The real part of $z^n$

Question

Prove that $${\displaystyle\lim \limits_{n \to +\infty}{|r^ncos(nθ)|}}=+\infty,$$

where $n$ is integer, $r>1$, $θ/π$ is irrational.

I got this problem from here $1+x+\ldots+x^n$ perfect square , I think it's true for the general situation.

PS: The original problem is to prove $${\displaystyle\lim \limits_{n \to +\infty}{|z^n+\overline{z}^n|}}=+\infty,$$ where $n$ is integer, $z=x+yi,\overline{z}=x-yi,x>y>0,x^2+y^2>1,i=\sqrt{-1}.$

But according to the answer given by @mrf, this is not true when $y/x=tan(π/8)$,so I add the condition that $θ/π$ is irrational.

Thanks in advance!

Answer 1

The statement can be false even when you assume $\frac{\theta}{\pi}$ is irrational.

Let $(s_k)_{k=1,2\ldots}$ be the sequence defined recursively by:

$$s_k = \begin{cases}1,&\quad\text{ for }k = 1\\2^{s_{k-1}},&\quad\text{ for }k > 1\end{cases}$$

and $s$ be the Liouville number $s = \sum_{k=1}^{\infty} 2^{-s_k}$. For $r = \sqrt{2}$, and $\theta = s\frac{\pi}{2}$, we have:

$$\begin{align}r^{2^{s_m}} |\cos( 2^{s_m} \theta )| = & 2^{(2^{s_m}/2)} \left|\cos\left(\frac{\pi}{2}(1 + \sum_{k=m+1}^{\infty} 2^{s_m-s_{k}})\right)\right|\\ \sim & \frac{\pi}{2}2^{(2^{s_m}/2) + s_m - s_{m+1}}\\ = & \frac{\pi}{2}2^{s_m - (s_{m+1}/2)} \end{align}$$

As this converges to $0$ as $m \to \infty$, $r^{n} |\cos(n\theta)|$ has a sub-sequence which converges to $0$ and hence it cannot diverges to $\infty$.

Answer 2

Use the polar representation of $z$, $z = re^{i\theta}$. By the assumptions, $r > 1$ and $0 < \theta < \pi/4$.

Computing, you get $z^n = r^n e^{in\theta}$ and $\bar z^n = r^n e^{-in\theta}$, so $$ z^n + \bar z^n = r^n \big( e^{in\theta} + e^{-in\theta} \big) = 2 r^n \cos(n\theta).$$

Here you can see that there's something wrong with your (supposed) conclusion. Take for example $z = 2e^{i\pi/8}$, then $z^n + \bar z^n = 0$ if $n = 8k+4$ for $k\in\mathbb{Z}$.