Solving recursion with characteristic equation and facing complex numbers

Question

So I have to solve this recursion $b_n = b_{n-1} - b_{n-2}$ where $b_0=1$ and $b_1=8$. The characteristic equation is $x^2-x+1$. Then I calculate roots which are unfortunately complex numbers: $x_1=\frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $x_2=\frac{1}{2} + \frac{\sqrt{3}}{2}i$. Then $b_n = A \cdot {x_1}^n + B \cdot {x_2}^n$. So I have: $$\begin{cases} 1=A+B\\ 8=A(\frac{1}{2} - \frac{\sqrt{3}}{2}i) + B(\frac{1}{2} + \frac{\sqrt{3}}{2}i)\end{cases}$$ Then after few transformations I have: $\frac{15}{\sqrt{3}}=(2B-1)i$.

But if I try to solve this equation I get $4B^2-4B+76=0$ which also yields negative delta... What now or what's wrong?

Answer 1

$$\frac{15}{\sqrt{3}}=(2B-1)i$$

then

$$-i\frac{15}{\sqrt{3}}=2B-1.$$

From $B$, draw $A$ and the rest follows.

Answer 2

$$ \begin{align} 1&=A+B\\ 8&=A\left(\frac{1-i\sqrt3}2\right)+B\left(\frac{1+i\sqrt3}2\right)\\ &=\frac12+(B-A)\frac{i\sqrt3}2\\ i\,5\sqrt3&=A-B\\ \end{align} $$ Solving for $A$ and $B$ is now easier.

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Another approach:

The characteristic equation $$ x^2-x+1=0 $$ Has roots $e^{\pi i/3}$ and $e^{-\pi i/3}$. Thus, the solution has the form $$ a\cos\left(\frac{\pi k}3\right)+b\sin\left(\frac{\pi k}3\right) $$ Since $k=0$ gives $1$, we must have $a=1$. Since $k=1$ gives $8$, $$ \frac12+b\frac{\sqrt3}2=8\implies b=5\sqrt3 $$