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Let $z_1$,$z_2$ e $\mathbb C$, $z_1 \not= z_2$ be two points in the complex plane.

Describe the set $S_1$ $=$ {$z$ e $\mathbb C$: $(z-z_1)^2$ + $(z-z_2)^2$ = $(z_1-z_2)^2$}

My attempt:

I expanded the above using remarkable identities, and got that:

$z^2-zz_1-zz_2 = - z_1z_2$ $z^2 - z(z_1+z_2)+z_1z_2=0$ Thus $z_1$ and $z_2$ are two distinct roots of the above quadratic equation, thus $z=z_1$ or $z=z_2$

Is it correct?

What happens if the parentheses were replaced by absolute value? I mean the set becomes: $S_2$ = {$z$ e $\mathbb C$: |$z-z_1|^2$ + $|z-z_2|^2$ = $|z_1-z_2|^2$}

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  • $\begingroup$ $S_2$ consists of all $z$ such that the angle at $z$ in the triangle formed by $z,z_1,z_2$ is $90^{0}$. $\endgroup$ – Kavi Rama Murthy Jan 28 at 9:13
  • $\begingroup$ The symbol $\in$ for set membership is not an "e" (although it is derived from $\epsilon$). You write it in $\mathrm{\LaTeX}$ as "\in". $\endgroup$ – Vsotvep Jan 28 at 11:30
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Your answer t the first is correct and in fact clever. By Pythagorous Theorem $S_2$ consists of all $z$ such that angle at $z$ in the triangle formed by $z,z_1,z_2$ is $\pi /2$.

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  • $\begingroup$ Can you show me how you got the answer for $S_2$ please? $\endgroup$ – JOJO Jan 28 at 9:27
  • $\begingroup$ absolute value is the equivalent of complex modulus. $\endgroup$ – user645636 Jan 28 at 9:36
  • $\begingroup$ @JOJO In the triangle formed by the three points $|z-z_1|^{2}$ is the square of the length of the side joining $z$ and $z_1$. etc. Just apply Pythagorous Theorem. $\endgroup$ – Kavi Rama Murthy Jan 28 at 9:47

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