Different approaches solving $z^{2n} = \bar{z}$

Question

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.

Answer 1

Equating the moduli,

$$r^{2n}=r$$ or $$r=0\lor r=1.$$

Equating the arguments,

$$2n\theta=-\theta+2k\pi$$ or $$\theta=\frac{2k\pi}{2n+1}.$$

There are $2n+2$ distinct solutions.

Answer 2

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. $$e^{(2n+1)i\theta}=1$$

Answer 3

Go to the $2n$-th root as late as possible.

In approach 1, you have to remember about $0$, which is a solution. If $z\ne0$, then you can write $z=r\cis\theta$ and then $$\DeclareMathOperator{\cis}{cis} r^{2n}\cis(2n\theta)=r\cis(-\theta) $$ and therefore $$ r^{2n-1}\cis((2n+1)\theta)=1 $$ Now we deduce $r=1$ and $$ (2n+1)\theta=2k\pi \qquad (k=0,1,\dots,2n) $$

For approach 2, do, instead $$ |z|^{2n}=|\bar{z}|=|z| $$ so either $z=0$ or $|z|=1$. In the latter case, $\bar{z}=z^{-1}$ and you get again $$ z^{2n+1}=1 $$