The equation I wanted to convert was $|z^2-1|=1$. This is a very easy example but I have no idea where I made my mistake.
Putting $z=re^{i\theta}$, we have $z^2=r^2e^{2i\theta}$. Square both sides on the equation above and use the fact that $|z|^2=z\bar{z},$ we have $$1=|r^2e^{2i\theta}-1|^2=(r^2e^{2i\theta}-1)(\overline{r^2e^{2i\theta}-1})=(r^2e^{2i\theta}-1)({r^2e^{-2i\theta}-1})=r^4-r^2e^{-2i\theta}-r^2e^{2i\theta}+1.$$
So this gives, (either $r=0$ or) $r^2=e^{-2i\theta}+e^{2i\theta}=2\cos(2\theta).$
However, naturally I though this should corresponds to ($r=0$ or) $r=2\cos(\theta).$ (which can be obtained via $z=x+iy$)
This should be relatively easy but I have no idea why two methods dont agree.
Many thanks in advance!
Alternatively, notice that ${|z^2 - 1|=1}$ would imply that ${z^2-1}$ lies on the complex unit circle. So ${\Re(z^2 - 1)=\cos(\theta)}$, and ${\Im(z^2-1)=\sin(\theta)}$ (since ${\cos(\theta) + i\sin(\theta)}$ describes the complex unit circle). You also know from simple algebraic manipulation that
$${z^2-1 = [\Re(z)^2 - \Im(z)^2 - 1] + [2\Re(z)\Im(z)]i}$$
And so you get a system of equations
$${\begin{array}{cc}\Re(z)^2 - \Im(z)^2 - 1 &=\cos(\theta)\\2\Re(z)\Im(z)&=\sin(\theta)\end{array}}$$
Which, for a given value of ${\theta}$ - is enough to deduce the value(s) of ${z \in \mathbb{C}}$.