$ 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.
 A: $$z^{n+1}=a_{n+1}+ib_{n+1}=zz^n=(1+2i)(a_n+ib_n)=(a_n-2b_n)+i(b_n+2a_n)$$
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
$$b_{n+2}-2b_{n+1}+5b_n=(b_{n+1}+2a_{n+1})-2(b_n+2a_n)+5b_n$$
$$=b_n+2a_n+2(a_n-2b_n)-2(b_n+2a_n)+5b_n$$
$$=0$$
A: If we let $\Im(z)$ denote the imaginary part of $z$ we find that your expression is equal to 
$$
\Im(z^{n+2})-2\Im(z^{n-1})+5\Im(z^n)=\Im\left(z^n\left(z^2-2z+5\right)\right).
$$
Since $1+2i$ is a root of $z^2-2z+5$, we find that this is equal to $0$.
A: 
Notice, when $z\in\mathbb{C}$ and $n\in\mathbb{R}^+$:
$$z^n=\left(|z|e^{\arg(z)i}\right)^n=|z|^ne^{n\arg(z)i}$$
Where $|z|=\sqrt{\Re^2[z]+\Im^2[z]}$ and $\arg(z)$ is the complex argument of $z$.


So, we get:
$$\text{Q}=(1+2i)^n=5^{\frac{n}{2}}e^{n\arctan(2)i}$$
Now, see:

*

*$$a_n=\Re[\text{Q}]=5^{\frac{n}{2}}\cos(n\arctan(2))$$

*$$b_n=\Im[\text{Q}]=5^{\frac{n}{2}}\sin(n\arctan(2))$$
So, we get that:
$$b_{n+2}-2b_{n+1}+5b_n=0\Longleftrightarrow$$
$$5^{\frac{n+2}{2}}\left(\sin((n+2)\arctan(2))+\sin(n\arctan(2))\right)-2\cdot5^{\frac{n+1}{2}}\sin((n+1)\arctan(2))=0\Longleftrightarrow$$
$$5^{\frac{n+2}{2}}\left(\sin((n+2)\arctan(2))+\sin(n\arctan(2))\right)=2\cdot5^{\frac{n+1}{2}}\sin((n+1)\arctan(2))\Longleftrightarrow$$
$$\frac{5^{\frac{n+2}{2}}}{2\cdot5^{\frac{n+1}{2}}}=\frac{\sin((n+1)\arctan(2))}{\sin((n+2)\arctan(2))+\sin(n\arctan(2))}\Longleftrightarrow$$
$$\frac{\sqrt{5}}{2}=\frac{\sqrt{5}}{2}$$
