We have the following equation:
$$z^3 = \overline{z} $$
I set z to be $z = a + ib$ and since I know that $ \overline{z} = a – ib$. I was trying to solve it by opening the left side of the equation.
$$ z^3 = (a+ib)^3 \Rightarrow $$
$$ [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow $$
$$ a^3 – b^2a – 2b^2a + i (2a^2b + b^2a – b^3) $$
but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it.
Best Answer
In polar form, on the modulus side, $$r^3=r,$$ hence $r=0\lor r=1$.
On the argument side,
$$3\theta=-\theta+2k\pi,$$ hence $\theta=k\pi/2$.
The solutions are $$0,1,i,-1,-i.$$