$ z = 1 + 2i $ - Prove that $ z^n \notin \mathbb{R} $

Question

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

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!

Answer 1

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 $\pi$.

Answer 2

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

Answer 3

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.

Answer 4

$b_{n+2} - 2b_{n+1} + 5b_n = 0$ 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.