I'm trying to solve this complex number equation: $$z^5=\bar z$$ As far as I understand, every complex number $z$ should be written in trigonometric form: $$r^5(\cos(5\phi) + i\sin(5\phi)) = r(\cos(-\phi) + i\sin(-\phi))$$
Unfortunately, at this moment, I'm stacked. Could you give some ideas that would help me to solve it?
On
We have that
$$z^5=\bar z \implies z^5z=\bar zz \implies z^6=|z|^2$$
which requires $|z|=0$ or $|z|=1$ and $z^6=1$.
On
$z=0$ is an obvious solution. Then the equation can be written
$$z^6=|z|^2$$ where the RHS is real. Hence $z$ is proportional to a sixth root of unity, let $\omega\ne1$.
Now $$z^6=|z|^6=|z|^2$$ requires $|z|=1$.
Finally
$$z=0\lor z=\omega^k$$ for integer $k\in[0,5]$.
On
The hard way:
$$(x+iy)^5=x^5+5ix^4y -10x^3y^2 -10ix^2y^3+ 5xy^4 +iy^5=x-iy$$
or
$$\begin{cases}x^5-10x^3y^2+5xy^4=x,\\5x^4y-10x^2y^3+y^5=-y.\end{cases}$$
We have the obvious solutions $x=0,y=0$ and $y=0,x=\pm1$.
Then
$$\begin{cases}x^4-10x^2y^2+5y^4-1=0,\\5x^4-10x^2y^2+y^4+1=0.\end{cases}$$
By elimination,
$$-24x^4+40x^2y^2-6=0$$
and
$$9-128x^4-256x^8=0.$$
The only real solutions are $x=\pm\dfrac12$, implying $y=\pm\dfrac{\sqrt 3}2$.
In total, $7$ solutions:
$$(0,0),(1,0),(-1,0),\tfrac12(1,\sqrt3),\tfrac12(1,-\sqrt3),\tfrac12(-1,\sqrt3),\tfrac12(-1,-\sqrt3).$$
Note that:
$$z = re^{i\phi}, z^5 = r^5 e^{i5\phi}~\text{and}~\bar{z} = re^{-i\phi},$$
where $r \geq 0$. Therefore:
$$z^5 = \bar{z} \Rightarrow \begin{cases} r^5 = r\\ 5\phi + 2k\pi= -\phi + 2h\pi \end{cases},$$
where $k, h \in \mathbb{Z}.$
The previous system can be rewritten as: $$ \begin{cases} r(r^4-1) = 0\\ 6\phi = 2s\pi \end{cases},$$
where $s = k-h \in \mathbb{Z}.$
The first equation has $3$ distinct roots: $r=-1$, $r=0$ and $r=1.$ Of course, $r=-1$ should be discarded. This means that for $r=0$, $z = 0$ is a solution, which obviously does not depend on the phase $\phi.$ Moreover, for $r=1$, the phase is important. Solving the second equation, we get:
$$\phi = \frac{s\pi}{3},$$
and hence $z = e^{\frac{is\pi}{3}}$ for $s \in \mathbb{Z}$, together with $z=0$ represent the solution of the equation.