Complex Roots to the Characteristic Equation
Lesson 4W
Complex Exponentiation
Every complex number can be written in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i=\sqrt{-1}\).
If you are solving a linear homogeneous differential equation with constant coefficients, \(e^{rt}\) is a solution if \(r\) is a root of the characteristic equation. But what if \(r\) is complex?
What is \(e^{(a+bi)t}\)? It is \[e^{at}\big(\cos(bt)+ i\sin(bt)\big).\] Here is why:
Practice by computing \(e^{i\pi}\) and \(e^{1+i}\).
Solving differential equations when there are complex roots to the characteristic equation.
Now we want to use this to solve a differential equation. The good news is, we can eliminate the imaginary numbers. Here is how it works: