Why do physicists like to put the imaginary unit $:i=sqrt{-1}:$ everywhere?

Question

There are many disagreements of convention between mathematicians and physicists, but a recurring theme seems to be that physicists tend to insert unnecessary factors of $i = sqrt{-1}$ into definitions.
I understand this is only convention, but I’m curious about why this seems so widespread. Does anybody know about the “etymological” reason(s) for physicists’ $i$-heavy conventions?
$newcommand{dd}{mathrm{d}}$

Thing
Mathematics convention
Physics convention

Lie algebra structure constants (Ref.)
$[L_a, L_b] = f_{ab}{}^cL_c$
$[L_a, L_b] = if_{ab}{}^cL_c$

Lie group transformations in terms of generators (Ref.)
$R_z(θ) = exp(θJ_z)$
$R_z(θ) = exp(-iθJ_z)$

Covariant derivative with $mathbb{C}$-valued connection 1-form
$nabla V = dd V + A V$
$nabla_μ V^a = ∂_μ V^a -iqA^a{}_{bμ} V^b$

Curvature of connection or gauge field strength (Ref., §7.4)
$F = dd A + A ∧ A$
$F_{μν} = ∂_μA_ν - ∂_νA_μ pm iq[A_μ, A_ν]$

I have a vague guess: physicists read and write $e^{iωt}$ a lot, and an exponential with an $i$ in it screams “rotation”. Fast forward to describing $mathrm{SO}(n)$ rotations in terms of matrix generators, and an expression like $e^{iθJ_z}$ just “feels more familiar” so much so that an extra $i$ is pulled out of the definition of $J_z$.
Can that guess be supported?
Not sure about the third and fourth rows, though.

fewfew4 · Accepted Answer

What you say is part of it. But I think a more important reason we have the $i$'s explicit is because we like to describe things with Hermitian operators. Taking the example of $SU(2)$, the Lie algebra in physicist's notation is
$$[L_i,L_j]=i varepsilon_{ijk}L_k$$
$L_3$ in physics is an observable, which describes the spin of a particle in the $z$ direction, which takes integer or half-integer values. Since this is an observable, we prefer it to be a real number. Hence why we want $L_3$ to be Hermitian.
More generally, we take generators of a symmetry group to be Hermitian (assuming we're dealing with a unitary representation), because they describe the observable "charges", which we want to be real.
The reason for the $i$'s in the covariant derivative and curvature $2$-form is similar. We want the connection to be Hermitian, since it describes an observable field that permeates space. Although in more advanced treatments we sometimes use the mathematicians notation in these cases.
In summary, physicists use this particular notation because these quantities represent something physical, whereas mathematicians use their notation because it is symbolically efficient. We have different priorities :)

J.G. · Answer

Real classical observables are quantized as Hermitian operators, and $partial_mu$ is anti-Hermitian. So:

Structure constants We seek Hermitian generators.
Transformations We seek real $theta$, unitary $R_z$ and Hermitian $J_z$.
Covariant derivative We seek Hermitian $A,,V$.
Curvature/field strength We seek Hermitian $F$.

Why do physicists like to put the imaginary unit $:i=sqrt{-1}:$ everywhere?

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Thing	Mathematics convention	Physics convention
Lie algebra structure constants (Ref.)	$[L_a, L_b] = f_{ab}{}^cL_c$	$[L_a, L_b] = if_{ab}{}^cL_c$
Lie group transformations in terms of generators (Ref.)	$R_z(θ) = exp(θJ_z)$	$R_z(θ) = exp(-iθJ_z)$
Covariant derivative with $mathbb{C}$-valued connection 1-form	$nabla V = dd V + A V$	$nabla_μ V^a = ∂_μ V^a -iqA^a{}_{bμ} V^b$
Curvature of connection or gauge field strength (Ref., §7.4)	$F = dd A + A ∧ A$	$F_{μν} = ∂_μA_ν – ∂_νA_μ pm iq[A_μ, A_ν]$