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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?

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    $\begingroup$ 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. $\endgroup$ – Brian Mar 28 at 2:39
  • $\begingroup$ Are you sure $|z|^2+1$ actually is some constant? As far as I can see it does not follow from the problem statement... $\endgroup$ – CiaPan Apr 1 at 13:29
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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

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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?

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

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

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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).

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