If I claim that the norm of a complex vector is to remain constant $|z|^2=(a+ib)(a-ib)=a^2+b^2$, then I obtain the U(1) symmetry (I think?).
My questions is; if I claim that the square of a complex vector is to remain constant $z^2= (a+ib)^2=a^2-b^2+2iab$, what is the resulting symmetry? In the latter case both the real and the imaginary part must be held constant: (1): $a^2-b^2$ and (2) $ab$.
So
- is it the case that the constraint $a^2+b^2=C$ is invariant with respect to U(1)?
- What symmetric group is produced by the constraints of $a^2-b^2=C$ and $ab=D$.
If the norm of a complex number remains constant, then $z$ lies on a circle, as you've stated: $|z|^2=a^2+b^2 = \textrm{const.}$
For $z^2$ to remain constant, it is the intersection of two hyperbolas, and is two points:
$z=z_0$ and $z=-z_0$.
(Every non-zero complex number has two square roots.)
Update (a few more details):
If $z^2= r_0 e^{i\theta_0}$, then $z \in \{ r_0^{1/2} e^{{i\theta_0}/{2}} , r_0^{1/2} e^{i(\theta_0/2+\pi)}\}$