= Solution
On $|z|=3$, the perturbation has modulus $|icz+8|\leq3c+8<11<27=|z^3|$. Thus <Rouché's theorem> gives \b[three roots inside $|z|<3$, counted with multiplicity].
For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt3$, $r_3=1-i\sqrt3$. Their pairwise distances are $2\sqrt3$. On the circle $|z-r_j|=1$,
$$
|P_0(z)|=\prod_{\ell=1}^3|z-r_\ell|\geq(2\sqrt3-1)^2>3,
\qquad |icz|\leq c(|r_j|+1)=3c<3.
$$
The three unit discs are disjoint, and <Rouché localization of polynomial roots> gives one root in each, counted with multiplicity. The disc around $1+i\sqrt3$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt3$ lies in the fourth quadrant; and that around $-2$ lies in $\operatorname{Re}z<-1$. Consequently \b[there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane]. The same count also shows that each root is simple.
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