<s>[INST] How to find all the $z$ that satisfy $(1+i)z^4=(1-i)|z|^2$? Would you please help me solve this? I need all the $z$ that satisfy the equality
$$(1+i)z^4=(1-i)|z|^2.$$
I tried doing this:
$$
\begin{aligned}
(1+i)z^4&= (1-i)z\overline z\\
(1+i)z^4 -(1-i)z\overline z &= 0\\
z[(1+i)z^3-(1-i)\overline z]&= 0
\end{aligned}
$$
Then $z= 0$ and 
$$(1+i)z^3-(1-i)\overline z= 0.$$
I don't know what to do with $\overline z$.
 [/INST] $$(1+i)z^4=(1-i)|z|^2\iff iz^4=z\bar z\iff\begin{cases}z=0\\iz^3=\bar z\end{cases}$$ $$iz^3=\bar z\iff i(a+ib)^3=(a-ib)\iff {ia^3 - 3 a^2 b -i 3 a b^2 + b^3}=(a-ib)\\\iff   b^3- 3 a^2 b  +i(a^3 - 3 a b^2)=a-ib\iff\begin{cases}b^3- 3 a^2 b=a\\a^3 - 3 a b^2=-b\end{cases}$$
We can also notice that $|b^3- 3 a^2 b  +i(a^3 - 3 a b^2)|=\sqrt{(a^2+b^2)^3}=\sqrt{a^2+b^2}=|a-ib|\implies
|z|=\begin{cases}1\\0\end{cases}$, so it comes with terms with @mfl's answer
</s>