$ z^n = a_n + b_ni $ Show that $ b_{n+2} - 2b_{n+1} + 5b_n = 0 $ (complex numbers) $$ z = 1+2i \ (complex \ number)
\\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*)
$$
Prove that $ b_{n+2} - 2b_{n+1} + 5b_n = 0$
How can I solve this? Thank you!
EDIT: Or please tell me your ideas.

$$z^{n+1}=a_{n+1}+ib_{n+1}=zz^n=(1+2i)(a_n+ib_n)=(a_n-2b_n)+i(b_n+2a_n)$$
So
$$a_{n+1}=a_n-2b_n$$
$$b_{n+1}=b_n+2a_n$$
Similary
$$a_{n+2}=a_{n+1}-2b_{n+1}$$
$$b_{n+2}=b_{n+1}+2a_{n+1}$$
So
$$b_{n+2}-2b_{n+1}+5b_n=(b_{n+1}+2a_{n+1})-2(b_n+2a_n)+5b_n$$
$$=b_n+2a_n+2(a_n-2b_n)-2(b_n+2a_n)+5b_n$$
$$=0$$
