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Please help me with the following problem:

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.

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    Is this really related with the differential equation? Isn't that a general fact of second degree equations? – 2017-02-19
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    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. – 2017-02-19

1 Answers 1

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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.