I am currently reading "Classical Electrodynamics" by Jackson and I couldn't reproduce a calculation made there.
I got $$\vec{A} = - ik \frac{e^{ikr}}{r} \vec{d}$$ and I want to calculate the following two things $$\vec{B} = \nabla \times \vec{A} \\ \vec{E} = \frac{i}{k} \nabla \times \left( \nabla \times \vec{A} \right)$$ Calculating $\vec{B}$ was pretty easy and straightforward by using $\nabla \times \left( f(r)\vec{d} \right) = f'(r) \frac{\vec{r}}{r} \times \vec{d}$ for a constant $\vec{d}$. I get $$\vec{B} = e^{ikr} \left( \frac{k^2}{r^2} + \frac{ik}{r^3} \right) \vec{r} \times \vec{d}$$ which is identical to the calculation in the book.
When I calculate $\vec{E}$ I don't get the book's solution, which is $$\vec{E} = k^2 \left(\vec{n} \times \vec{d} \right) \times \vec{n} \frac{e^{ikr}}{r} + \left[ 3 \vec{n} \left( \vec{n} \cdot \vec{d} \right) - \vec{d} \right] \left[ \frac{1}{r^3} - \frac{ik}{r^2} \right] e^{ikr} $$
I tried using the identity $$\nabla \times \left( \nabla \times \vec{A} \right) = \nabla \left( \nabla \cdot \vec{A} \right) - \Delta \vec{A},$$ but I couldn't find an elegant way of calculating $\Delta \vec{A}$. I think it should be possible by using the vector laplacian as described here http://mathworld.wolfram.com/VectorLaplacian.html and just calculating it but isn't there a better way to do this?