linear algebra - complex analysis(introduction) - locus on Argand diagram satisfying $z^{n-1}=bar{z}$ ($zinmathbb{C}, ninmathbb{N}setminus{1}$)

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.

Answer

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).