$$ z = 1+2i \ (complex \ number) \\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) $$ 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.

  • $\begingroup$ Is $n$ also a real integer (positive or negative)? $\endgroup$ – Jan Jul 12 '16 at 11:48
  • $\begingroup$ See en.wikipedia.org/wiki/Recurrence_relation#Solving $\endgroup$ – lab bhattacharjee Jul 12 '16 at 11:48
  • 1
    $\begingroup$ @JanEerland Thank you. Edited question: $ n \in \mathbb{N}^* $ $\endgroup$ – MM PP Jul 12 '16 at 11:51
  • $\begingroup$ Have you tried induction? $\endgroup$ – alphacapture Jul 12 '16 at 11:54

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

  • $\begingroup$ Thank you very much! I tried with binomial theorem but I was stupid. $\endgroup$ – MM PP Jul 12 '16 at 12:03

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 $0$.

  • $\begingroup$ @MMPP You're welcome :). If you replace $\Im$ with $\Re$ this shows that the same relation holds for $a_n$. $\endgroup$ – Pjotr5 Jul 12 '16 at 12:37

Notice, when $z\in\mathbb{C}$ and $n\in\mathbb{R}^+$:


Where $|z|=\sqrt{\Re^2[z]+\Im^2[z]}$ and $\arg(z)$ is the complex argument of $z$.

So, we get:


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:

$$b_{n+2}-2b_{n+1}+5b_n=0\Longleftrightarrow$$ $$5^{\frac{n+2}{2}}\left(\sin((n+2)\arctan(2))+\sin(n\arctan(2))\right)-2\cdot5^{\frac{n+1}{2}}\sin((n+1)\arctan(2))=0\Longleftrightarrow$$ $$5^{\frac{n+2}{2}}\left(\sin((n+2)\arctan(2))+\sin(n\arctan(2))\right)=2\cdot5^{\frac{n+1}{2}}\sin((n+1)\arctan(2))\Longleftrightarrow$$ $$\frac{5^{\frac{n+2}{2}}}{2\cdot5^{\frac{n+1}{2}}}=\frac{\sin((n+1)\arctan(2))}{\sin((n+2)\arctan(2))+\sin(n\arctan(2))}\Longleftrightarrow$$ $$\frac{\sqrt{5}}{2}=\frac{\sqrt{5}}{2}$$

  • $\begingroup$ Wow! Thank you, but didn't think until now a simple problem can be proved as difficult as here. You had really a good idea. Very interesting! $\endgroup$ – MM PP Jul 12 '16 at 12:09
  • $\begingroup$ @MMPP I was solving your question the same way as 'AlphaGo' did, so I switched to a more difficult but elegant way of solving your problem $\endgroup$ – Jan Jul 12 '16 at 12:13
  • $\begingroup$ @MMPP You're welcome. $\endgroup$ – Jan Jul 12 '16 at 12:21

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