# 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)?

• @Asked: without more context it's hard to say. If the complex numbers were introduced in order to simplify calculations -- exponentials in place of trigonometric functions -- and if the equations are linear, then the real and the imaginary parts are both solutions. But if the imaginary represents losses, the – Peter Diehr Mar 5 '16 at 12:18
• ... then it is part of the answer; this happens when the imaginary part represents the absorption in optics. One cannot tell by simply looking at an answer; the problem setup is required. – Peter Diehr Mar 5 '16 at 12:20
• The complex numbers were not introduced to simplify calculations. The problem is the resonant frequency of bubbles in water. – sonicboom Mar 5 '16 at 17:04
• then you already have your answer, and it is complex. Here is a paper that uses the complex angular resonant frequency - I'm more familiar with optical applications than with bubbles, but the complex value simply needs to be decoded into two independent parameters; the imaginary part is usually related to damping, as in the reference paper. – Peter Diehr Mar 5 '16 at 20:13
• ncbi.nlm.nih.gov/pmc/articles/PMC3855064; I need new fingertips! – Peter Diehr Mar 5 '16 at 21:07

• If you had a pole on the imaginary axis which implies no damping then when you project it onto the real axis you get $0$ which seems incorrect. Did you mean to say poles with zero damping are those on the real axis? Because those ones will not change under projection onto the real axis? – sonicboom Mar 15 '16 at 10:55
• Yes but then for undamped systems we can only have a damped natural frequency of zero? I would have thought that we could have a frequency at any $\omega \in \mathbb{R}$, the natural resonant frequency with no decrease in oscillations, if there was no damping but you seem to be saying $\omega$ has to be zero in this case? – sonicboom Mar 15 '16 at 11:21