Equation 13

Yes, I overlooked that. Thank you.

Mondo

No that doesn't seem right to me.

I would do it like this:

Since $H=u+iv$ it follows that

$\partial_xH=\partial_xu+i\partial_xv$ and $\partial_yH=\partial_yu+i\partial_yv$

Substituting into the left hand side of (13) we have

$i\partial_xH-\partial_yH=i(\partial_xu+i\partial_xv)-(\partial_yu+i\partial_yv)=i\partial_xu-\partial_xv-\partial_yu-i\partial_yv$

Rearranging this we obtain

$i\partial_xH-\partial_yH=-(\partial_xv+\partial_yu)+i(\partial_xu-\partial_yv)$

Substituting using the two equations just before (13) we find that

$i\partial_xH-\partial_yH=\boldsymbol{\nabla\times \overline{H}}+i\text{ }\boldsymbol{\nabla\cdot \overline{H}}$

which is the right hand side of (13).

Vasco

Vasco,
thank you. That of course is correct. What I wanted to show in my post, is that we can also get it by gematrical/visual means. Let's take $i(\partial_x{u} - \partial_y{v})$, $u$ and $v$ are projections of vector field $H$ on real and imaginary axis respectively. If you look at my diagram in reply #3 we can see that addition of these two gives $H$ while substitution gives $\overline{H}$. I think it is correct too.

Sure, I have prepared a more clear diagram:


So my point is $u + v = H$ hence $i(\partial_x{u}+ \partial_y{v}) = i\nabla \cdot H$. However if we do $u - v = \overline{H}$ and hence $i(\partial_x{u}- \partial_y{v}) = i\nabla \cdot \overline{H}$ Does it make sense to you?

From vector addition. $u$ and $v$ are projections of $H$ on a real and imaginary axis.

If $u$ and $v$ are both scalars then $H(z) = u + iv$ should also be a scalar.  Ok, that's the defining property of a complex number so this is not true. However, if they were scalars then $\partial_x{u}$ would always be $0$. Likewise partials in regards to $y$.
I agree with you that saying $u + v$ is a vector addition is not true but we can treat them as such by defining $u = [x,0]$ and $v = [0,y]$ now my addition makes sense.