complex analysis(introduction) - locus on Argand diagram satisfying $z^{n-1}=\bar{z}$ ($z\in\mathbb{C}, n\in\mathbb{N}\setminus\{1\}$) Question: Determine on Argand diagram the locus in $\mathbb{C}^{2}$ satisfying: $z^{n-1}= \bar{z} \space (n\in\mathbb{N}\setminus\{1\})$
My attempt: Let: $z:=r(\cos\theta+i\sin\theta) (r:=|z|, \theta:=argz)$ Then:
$z^{n-1}= \bar{z} \iff r^{n-1}(\cos (n-1)\theta+i\sin(n-1)\theta)=r\cos\theta-ir\sin n \theta\iff \begin{cases}r^{n-1}\cos(n-1)\theta =r\cos\theta && (1)\\
r^{n-1}\sin(n-1)\theta=-r\sin\theta  && (2)\end{cases}$
If: $\cos(n-1)\theta \neq 0$ then: dividing $(2)$ by $(1)$ yields: $\tan(n-1)\theta=\tan(-\theta) \implies (n-1)\theta=-\theta+k\pi \space(k\in\mathbb{Z}) \implies \theta = \frac{k\pi}{n}$ In this case the locus is set of rays going from $(0,0)$ satisfying: $\theta=\frac{k\pi}{n} $($\theta$-angle between real line of Argand diagram and given ray)
If: $\cos(n-1)\theta=0$,then: 
$\begin{cases} 
r\cos(\theta)=0 && (3)\\ 
r^{n-2}\sin(n-1)\theta=-\sin\theta && (4)
\end{cases}$
$(3)$implies: $\theta=\frac{(2k+1)\pi}{2}\space(k\in\mathbb{Z})$ implying that $(4)$ then yields: $r=1$ (since then sine functions are equal to $(-1)^{l},(-1)^{m}$, where: $l,m\in\mathbb{Z}$ but modulus of complex number is positive). In this case the locus is set of point lying on unit circle centred at $(0,0)$, where angle between real line and segment joining origin and k-th distinct point is $\frac{(2k+1)\pi}{2}$
Troubles: My analysis is incorrect since I think that I should obtain only set of points, while at some point set of rays pops out. Besides, I intuitively feel that my reasoning is incomplete...
Request: Could anyone state gaps in my reasoning, please?
Thanks for help in advance.
 A: Here is a different solution to the problem -- if I have more time later, I will try to correct your solution.
We want to solve $z^{n-1} = \bar z$. Multiply both sides by $z$ to get
$$z^n = z \bar z = \left|z\right|^2,$$
so then $\left|z^n\right| = \left|z\right|^2$, so $\left|z\right|^n = \left|z\right|^2$. If $\left|z\right| \neq 0$, then by taking logs we get
$$n \log \left|z\right| = 2 \log \left|z\right|,$$
and if $\left|z\right| \neq 1$ then $\log\left|z\right| \neq 0$, so $n = 2$ by cancellation. In this case, then we are solving $z^{2-1} = z = \bar z$, which has the entire real line as solutions.
Otherwise (if $\left|z\right| = 1$) then we want to solve $z^n = \left|z\right|^2 = 1$; here we see that $z$ must be an $n$-th root of unity.
Thus here are the possible solutions:


*

*If $n = 1$, then $z^{n-1} = \bar z$ if and only if $z = 1$.

*If $n = 2$, then $z^{n-1} = \bar z$ if and only if $z \in \mathbb R$.

*If $n > 2$, then $z^{n-1} = \bar z$ if and only if


*

*$z = 0$, or

*$z^n = 1$ ($z$ is an $n$-th root of unity).


A: Your analysis seems to be correct as is, but you have not proven what you actually want to prove. What you have shown is that if $z^{n-1} = \bar z$, then either


*

*$z$ lies in a ray extending from $0 + 0i$ with angle $\theta = \frac{k\pi}{n}$, or

*$z$ has angle ("argument") $\theta = \left(k + \frac{1}{2}\right)\pi$ and distance from the origin $\left|z\right| = 1$. (Note that then $z$ must be $i$ or $-i$, since only the angle mod $2\pi$ matters.)


This is, in fact, true, but we want a converse. For example, if $n = 4$ then we may consider $z = \frac{1}{2}i$; it certainly lies on the ray from the origin with angle $\theta = \frac{2\pi}{4}$ ($k = 2$), but it is not a solution to $z^{n-1} = \bar z$, since $z^3 = -\frac{1}{8}i \neq -\frac{1}{2}i = \bar z$. What I did in my other answer is show a bidirectional implication: $z$ is a solution to $z^{n-1} = \bar z$ if and only if certain conditions apply to $z$. You proved a half of this: that if $z$ is a solution to $z^{n-1} = \bar z$ then certain conditions apply. You need the other half: if certain conditions apply to $z$, then $z$ is a solution.
