Complex resonant frequency not resonant without imaginary part. So can I still just take real part as solution? I am working with a matrix on a harmonic oscillator problem and the lowest (absolute) frequency $\omega_0$ where the matrix becomes singular is the resonant frequency.
Now I obtained this frequency by calculating it numerically at let's say it is $\omega_0 = 0.3302 - 0.0121i$, ie. a complex frequency.
My question is what is the actual resonant frequency considering there is no such thing as a complex resonant frequency in real life? I know that in various situations you work with complex numbers for convenience and then take the real part at the end to obtain a solution.
However the imaginary part of $\omega_0$ here is necessary to make the matrix singular, if I just take $\omega_0 = 0.3302$ the matrix is not singular and $w_0$ will not be a resonant frequency.
Or have I got it wrong and is it actually fine to just take the real part (and the imaginary part can be sort of considered to be implied)?
 A: If you were to plot the eigenvalues of your system matrix in the complex plane (real x-axis, y imaginary axis), your harmonics would appear as complex pairs in the left half plane, reflected from one another by the real axis. These are called poles of the system.
Draw a vector to either of these poles originating at the origin. The magnitude of this vector is the natural frequency of the harmonic. The angle the vector makes with the imaginary axis is a measure of loss due to damping. Poles with zero damping exist at an angle of ninety degrees - on the imaginary axis.
For complex poles with any degree of damping, the projection of the vector onto the imaginary axis provides the damped natural frequency. This is the frequency you would measure in the real world with whatever damping exists. With damping of zero the projection is equal to the magnitude of the vector. With increasing damping the projection gets smaller and smaller so the period between the settling oscillations gets further apart.
