4 - 4x^3 + x^2 + 4x$
Using the roots $0, 4, $and $i$, I found the factors to be $(x)(x-4)(x-i)$, which simplifies to choice b, $x^3 - ( 4+i )x^2 + 4ix$. However, the answer in the book states:
C Complex roots occur in conjugate pairs. If $i$ is a root of the polynomial, then $-i$ is also a root. Use the four roots to determine the factors of the polynomial. Then multiply to get the polynomial.
$x^4 - 4x^3 + x^2 - 4x$
I'm not sure how they can assume that their are complex conjugate pairs, since there are no constraints that the coefficients be real.