(a)

Since roots come in conjugate pairs, -3 - 2i is also a root.

(b)

We have:

\displaystyle a = -(4 + 2i - 3 - 3 - 2i) = 2

and:

\begin{align*}b &= 4(-3 + 2i) + 4(-3 - 2i) + (-3 + 2i)(-3 - 2i) \\ & = -12 + 8i - 12 - 8i + 3^2 + 2^2 \\ & = -11\end{align*}

by Vieta’s relations. (wasn’t covered in FP1 but is now covered in AS FM so I include it here)

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Very nice