Locus of a complex number $z$ when locus of $z^2$ is known

We know $|z|^2 = |z^2|$. Let's substitute $t=z^2$ then, and we get an equation $$|t-1| = |t| + 1.$$ If we denote the complex plane's origin with $O$ and the point $(1,0)$ with $U$, the above equation can be expressed with line segments' lengths as $$Ut = Ot + OU$$ which is a triangle inequality for the triangle $triangle OUt$ degenerated to a segment. So $O$ must lie between $U$ and $t$, hence $t$ is a non-positive real number.

As a result, $z$ is an imaginary number (including zero).

Note that for $u,v in mathbb{C}$ you have

• $|u+v| leq |u| + |v|$ and equality holds if and only if $operatorname{Re}(ubar v)=|u||v|$.

Applying this to the given equation you get

$$|z^2 + (-1)| = |z|^2 +1 Leftrightarrow operatorname{Re}(z^2cdot (-1)) = |z|^2 Leftrightarrow boxed{operatorname{Re}(z^2) = -|z|^2}$$

With $z= x+iy$ you get immediately $$boxed{operatorname{Re}(z^2) = -|z|^2} Leftrightarrow x^2-y^2 = -(x^2+y^2) Leftrightarrow x = 0, ; y in mathbb{R} Leftrightarrow boxed{z = iy}$$

This is a vertical line through the origin.

You can do the algebra (in polar or Cartesian coordinates) ... a bit tedious.

Here is a simple demonstration that every point on this line satisfies the equation:

Consider ${z:z=x+iyinmathbb{C}, textrm{with } x=0 } $.

Then $|z^2 -1 | = |-y^2 -1| = y^2+1 = |z|^2+1.$

WLOG

$z=r(cos t+isin t)$ where $r>0,t$ are real

$|z|=r,|z^2-1|=sqrt{r^2+1-2rcos2 t}$

Can you take it from here?