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.
Hint
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 $n$).
I am sure that you can take from here.