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While working through some exercises in Walter Ender's Applied Econometric Time Series, I came across an assertion which I struggle to understand.

(for those familiar with the book, I'd like to refer you to "Appendix 1.1: Imaginary Roots and de Moivre’s Theorem").

It is stated, that a homogenous solution to a second order difference equation $y_t=a_1y_{t-1}-a_2y_{t-2}$ can be expressed as $y_t^h = A_1a_1^t + A_2a_2^t$

Characteristic roots are next established using standard procedures.

In this case the discriminant is assumed to be negative. As a result we arrive at the solution using trigonometric identities.

Characteristic roots are found to be $a_1^t=r^t[cos(tθ)+isin(tθ)]$ and $a_2^t=r^t[cos(tθ)-isin(tθ)]$.

Next, it is stated (the part I don't understand) that although $A_1$ and $A_2$ are arbitrary numbers, they must have the form: $A_1 = B_1[ cos(B_2) + i sin(B_2)]$ and $A_2 = B_1[ cos(B_2) – i sin(B_2) ]$ where $B_1$ and $B_2$ are arbitrary real numbers measured in radians.

How come $A_1$ and $A_2$ "must have" this form? Can't they simply remain $A_1$ and $A_2$, as it is in the case when the discriminant is positive?

goose
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Yes they can, but they are now complex but not independent. Indeed, complex numbers carry two constants (the real and the imaginary part), for a total of four, while the solution of a second order difference equation only depends on two.

For the solution to be real, it must equal its complex conjugate, so that

$$A_1r^t[\cos(tθ)+i\sin(tθ)]+A_2r^t[\cos(tθ)-i\sin(tθ)]=\\(A_1r^t[\cos(tθ)+i\sin(tθ)]+A_2r^t[\cos(tθ)-i\sin(tθ)])^*= \\A_1^*r^t[\cos(tθ)-i\sin(tθ)]+A_2^*r^t[\cos(tθ)+i\sin(tθ)].$$

By identification, $A_1=A_2^*,A_2=A_1^*$ hence the two complex constants are conjugate of each other. The $B$ representation is just their polar form.

  • But why $A1$ and $A2$ must now be complex? – goose Dec 20 '16 at 09:38
  • @goose: because reals would give the solution $(A_1+A_2)r^t\cos(t\theta)$, which is not general enough. –  Dec 20 '16 at 09:42
  • Does this mean that $A1$ and $A2$ can, in principle, be reals, but for the sake of reaching a general solution, we choose to assume they are complex? – goose Dec 20 '16 at 10:55
  • @goose: right, otherwise you miss the sine component. –  Dec 20 '16 at 10:59