So I go this formula: $ z^{2n} = \bar{z} $, of course $z \in \mathbb{C}$.
Me and a couple of friends tried solving for $z$, but came with different solutions, and we can't figure out exactly who's right and why.
Approach one
Using trigonometric notation:
$$\DeclareMathOperator{\cis}{cis}
(r\cis(\theta) )^{2n} = r\cis(-\theta)
$$
And using De-Moivre:
$$r\cis(\theta) = \sqrt[2n]{r\cis(-\theta)} = r^{1\over2n}\cis({2\pi k -\theta\over2n}), k=0,\dots,2n-1$$
Second approach
Multiply by z:
$$z^{2n+1} = |z|^2$$
When $|z|^2=r^2$ in the trig notation
$$z= \sqrt[2n]{r^2}=\sqrt[2n]{r^2\cis(0)}=r^{2\over2n+1}\cis({2\pi k \over2n+1}), k=0,\dots,2n$$
And of course also 0 is a solution.
Second method gives you one more solution, and none of those look the same.
Best Answer
Let me offer you a third approach to compare to.
$z^{2n}=r^{2n}e^{2ni\theta}=re^{-i\theta}$
And so $r=1$ clearly. Now solve for values of $\theta$. i.e. $$e^{(2n+1)i\theta}=1$$