= Solution
For a purely imaginary candidate $z=it$, $t\in\mathbb R$, the imaginary part of the <polynomial> is $2t(t^2-4)$. The candidate $t=0$ is not a <root>, while $t=\pm2$ both make the real part zero. The resulting <factorization> into <quadratic polynomials> gives
$$
f(z)=(z^2+4)(z^2-2z+4),\qquad
\boxed{z=2i,-2i,1+i\sqrt3,1-i\sqrt3.}
$$
The <roots of a polynomial> sum to $0+2=2$ and have product $4\cdot4=16$, as required by <Vieta's formulas> for a monic quartic with cubic coefficient $-2$ and constant coefficient $16$.
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