Complex Roots

Complex roots are simply roots of the characteristic equation that come out with an $i$ in them, which happens when the quadratic formula asks for the square root of a negative number. ("Complex" here doesn't mean complicated; it's the name for numbers with a real part and an imaginary part.) All we're really saying here is that when that happens, the solution swings back and forth instead of just growing or shrinking, which is why this case describes anything that vibrates or rings. The formula $y = e^{\lambda t}\left(c_1\cos(\mu t) + c_2\sin(\mu t)\right)$ is only two parts multiplied together: $e^{\lambda t}$ (an envelope that grows, shrinks or stays the same size) and the cosine-plus-sine (a plain wave, wiggling $\mu$ radians per unit time). The $i$ in the roots looks scary, but it never shows up in the answer. Bear with the next couple of paragraphs; they explain why.

Why sines and cosines? Start from what always works: $y = e^{rt}$ turns $ay'' + by' + cy = 0$ into $(ar^2 + br + c)e^{rt} = 0$. When $b^2 - 4ac < 0$ the roots are complex, $r = \lambda \pm \mu i$, with

Nothing in that algebra cared whether $r$ was real, so $e^{(\lambda + \mu i)t}$ still solves the equation. The real question is what $e$ to an imaginary power even means.

Euler's formula, straight from Taylor series. The Taylor series of $e^x$ is $\sum\frac{x^n}{n!}$, and it makes sense for any number, real or complex. Plug in $x = i\theta$. The powers of $i$ cycle, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, so the even powers come out real and the odd powers carry an $i$:

Those two sums are exactly the Taylor series of $\cos\theta$ and $\sin\theta$. So $e^{i\theta} = \cos\theta + i\sin\theta$: as $\theta$ grows, $e^{i\theta}$ walks around the unit circle.

From complex to real. By Euler's formula,

The equation has real coefficients, so if a complex function $u + iv$ solves it, the left side becomes $L[u] + i\,L[v] = 0$, and a complex number is zero only when its real and imaginary parts both are. So $u$ and $v$ are each solutions, which hands us two real ones:

The other root, $\lambda - \mu i$, gives the same pair back, so it adds nothing new. Their Wronskian is $W = \mu e^{2\lambda t}$, which is never zero because $\mu \ne 0$, so they form a fundamental set and every solution is $c_1y_1 + c_2y_2$. That's the formula at the top, proved.

What each piece does. The factor $e^{\lambda t}$ is always positive: it's the envelope. If $\lambda < 0$ the oscillation dies out, if $\lambda = 0$ it keeps the same size forever, and if $\lambda > 0$ it grows. The cosine and sine swing between $-1$ and $1$. That's the oscillation itself, at $\mu$ radians per unit time, one full cycle every $\frac{2\pi}{\mu}$. There's a second animation further down that shows the whole story: the complex solution spirals around the complex plane, and its shadow on the real axis is the real solution you actually see.

Reading the roots. For $y'' + 2y' + 10y = 0$, the roots are $-1 \pm 3i$. So without solving anything else, you already know the answer oscillates at 3 rad/s inside an envelope $e^{-t}$ that shrinks.

A common mistake. It's tempting to think the $i$ means the solution is complex, so it can't describe a real spring. It isn't: the $i$ only lives in the roots. The real and imaginary parts of $e^{(\lambda + \mu i)t}$ are each real solutions on their own, and those are the only ones you write down.

A guitar string, a car bouncing on its springs, and a tuned circuit in a radio all have complex roots: they ring at a frequency set by $\mu$, fading at a rate set by $\lambda$.