# Algebraic complex numbers $z$ satisfying the equations $z^n+\bar{z}^n = z+\bar{z}$ for all positive $n$

Is there a complex number $z$ such that $z\neq 0,1$ and $z^n +\bar{z}^n = z+\bar{z}$ for all positive integers $n$?

Is there an algebraic complex number $z$ such that the above properties hold?

The existence of such a complex number would amaze me.

-

Nope. Among slick algebraic proofs while will no doubt appear momentarily from others:

First, it's easy to check those are the only real solutions: If $z$ is real, then the equation requires $2z^n=2z$ for all $n$, which gives $z=0$ or $z=1$. Now we turn to solving your equation $\Re(z^n)=\Re(z)$ if $z$ is not real. Since $z$ isn't real, then certainly $\arg(z)\neq 0$, and so it's easy to argue that the powers $z^n$ oscillate between the left- and right- halves of the complex plane. So $\Re(z^n)$ oscillates between and positive and negative (and/or zero), where as $\Re(z)$ does not.

-

If you write $z = re^{i\theta}$ then you seek $z$ for which $r^n\cos(n\theta) = r\cos(\theta)$ for all $n$. If $0 < r < 1$ then $|r^n\cos(n\theta)| \rightarrow 0$ but not all $r^n\cos(n\theta)$ are zero, a contradiction since they are all the same, while if $r > 1$ then a subsequence of $\{|r^n\cos(n\theta)|\}$ goes to infinity. So you can only have $r = 0$ or $r = 1$.

$r = 0$ solves (here $z = 0$), so it remains to look at $r = 1$. In this case unless $\theta = 0$, $\cos(n\theta)$ diverges (oscillates) as $n \rightarrow \infty$, so $\theta$ must be zero here. In this case $z = re^{i\theta} = 1$.

Hence you have two solutions, $z = 0$ and $z = 1$.

-

Here's a purely algebraic derivation. Write $\ell=z+\bar{z}$. Investigate squares and cubes:

$$\color{Blue}{\ell^2=}(z+\bar{z})^2=z^2+2z\bar{z}+\bar{z}^2=\color{Blue}{\ell+2}\color{Red}{r^2} \tag{1}$$

$$\ell=z^3+\bar{z}^3=\ell(z^2-z\bar{z}+\bar{z}^2)=\ell(\ell-r^2)$$

$$\implies \color{Red}{r^2=\ell-1} \tag{2}$$

$$\color{Blue}{\ell^2=\ell+2}\color{Red}{(\ell-1)} \implies \ell\in\{1,2\} \tag{1+2}$$

Now $\ell=1\implies r=0$ corresponds to $z=0$, and $\ell=2\implies r=1$ corresponds to $z=1$ because it is the only number with both magnitude and real part equal to one.

-
+1 for pretty colors :) –  Zev Chonoles Feb 3 '12 at 6:23