= Solution
At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2)^2=12\ne0$. The <holomorphic implicit function theorem> states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne0$, there is a unique holomorphic function $z=\gamma(c)$ near $c_0$, with $\gamma(c_0)=z_0$ and $P(\gamma(c),c)=0$. It therefore supplies an analytic root branch $\gamma(c)$ near zero. Differentiate its equation:
$$
(3\gamma(c)^2+ic)\gamma'(c)+i\gamma(c)=0.
$$
It follows that $\gamma'(0)=i/6$ and
$$
\boxed{\gamma(c)=-2+\frac{i}{6}c+O(c^2).}
$$
For sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus \b[that root lies above the real axis for all sufficiently small positive $c$].
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