# Imaginary solutions of a recurrence relation

How to solve this recurrence relation using characteristic equation and imaginary numbers?

We have $a_0 = 0$ and $a_1 = 1$ , and for all $j\in\mathbb N$:

$$a_{j+2} = 6a_{j+1} - 10a_j$$

I would really appreciate it if you can guide me through the taken steps.

This is how far I have gotten:

$$a^2 - 6a + 10 = 0$$

Can't solve it normally, even with quadratic function. But then someone told me that these are the solutions: $$a_0 = 3 - i$$ $$a_1 = 3 + i$$

From there I can just instantiate the findings in the general solution and find the constants.

• You state $i \in \mathbb{N}$? Where do the imaginary numbers come in? – Chantry Cargill Oct 22 '14 at 6:25
• Do you know the standard method of solving this kind of equations? Constructing the characteristic polynomial, finding its roots and so on... If yes, then please first attempt to solve the problem yourself, edit your question with your own work and tell us where the problem occurs. – 5xum Oct 22 '14 at 6:28
• Owkey... edited. I actually need the part of finding those first two solutions... – Freakenknight Oct 22 '14 at 7:29
• SO, basically you're asking to solve a quadratic equation? – TZakrevskiy Oct 22 '14 at 9:24
• Nvm, I figured it out... – Freakenknight Oct 22 '14 at 9:33

Since you seem to have found the roots of the characteristic equation, then you have, as usual, $$a_n=A(3-i)^n+B(3+i)^n$$ and the coefficients $A,B$ have to be determined from the first two terms. So, $$a_0=A+B=0$$ $$a_1=A(3-i)+B(3+i)=1$$ from which $A=\frac{i}{2}$ and $B=-\frac{i}{2}$.
This makes $$a_n=\frac{1}{2} i \left((3-i)^n-(3+i)^n\right)$$ Now, you must use the transformation of complex numbers to rationalize the result and get a nice and simple expression (as you can expect, the result is a real for any $n$).