Hopefully this is just some minor confusion...The first exercise wants us to show that $$\vec \epsilon(t,\vec x)=\vec Ee^{-i(wt-\vec k \cdot\vec x )}$$ satisfies the vacuum Maxwell equations where $\vec k \in \mathbb{R}^3$ $w=|\vec k|$ and $\vec E$ is fixed in $\mathbb{C}^3$. We also have that $\vec k \cdot \vec E =0$ and $i\vec k \times \vec E=w\vec E$. Some of the givens I've included for completeness, but really I'm just confused about computing $\nabla\cdot \epsilon)$. I want to be computing 4 partials with respect to time and the x,y,z coordinates, but $\vec E$ is a vector in $\mathbb{C}^3$, and I'm not sure how to do that. If it makes a difference, the divergence should end up being zero.
Feel free to adjust the tags if I've abused them.
The exponential factor is a scalar valued function. When we have a scalar valued function $\phi$ and a vector valued function $F$, the divergence satisfies the following property: $$\nabla\cdot(\phi \vec{F}) = \nabla(\phi)\cdot \vec{F} + \phi (\nabla \cdot \vec{F})$$ In our case, we have $$\nabla \cdot (e^{i(\vec{k} \cdot \vec{x}-wt)}\vec{E}) = \nabla(e^{i(\vec{k} \cdot \vec{x}-wt)})\cdot \vec{E} + e^{i(\vec{k} \cdot \vec{x}-wt)} (\nabla \cdot\vec{E}) \qquad \qquad \qquad (1)$$ The gradient of the exponential is easy to calculate, $$\nabla (e^{i(\vec{k} \cdot \vec{x}-wt)}) = i(k_x,k_y,k_z)e^{i(\vec{k} \cdot \vec{x}-wt)} = ie^{i(\vec{k} \cdot \vec{x}-wt)} \vec{k}$$ where $\vec{k} = (k_x,k_y,k_z)$. Then, since $\vec{k} \cdot \vec{E} = 0$, the first term in $(1)$ vanishes. On the other hand, we know that $\vec{E}$ is a constant vector in $\mathbb{C}^3$, so $\nabla \cdot \vec{E} = 0$, and so the second term vanishes as well.