What is the physical meaning of a pure imaginary force? I am reading an article (this) and the equations result in an inertial force $F_g$ (which I understand as a fictitious force (?)) that is purely imaginary. The system consists of a gyroscope fixed to the floor and attached to a mass (in a lattice of identical masses-gyroscopes) through which a wave passes that would cause a rotational movement on the gyroscope (as detailed below). I understand that due to the nutation of the gyroscope the mass can change its coordinate in z, but I do not understand the physical meaning that it is purely imaginary. As the author details at the end, this indicates "This imaginary nature of the gyroscopic inertial effect indicates directional phase shifts between two independent directions of the tip displacements, which breaks time-reversal symmetry", but I do not understand it at all.


 A: The physical force is not imaginary, it is given by the real part of the expression. That expression is not pure imaginary because each component of ${\bf U}_{\rm tip}$ is itself a complex number here. So its real part (which is the force) will not be zero. The presence of $i$ here is telling you that the oscillations of the force are a quarter-cycle out of phase with the oscillations of the velocity.
This whole calculation is an example of the use of complex numbers to analyze linear sets of equations. It is a method that is particularly helpful when oscillation is involved, because the function $e^{i \omega t}$ is easier to work with than combinations of $\cos(\omega t)$ and $\sin(\omega t)$.
In general if one has a physical quantity $x$ satisfying some equation $\hat{D} x = 0$ then one can introduce a complex number $z$ satisfying $\hat{D} z = 0$, where $\hat{D}$ is some bunch of algebraic operations and differential operations, all real-valued. But for a complex number to be zero, both its real and imaginary parts are zero, so the real part of $z$ here satisfies the same equation as $x$. The method consists in solving $\hat{D} z = 0$ to obtain $z$, and then you finish by asserting $x = {\rm Re}[z]$.
