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$$ z = 1 + 2i \ (complex \ number) \\ z^n = a_n + b_ni \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) \\ We \ know \ that: \ b_{n+2} - 2b_{n+1} + 5b_n = 0 \\ a_{n+1}=a_n-2b_n \\ b_{n+1}=b_n+2a_n $$

Prove that $ z^n \notin \mathbb{R} $

What I tried:

We know that $ b_{n+2} - 2b_{n+1} + 5b_n = 0$

If $ z^n \in \mathbb{R} $ it means that $ b_n = 0 $

So, $ b_{n+2} - 2b_{n+1} = 0 $ and I tried to get a contradiction but I can't because I get $ -3b_n = 0 $.

How can I solve this problem? Thank you!

ITChristian
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    http://math.stackexchange.com/questions/1856963/zn-a-n-b-ni-show-that-b-n2-2b-n1-5b-n-0-complex-number – lab bhattacharjee Jul 12 '16 at 16:12
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    Rather mischievous to not even mention the other question, which indicates a direct solution. That, or you are lost about its relevance... – Did Jul 12 '16 at 16:58
  • @Did yes, it's true. I don't know why the question before is helping me. Can you please explain? THANK YOU! – ITChristian Jul 12 '16 at 17:05
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    Sure -- as soon as you add some personal input to the question. (I just had a look at your previous posts and I am sorry to say that they are an awful sample of how NOT to ask questions on the site...) – Did Jul 12 '16 at 17:08
  • @Did I edited my question. It is ok now? Thank you! – ITChristian Jul 12 '16 at 17:24
  • Hint: $b_{n+2}=2b_{n+1}$ mod $5$ and $b_1=2$ hence $b_n=2,4,3,1,2,4,3,1,2,\ldots$ mod $5$, in particular $b_n\ne0$ mod $5$ hence $b_n\ne0$, qed. – Did Jul 13 '16 at 13:08

4 Answers4

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Put $A=\mathbb{Z}[i]$. It is well known that $A$ is an euclidian, hence a unique factorisation ring, with units $\pm 1$ and $\pm i$. Now $1+2i$ and $1-2i$ are primes in $A$ (if $1+2i=ab$, with $a, b\in A$ we have $5=|a|^2|b|^2$, hence $a$ or $b$ is a unit) They are not associated, as $(1-2i)=\pm (1+2i)$ or $(1-2i)=\pm i (1+2i)$ are not true

Now suppose that $n\geq 1$ and $(1+2i)^{n}\in \mathbb{R}$. Taking the conjugate give $(1+2i)^n=(1-2i)^n$, a contradiction with the unique factorisation in prime elements.

Kelenner
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Recall that the argument of a complex number is the angle between the positive real line and the number plotted in the complex plane. If the real and imaginary parts of the number are both positive, then the argument is $$\arg(z)=\arctan(\Im(z)/\Re(z)).$$ When you multiply complex numbers, their arguments add. Thus, $\arg(z^n)=n\arg(z)$. A real number has an argument that is a multiple of $\pi$. Show that for your complex number, $n\arg(z)$ can never be an integer multiple of $\pi$.

Plutoro
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$b_{n+2} - 2b_{n+1} + 5b_n = 0$ implies $b_{n+2} \equiv 2b_{n+1} \bmod 5$.

Since $b_1 = 2 \not\equiv 0 \bmod 5$, we never have $b_n \equiv 0 \bmod 5$.

In particular, $b_n$ is never zero.

lhf
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In polar form, we can write that $$z=a+ib=r(\cos \theta+ i \sin \theta)$$ where $r=\sqrt{a^2+b^2}$ and $\theta=\arctan (\frac{b}{a})$

So for the given problem, $$z=1+2i=\sqrt5(\cos \theta+ i \sin \theta)$$ where $\theta=\arctan 2$

Using De-Moivre's theorem, $$z=[\sqrt5(\cos \theta+ i \sin \theta)]^n=z=1+2i=\sqrt{5^n}(\cos n\theta+ i \sin n\theta)$$

Now, if $z\in \mathbb{R}$, then $\sin n\theta=0 \implies n\theta=k\pi \implies \theta=\frac{k}{n}\pi=t\pi$ where $t=\frac{k}{n}$

But $t\pi \not = \arctan 2$ for any rational value of $t$.(In case you want to know why this is true, check this link.)

So, $$\color{blue}{z \not \in \mathbb{Q}.}$$

Community
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SchrodingersCat
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