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

If $$|z^2 -1| = |z|^2 +1$$ then $$z$$ lies on a:
a) circle.
b) parabola.
c) ellipse.
d) straight line.

My attempt: Since $$|z|^2 +1$$ is some constant value hence the locus of $$z^2$$ is a circle with centre at $$1+i0$$ but how do I find the locus of $$z$$ with this?

• Thanks for transcribing your problem instead of link a picture! I have added some additional formatting to your post using MathJax. This page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. – Brian Mar 28 at 2:39
• Are you sure $|z|^2+1$ actually is some constant? As far as I can see it does not follow from the problem statement... – CiaPan Apr 1 at 13:29

You have: $$|z^2-1|=|z|^2+1$$.
Squaring both sides: $$|z^2-1|^2=(|z|^2+1)^2$$.
Since:$$|a-b|^2=|a|^2+|b|^2-a\bar b-\bar a b$$ Now You have: $$|z|^4+1-z^2-\bar z^2=|z|^4+2|z|^2+1$$
Rearranging and cancelling terms:
$$2|z|^2+z^2+\bar z^2=0$$ Now, $$|z|^2=z\bar z$$ So, you get $$(z+\bar z)^2=0$$ i.e., $$z+\bar z=0$$ z is the set of purely imaginary numbers

WLOG

$$z=r(\cos t+i\sin t)$$ where $$r>0,t$$ are real

$$|z|=r,|z^2-1|=\sqrt{r^2+1-2r\cos2 t}$$

Can you take it from here?

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 $$\{z:z=x+iy\in\mathbb{C}, \textrm{with } x=0 \}$$.

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

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

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

Applying this to the given equation you get

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

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

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).