Must imaginary roots come in conjugate pairs?

The book's answer is nonsense. Your calculation gives a polynomial with roots at $0,\,4,\,i$ as desired and is a very direct way to find the minimal polynomial that will have a root at all of those places. More simply, one can see this phenomenon since $x-i=0$ is a polynomial with a single root at $i$. Complex roots only necessarily come in conjugate pairs for polynomials with real coefficients. As you note, the book's answer would be correct if you were asked to find a real polynomial with those roots.

(Not to mention that this question is terribly ambiguous, since it doesn't seem clear whether the polynomial should have only those three roots, or those three should be among the roots, in which case both (b) and (c) would work)

While Milo Brandt's answer is perfect, I want to give a quick demonstration that the non-real roots of a real polynomial must come in conjugate pairs. Given any complex $x$ I shall use "$x^*$" to denote the complex conjugate of $x$. Also, given any complex polynomial $f$ I shall use "$f^*$" to denote $f$ with each coefficient replaced by its conjugate.

Take any real polynomial $f$ and any complex root $r$ of $f$.

Then by long division we can find a complex polynomial $g$ and a complex constant $c$ such that for all complex $x$ we have $f(x) = g(x) \cdot (x-r) + c$. Then $0 = f(r) = g(r) \cdot 0 + c = c$.

Now simply take the conjugate on both sides of the identity and use the basic properties of complex conjugation to get $f^*(x^*) = g^*(x^*) \cdot (x^*-r^*)$ for any complex $x$.

But $f^* = f$ since $f$ has only real coefficients, and we can apply the above identity to the case where $x = r$.

Thus $f(r^*) = g^*(r^*) \cdot 0 = 0$.

Therefore $r^*$ is a root of $f$.