Complex roots of $z^3 + \bar{z} = 0$ I'm trying to find the complex roots of $z^3 + \bar{z} = 0$ using De Moivre.
Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know which root was the extra ).
I noticed that $z=a+bi$ and there exists $\theta$ such that the trigonometric representation of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$ .
It seems that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$
However, my trig is pretty rusty and I'm not quite sure where to go from here.

OK, I'm going to take a stab at this.
Given:  $z^3 + \bar{z} = 0$
Therefore $z^3 = -\bar{z}$ and $|z^3| = |-\bar{z}|$ .  
We have that $|-\bar{z}|=|\bar{z}|=|z|$ therefore $|z^3|=|z|$ and $|z||z||z|=|z|$.  
Therefore $|z|=0$ or $|z|=1$.
Case 1: Assume $|z|=0$ then if
$z=a+bi\leftrightarrow|z|=\sqrt{a^2+b^2}=0\leftrightarrow a=0 \wedge b=0\leftrightarrow a+bi=0$ 
thus $z = 0$.
Case 2: Assume $|z|=1$ 
Let's multiply by $z$ and we get $z^4 = -z\bar{z}$.  We see that $z\bar{z} = |z|^2$ so we get $z^4 = -( |z|^2 )$ so from the above either:
therefore $z^4 = -|z| = -1$
The trigonometric representation of $-1$ is $1*( \cos \pi + i \sin \pi )$ so according to De Moivre:
$z^4 = r^4(\cos 4\theta + i \sin 4\theta ) = 1*( \cos \pi + i \sin \pi )$
These are two complex numbers in trigonometric form so:
$r^4 = 1$ and $4\theta = \pi + 2\pi*k$ or
$r=1$ and $\theta = \frac{\pi + 2\pi*k}{4}$ and each solution has the form:
$z_k = \cos( \frac{\pi + 2\pi*k}{4} ) + i \sin (\frac{\pi + 2\pi*k}{4})$
for $0\leq k \leq 3$.
Which together with Case 1 gives the following values for $z$:
$0,\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$
