<s>[INST] $z^4=\bar z$ 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 $4\varphi =-\varphi \Rightarrow \varphi=0$?
 [/INST] 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 $y$.
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