= Solution
The coefficients are real, so the supplied <polynomial root> $2+2i$ has its <complex conjugate> $2-2i$ as another <polynomial root>. Their monic <polynomial factor> of degree two is
$$
(z-2-2i)(z-2+2i)=(z-2)^2+4=z^2-4z+8.
$$
Polynomial division then gives
$$
\begin{aligned}
z^5-5z^4+11z^3-3z^2-12z+8
&=(z^2-4z+8)(z^3-z^2-z+1)\\
&=(z^2-4z+8)(z-1)^2(z+1).
\end{aligned}
$$
Thus all five <polynomial roots>, with <multiplicity of a root>, are
$$
\boxed{2+2i,\qquad2-2i,\qquad1,\qquad1,\qquad-1.}
$$
Their sum is $5$ and their product is $8\cdot1\cdot(-1)=-8$, agreeing with <9/c/i/solution> and <9/c/ii/solution>.
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