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Frequency = Imaginary Rate?

Suppose we have

$$q(t) = q_0 e^{\lambda t} .$$

It is easy to take the $n^{\rm th}$ time derivative of this function - we just "pull out a factor $\lambda$" n times. For n=2 we get $\ddot{q} = \lambda^2 q_0 e^{\lambda t}$ or just

$$\ddot{q} = \lambda^2 q .$$ (12.10)

Now go back to the example "solution" in Eq. (5), which turned out to be equivalent to HOOKES'S LAW [Eq. (6)]: $\ddot{x} = -\omega^2 \; x$, where $\omega = \sqrt{k/m}$ and k and m are the "spring constant" and the mass, respectively.

Equations (10) and (6) would be the same equation if only we could let $q \equiv x$ and $\lambda^2 = -\omega^2$. Unfortunately, there is no real number whose square is negative. Too bad. It would be awfully nice if we could just re-use that familiar exponential function to solve mass-on-a-spring problems too . . . . If we just use a little imagination, maybe we can find a $\lambda$ whose square is negative. This would require having a number whose square is -1, which takes so much imagination that we might as well call it i. If there were such a number, then we could just write

$$\lambda = i \omega .$$ (12.11)

That is, the rate $\lambda$ in the exponential formula would have to be an "imaginary" version of the frequency $\omega$ in the oscillatory version, which would mean (if the solution is to be unique) that

$$e^{i\omega t} = \cos{\omega t} .$$

It's not.

Oh well, maybe later . . . .