# Show that $α - iβ$ is also a root for $y'' + a_1y' + a_2y = 0$.

Consider the equation $y'' + a_1y' + a_2y = 0$, where the constants $a_1, a_2$ are real. Suppose $α + iβ$ is a complex root of the characteristic polynomial, where $α, β$ are real, $β≠0$.

Show that $α - iβ$ is also a root.

• Is this really related with the differential equation? Isn't that a general fact of second degree equations? Feb 19, 2017 at 16:11
• This is trivial. The characteristic equation is a second degree equations with real coefficients, one it's root is complex, so other is it's conjugate. Feb 19, 2017 at 16:26

The characteristic polynomial is of the form $$k^2+a_1 k + a_2 = 0.$$ $\alpha+i\beta$ is a root, so $$(\alpha+i\beta)^2+a_1 (\alpha+i\beta) + a_2 = 0.$$ There are now two ways to proceed:
1. If you are comfortable taking complex conjugates, you have $$0 = \overline{(\alpha+i\beta)^2+a_1 (\alpha+i\beta) + a_2} = (\alpha-i\beta)^2+a_1 (\alpha-i\beta) + a_2,$$ since $\alpha,\beta,a_1,a_2$ are real, and hence $\alpha-i\beta$ is a root.
2. Alternatively, we can take real and imaginary parts: $$0 = \alpha^2-\beta^2 +a_1\alpha + a_2, \\ 0 = 2\alpha\beta + a_1 \beta$$ These equations are still both satisfied if we replace $\beta$ by $-\beta$, so $\alpha-i\beta$ is a root.