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In this article (Link to the article), the author uses a vector identity to prove the following (equation 3.5 in the article)

$\mathrm{Im}\left(\mathbf{E}^*\times (\nabla\times \mathbf{E})\right)=\mathrm{Im}\left(\mathbf{E}^*.(\nabla)\mathbf{E}\right)+\frac{1}{2}\nabla \times \mathrm{Im}(\mathbf{E}^*\times \mathbf{E})$

I can't find which vector identities are used and how to prove this. Can someone help me?

Note that $(\mathbf{A}.(\nabla)\mathbf{B})_i=\sum_{j=1}^{3}A_j\frac{\partial}{\partial x_i}B_j$

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    $\begingroup$ Hi, welcome to Math SE. How is $A\cdot\nabla B$ defined? But the proof should be easy enough if you work in index notation. $\endgroup$
    – J.G.
    Commented Jun 11 at 7:55
  • $\begingroup$ Hi, Thanks J.G, you are right: I think I will let Mathematica do the hard work and check if it works. $\endgroup$
    – Cuki79
    Commented Jun 11 at 10:07

1 Answer 1

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Ok got it (thanks J.G and Mathematica)

$\mathrm{Im}\left(E^*\times(\nabla\times E\right)= \frac{1}{2}\mathrm{Im}\left(E^*\times E\right) +\mathrm{Im}\left\{E^*.(\nabla) E\right\} +\mathrm{Im}\left\{(\nabla.E^*) E\right\} $

the $\frac{1}{2}$ simplifies because the following term writes $a-a^*=2\mathrm{Im}(a)$ and the last term is equal to zero thanks to Maxwell-Gauss equation in dielectric medium.

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