# Complex Roots of The Characteristic Equation

Consider the following second order linear homogenous differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ where $a$, $b$, and $c$ are constants. Recall that the characteristic equation for this differential equation is the quadratic polynomial $ar^2 + br + c = 0$ and that the roots of this polynomial, $r_1$ and $r_2$ are either both real and distinct, both complex and conjugates of each other, or are the same real root repeated twice.

In the case when $r_1$ and $r_2$ were two distinct roots, we saw that the general solution to the second order linear homogenous differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ was $y = Ce^{r_1 t} + De^{r_2 t}$ where $C$ and $D$ are constants. We will now look at the case where the roots $r_1$ and $r_2$ are complex numbers.

Suppose that $r_1$ and $r_2$ are both complex number roots of the characteristic equation $ar^2 + br + c = 0$. One important theorem regarding polynomials is that a polynomial with real coefficients has the property that if $r_1$ is a complex root of the polynomial, then the complex conjugate of $r_1$ is also a root of the polynomial. In this case, since $r_1$ is a complex root of the characteristic equation, we must have that $r_2$, the second root, must be the complex conjugate of $r_1$, that is $r_1 = \overline{r_2}$. We thus have that the roots $r_1$ and $r_2$ can both be written as $r_1 = \lambda + \mu i$ and $r_2 = \lambda – \mu i$ where $\lambda, \mu \in \mathbb{R}$.

Therefore we have that two solutions for our differential equation are:

(1)

As of the moment, the solutions given above are not that useful to us, so we will make use of perhaps one of the most famous formulas in mathematics known as **Euler’s Formula** which is:

(2)

With this in hand, we see that the our solution $y_1 = e^{(\lambda + \mu i)t}$ can be rewritten as:

(3)

Furthermore, noting that $\cos (-\mu t) = \cos (\mu t )$ and $\sin (- \mu t) = – \sin (\mu t)$ we have that our solution $y_2 = e^{(\lambda – \mu i)t}$ can be rewritten as:

(4)

Once again, the solutions above are nicer, though they still include the complex number $i$, while our original differential equation has only real coefficients. Fortunately, the following theorem will allow us to write our solutions in terms of real-valued functions.

Theorem 1: If $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ is a second order linear homogenous differential equation where $p$ and $q$ are continuous real-valued functions and if $y = u(t) + v(t) i$ is a complex-valued solution to the differential equation above, then $y = u(t)$ and $y = v(t)$ are also solutions to this differential equation. |

**Proof:**Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous, and suppose that $y = u(t) + v(t) i$ is a complex-valued solution. Plugging in $y = u(t) + v(t)i$ into our differential equation and we get that:

(5)

- Note that the righthand side of the equation above is a complex number for each $t$. However, a complex number $z$ is equal to $0$ if and only if $\mathrm{Re} (z) = 0$ and $\mathrm{Im} (z) = 0$. From above, this implies that both:

(6)

- Therefore $y = u(t)$ and $y = v(t)$ are both solutions to our second order linear homogenous differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$. $\blacksquare$

From Theorem 1 above, we thus see that $y_1 = e^{\lambda t} \cos (\mu t)$ and $y_2 = e^{\lambda t} \sin (\mu t)$ are both solutions to our second order linear homogenous differential equation. Therefore if the roots of the characteristic equation $ar^2 + br + c = 0$, then for $C$ and $D$ as constants, the general solution to our second order linear homogenous differential equation is given by:

(7)