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 $\mathbb C$: |$z-z_1|^2$ + $|z-z_2|^2$ = $|z_1-z_2|^2$}

  • $\begingroup$ $S_2$ 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

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 $\pi /2$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.