# Proof of $\mathbf{E}^*\times (\nabla\times \mathbf{E}) =\mathbf{E}^*.(\nabla)\mathbf{E}+\frac{1}{2}\nabla \times \mathbf{E}^*\times \mathbf{E}$

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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• 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.
– J.G.
Commented Jun 11 at 7:55
• Hi, Thanks J.G, you are right: I think I will let Mathematica do the hard work and check if it works. Commented Jun 11 at 10:07

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