# Describing a set of elements in a complex plane

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

• $S_2$ consists of all $z$ such that the angle at $z$ in the triangle formed by $z,z_1,z_2$ is $90^{0}$. – Kavi Rama Murthy Jan 28 at 9:13
• The symbol $\in$ for set membership is not an "e" (although it is derived from $\epsilon$). You write it in $\mathrm{\LaTeX}$ as "\in". – 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 $$\pi /2$$.
• Can you show me how you got the answer for $S_2$ please? – JOJO Jan 28 at 9:27
• @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. – Kavi Rama Murthy Jan 28 at 9:47