Are there complex solutions for $z^3=\bar z$ I'm asked to solve $z^3=\bar z$. I got $z=0, 1, -1$. Are there any complex solutions $a+bi$  to this though? 

$(a+bi)^3 = (a^3 - 3ab^2) + (3a^2b - b^3)i = a - bi$
$a^3 - 3ab^2 = a, 3a^2b - b^3 = -b$
If $a = 0$, then $-b^3 = -b$, and the only reals which statisfy that is $b = 0, 1, -1$
If $b = 0$, then similarly $a^3 = a$ so $a = 0, 1, -1$
So far this gives $0, -1, 1, i, -i$
If $a, b \neq 0$,
$a^2 - 3b^2 - 1 = 0, b^2 - 3a^2 - 1 = 0$. This implies $a = b$, so $a^2 - 3a^2 - 1 = 0 \rightarrow 2a^2 = -1$ which is impossible in the reals.
