$z^4=\bar z$ complex number equation Find all $z\in\mathbb{C}$ that satisfy $z^4=\bar z$ and display them on the complex plane.
I started with the exponential forms of the two and got to $r^3e^{i4\varphi}=e^{-i\varphi}$. Is it wrong to assume, that $r=1$ and $4\varphi =-\varphi \Rightarrow \varphi=0$?
 A: Multiplying both sides by $z$ yields $z^5=|z|^2$, and by taking modulus on both sides we get $|z|^5=|z|^2$, therefore $|z|=0$ or $1$. The former gives $z=0$, while the latter implies $z^5=1$. Writing this in the exponential forms $\Rightarrow \mathrm{Arg}(z)=\dfrac{2k\pi}{5} (k=0,1,2,3,4)$, thus the final results are $\cos\dfrac{2k\pi}{5}+\mathrm{i}\cdot\sin\dfrac{2k\pi}{5} (k=0,1,2,3,4)$ and $0$.
A: An alternative route
Let $z=x+yi$. Then, $$(x+yi)^4=x-yi.$$ Expansion on the left side yields $$x^4-6x^2y^2+y^4+(4x^3y-4xy^3)i=x-yi$$. This gives us two equations:
$$x^4-x-6x^2y^2+y^4=0$$ and $$y+4x^3y-4xy^3=0$$
The bottom equation simplifies when you divide by $y$ (assuming $y\ne 0$):
$$1+4x^3-4xy^2=0.$$
This is quadratic in $y$: $y^2=\frac{-4x^3-1}{-4x}.$
Substituting this into the original top equation:
$$x^4-x-6x^2\frac{-4x^3-1}{-4x}+\frac{(-4x^3-1)^2}{(-4x)^2}=0.$$
Assuming $(-4x)^2\ne 0$, we can remove the denominators:
$$16x^6-6x^2(-4x^3-1)+(-4x^3-1)^2=0.$$
Simplifying:
$$32x^6+24x^5+8x^3+6x^2+1=0.$$
From this, you can find approximate values of $x$ and subsequently, using the formula above for $y$ in terms of $x$, approximate values of $y$.
A: We may also pair the original equation with its complex conjufate:
$z^4=\overline{z}$
$z=\overline{z}^4$
Multiplying these equations together gives
$z^5=\overline{z}^5,$
so $z^5$ is a real number. At the same time, substituting the expression for $\overline{z}$ from the first equation into the second gives
$z=z^{16},$
from which either $z=0$ or $z^{15}=1$. The only candidates that match both results are zero and the fifth roots of unity; all six can be checked for correctness in the original equation.
A: One root is clearly $z=0$. For the other roots, let $z=re^{i\theta}$, $r\in\Bbb{R}^{+}$. Then $z^4=\overline{z}$ gives $r^3e^{5\theta i}=1$. Taking the modulus, $r=1$ and so $e^{5\theta i}=1\implies \theta=\frac{2\pi k}{5}$ where $k=0,1,2,3,4$.
So the solution set contains six numbers: $z=0$ and $z_k=\cos(\frac{2\pi k}{5})+i\sin(\frac{2\pi k}{5})$, $k=0,1,2,3,4$.
