No explicit complexification is needed to derive this breakdown of Maxwell's equations. This can be understood wholly through the real vector space of special relativity.
Let's start with Maxwell's equations for the EM field, in the clifford algebra language called STA: the spacetime algebra. Maxwell's equations take the form
$$\nabla F = -J$$
where $\nabla F = \nabla \cdot F + \nabla \wedge F$, $F = e_0 E + B \epsilon_3$, in the $(-, +, +, +)$ sign convention.
Let $x$ be the spacetime position vector. It's generally true that, for a vector $v$ and a constant bivector $C$,
$$\nabla (C \cdot x) = -2 C, \quad \nabla (C \wedge x) = 2C \implies \nabla (Cx) = 0$$
One can then evaluate the expression
$$\nabla (Fx) = (\nabla F)x + \dot \nabla (F \dot x)$$
where the overdot means that only $x$ is differentiated in the second term; using the product rule, $F$ is "held constant" and so the above formulas apply. We just argued that the second term is zero, so we get $\nabla (Fx) = (\nabla F) x$. Thus, we arrive at the following transformation of Maxwell's equations:
$$\nabla (Fx) = -Jx$$
Now, we could always write $F$ as a "complex bivector" in the sense that, using $\epsilon = e_0 \epsilon_3$, and $\epsilon \epsilon = -1$, we have
$$F = e_0 E - B \epsilon_3 e_0 \epsilon_3 \epsilon = e_0 (E + \epsilon B)$$
It's crucial to note that $\epsilon$ does not commute with any vector.
What are the components of $Fx$? Write $x = t e_0 + r$ and we can write them as
$$Fx = e_0 (Ex + \epsilon Bx) = e_0 (E \cdot r + E \wedge r - e_0 Et + \epsilon B \cdot r - e_0 B \times r + \epsilon B t e_0)$$
This too can be written in a "complex" form:
$$Fx = (e_0 E \cdot r + Et + B \times r) + \epsilon (E \times r + e_0 B\cdot r + Bt)$$
We seem to differ on some signs, but this is recognizably the same quantity you have called $G$.
Now, to talk about how these equations break down, let's write $G = G_1 + G_3$, where $G_1 = (e_0 E \cdot r + \ldots)$ and $G_3 = \epsilon (E \times r + \ldots)$. Let's also write for $R = Jx = R_0 + R_2$.
Maxwell's equations then become
$$\nabla \cdot G_1= R_0, \quad \nabla \wedge G_1 + \nabla \cdot G_3 = R_2, \quad \nabla \wedge G_3 = 0$$
The first and third equations are the components of the Gauss dipole; the second equation is the Ampere-Faraday dipole equation.
Now, what does it all mean? The expression for $G = Fx$ includes both rotational moments of the EM field as well as some dot products, so it measures both how much the spacetime position is in the same plane as the EM field as well as how much the spacetime position is out of the plane.
It's probably more instructive to look at the source term $-Jx$. This tells us both about the moments of the four-current as well as how it goes toward or away from the coordinate origin. The description for the moments is wholly in the Ampere-Faraday dipole equation. What kinds of moments would this describe? A pair of two opposite point charges at rest, separated by a spatial vector $2 \hat v$ and centered on the origin, each with current at rest $j_0$, would create a $R = Jx = + j_0 e_t \hat v - j_0 e_t (-\hat v) = 2 j_0 e_t \hat v$, so this would be described wholly by the A-F dipole equation.
That's at time zero, however. At later times, $R$ will pick up these weird time terms. Say we're at time $\tau$. Then $R = 2 j_0 e_t \hat v + j_0 e_t (\tau e_t) - j_0 e_t (\tau e_t)$. So for this case, there's no problem: the extra stuff will just cancel. A single charge, however, would start picking up this term.
In a few words, these equations are weird.
Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units
$$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$
$$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$
If we let $\mu_0=1$, $\epsilon_0 =1$ (effectively saying we are adopting a system of units where $c=1$, then these equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ except for the minus sign in Faraday's law. They are symmetric to a rotation (see below).
If the source terms are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations can be written symmetrically. We suppose a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$, then we write
$$ \nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$
$$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$
With these definitions, Maxwell's equations acquire symmetry to duality transformations.
If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form
$$ \left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$
where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.
Whilst one can argue then about what we define as electric and magnetic charges, it is an empirical fact at present that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations) all particles appear to have the same ratio, so we choose to fix it that one of the charge types is always zero - i.e. no magnetic monopoles.
I mention all this really as a curiosity. It seems to me that the real symmetries of Maxwell's equations only emerge when one considers the electromagnetic potentials.
e.g. if we insert $B = \nabla \times {\bf A}$ and $E= -{\bf \nabla V} - \partial {\bf A}/\partial t$ into our Ampere's law
$$\nabla \times (\nabla \times {\bf A}) = \frac{\partial}{\partial t} \left({\bf -\nabla V} - \frac{\partial {\bf A}}{\partial t}\right) +{\bf J}, $$
$$-\nabla^2 {\bf A} +\nabla(\nabla \cdot {\bf A}) = -\nabla \frac{\partial V}{\partial t} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J}.$$
Then using the Lorenz gauge
$$\nabla \cdot {\bf A} + \frac{\partial V}{\partial t} = 0$$
we can get
$$ \nabla^2 {\bf A} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J} = 0$$
A so-called inhomogeneous wave equation.
A similar set of operations on Gauss's law yields
$$ \nabla^2 V - \frac{\partial^2 V}{\partial t^2} + \rho= 0$$
These remarkably symmetric equations betray the close connection between relativity and electromagnetism and that electric and magnetic fields are actually part of the electromagnetic field. Whether one observes $\rho$ or ${\bf J}$; ${\bf E}$ or ${\bf B}$, is entirely dependent on frame of reference.
Best Answer
In plain English, it is just Lenz’s law :
It is the basic principle behind all electric motors and dynamos, alternators, etc.
Wikipedia: Lenz's Law