Well, I am not sure how used you are in Classical Mechanics (waves) and/or Electromagnetism?
As a general setup, complex numbers never arise in describing physical observables not even in QM. In QM the observables are the eigenvalues of Hermitian operators , and Hermitian operators have real-valued eigenvalues.
In classical mechanics you also use complex numbers. For example when you describe the solution to the problem:
\frac{d^2 x}{dt^2} + \gamma \frac{dx}{dt} + cx =0
Harmonic oscillator with damping... in fact complex numbers can be used anytime you come across trigonometric functions thanks to the Euler's formula (e^{\pm ia}= \cos (a) \pm i \sin (a)). Except for this mechanical problem which complex numbers can help you solve, you can see them in electrodynamics, since the mechanical differential equations also apply in electromagnetism (in similar forms, maybe with different physical reasonings/explanations). In wave equations you use them to find easily the solutions which afterwards you take their real parts, and so on...
So nothing is special about complex numbers. They just make your life easier.
Now QM use a lot the wave mechanics, so it's not a big deal that you have complex numbers playing a role in your calculations. But anything that you can actually observe and use for testing the theory/experimenting, is absolutely real: It's the same reason you can never measure a vector in an experiment as a vector, but only its components that are "real", a complex number can be seen as a vector \vec{z}= \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix} on the complex plane, since any complex number can be written as z= r e^{i \theta}. There is a small difference in inner products, but that's why instead of using the transpose you use the dagger (complex conjugate of the transpose).
Can a complex number carry physical significance? Yes, it depends on the context. For a fast classical example, if you take the refractive index of a wave to be complex n= a + i b, then your wave will not only receive refraction but the imaginary part will give attenuation. Both you can measure as a and b seperately, but you won't measure some i in there...
