In a complex methods course I am taking, we were given an equation for a particular driven harmonic oscillator where the driving force is trigonometric. I have worked out the math and obtained an equation that tells me that the driving frequency at resonance is the natural frequency multiplied by i. My tutor tells me that this is a 90 degree phase shift, but I don't really understand why. Isn't a phase shift obtained by adding or subtracting 90 degrees? And how can a frequency, which is a measurable physical value, take on imaginary values? I would understand if we were talking about velocity. Because velocity has a direction, addition or scalar multiplication by a real value would not describe a 90 degree rotation of the vector. But frequency is a scalar quantity. What does it mean to have an imaginary frequency?

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    $\begingroup$ Hint: $e^{i \theta} = \cos \theta + i \sin \theta$ (Euler's formula) $\endgroup$
    – K_inverse
    Sep 12, 2018 at 2:39
  • $\begingroup$ I realize that, but multiplying i to a real value is not the same as adding pi/2 to the same real value. Since the natural frequency is real valued, even using Euler's identity, I can only at most add to the power. Unless the power of the frequency is the phase, that would not explain the phase shift. Even if it did, it doesn't answer the question of what an imaginary frequency means. I cannot create a driving force with a frequency of i... $\endgroup$
    – Clovis Nyu
    Sep 12, 2018 at 2:49
  • $\begingroup$ Building onto K_inverse's comment, I don't know the exact context, but pertaining to $i$, this is 90 degrees (or an angle of $\pi / 2$) in the complex plane. As for the units curiosity, remember degrees, or radians, aren't really a physical unit. So if your function is rolling off of say $\omega t$, multiplying by $-1$ throws it $180$ degrees out of phase just as $i$ to $90$ degrees. Could you add the particular oscillator to the question? $\endgroup$ Sep 12, 2018 at 2:51

3 Answers 3


If your oscillating function is of the form $e^{i\omega t}$, a phase shift looks like $e^{i(\omega t+\phi)}$, which can be rewritten as $e^{i\omega t}e^{i\phi}$.

Now, recall that $e^{i\phi}=\cos\phi + i\sin\phi$. A 90 degree phase shift corresponds to $\phi=\frac{\pi}{2}$.


$$e^{i\frac{\pi}{2}}=\cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i = i.$$

So finally we have,

$$e^{i(\omega t+\frac{\pi}{2})}=e^{i\omega t}e^{i\frac{\pi}{2}}=ie^{i\omega t}.$$

So we see that a phase shift of 90 degrees corresponds to multiplication by $i$.


It is a phase shift by 90 degrees if multiplied by $i$ indeed.

Note that $i=e^{i\pi/2}$. Writing whatever driving signal in complex form, since it is sinusoidally driven, it will have an $e^{i\omega t}$ in it, multiplying by $i$ multiplies by $e^{i\pi/2}$, and when you multiply the exponentials you add the exponents to get $e^{i(\omega t+\pi/2)}$.

Taking the real part to get an answer that actually makes sense physically, you would have a $\cos(\omega t+\pi/2)$ dependency in your driving.

I think this is what you are asking, hope this helps.


There is an article here: (the optimal driving force is shown to be $90^{\circ}$ out of phase of the motion)

Phase difference of driving frequency and oscillating frequency

Also any vector like $4j + 3i$ can be expressed in phasor form as $5 \angle 41^{\circ}$ or in complex form $4+3i$. Adding $90^{\circ}$ is just a vector of same amplitude at $90^{\circ}$ to the original.


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