What is the fundamental reason for the imaginary unit in Heisenberg's commutator relations? The well known Heisenberg commutator relation
$$[p,q]=\cfrac{\hbar}{i} \cdot \mathbb{I}$$
introduces the imaginary unit $i$ into quantum mechanics. I ask for the deeper reason:
Why does the correspondence with real coordinates q and p introduces complex numbers for the commutator? Is the reason from physics or from mathematics?
Aside: I'm familiar with complex numbers and with the fact, that some results from the real domain find a satisfactory explanation not until generalization to the complex domain.
 A: Because operators $p$ and $q$ represent physical observables
(i.e. they have real eigen-values), they need to be Hermitean
(i.e. $p^\dagger=p$ and $q^\dagger=q$).
From this it is easy to show that their commutator $[p,q]$ is
anti-Hermitean.
$$[p,q]^\dagger
= (pq-qp)^\dagger
= (pq)^\dagger-(qp)^\dagger
= q^\dagger p^\dagger-p^\dagger q^\dagger
= qp-pq
= -[p,q]$$
You can get a Hermitean operator from this anti-Hermitean $[p,q]$
only by multiplying it with $i$.
$$(i[p,q])^\dagger = i[p,q]$$
So you can write Heisenberg's commutator relation also as
$$i[p,q]=\hbar\mathbb{I}$$
with Hermitean operators on both sides.
The operator on the right side corresponds to the very
trivial physical observable, which always gives the
same measurement value $\hbar$.
A: 
Is the reason from physics or from mathematics?

You can argue it either way. To deem it mathematical, see @ThomasFritsch's answer. But here's a physical insight, even if it requires mathematics to explain it. I'll work in $1$ dimension for simplicity. The real classical observable $p$ corresponds to a Hermitian $\hat{p}$, and leads to $e^{ipq/\hbar}$ factors of eigenvalue $p$. Why that factor, though? Because we need to tie momentum to arbitrary space translations ($[p,\,q]=-i\hbar\mathbb{I}$ is related to the Poisson bracket $\{p,\,q\}=-1$), quantum mechanics needs to be able to e.g. square-root unitary transformations (if I can move something a metre I can move it a half-metre, then another half-metre). This is arguably the main reason QM involves complex numbers. Luckily, if $\hat{O}$ is a dimensionless Hermitian operator, $\exp(i\hat{O})$ is unitary.
A: This, bizarrely, is something I lectured on yesterday.  The i emerges from the Fourier transformations and we don't need to know anything about position, momentum, or the wavefunction, for it to emerge completely naturally from the mathematics where it comes from a differential of position.  I include the two pages of my lecture notes below:


