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Let $z_1,z_2,z_3 \in \mathbb{C}$ with $z_i=1$ for $i=1,2,3$ and $z_{1}+z_{2}+z_3=0$. Show that $z_i$ are vertices for a equilateral triangle.

Tip: Think about the case $z_3=1$. What then follows the general case?

My attempt:

Since $z_3=1=1+0i$, it must be $z_1+z_2=-1$ with $z_1:=a_1+ib_1$ and $z_2:=a_2+ib_2$. Adding these equations leads to $(a_1+a_2)+i(b_1+b_2)=-1+0i$.

So $a_1+a_2=-1$ and $b_1+b_2=0$. Now I am stuck with my argumentation, there are just too many variables...

Any Ideas?

2 Answers2

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The answer to your question is $$ z_1=1\\ z_2=-\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot i\\ z_3=-\frac{1}{2}-\frac{\sqrt{3}}{2}\cdot i $$ enter image description here

If $\vec{z}_1=(x,y)$, $\vec{z}_2=(z,w)$ and $\vec{z}_3=(u,v)$ then by figure \begin{align} (z-u)^2+(w-v)^2=&(z-x)^2+(w-y)^2\\ (z-x)^2+(w-y)^2=&(u-x)^2+(v-y)^2\\ (z-u)^2+(w-v)^2=&(u-x)^2+(v-y)^2\\ \end{align} By $\|z_i\|=1$, $i=1,2,3$, we have \begin{align} z^2+w^2=&1\\ u^2+v^2=&1\\ x^2+y^2=&1\\ \end{align}

By $z_1+z_2+z_3=0$ we have \begin{align} u+x+z=&0\\ v+y+w=&0\\ \end{align} Now use those $ 8 $ equations to get the answer that was given.

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Elias Costa
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  • Sorry, It must be $|z_i|=1$ –  May 19 '19 at 18:46
  • And how? How did you find $a,b$? –  May 19 '19 at 18:47
  • I got it. :-) Thanks –  May 19 '19 at 19:34
  • Woah, thanks for your effort. But I solved it way easier. Look at my attempt. The missing two equations necessary to solve for all variables are $a_1+a_2=-1$, $b_1+b_2=0$, $a_1^2+b_1^2=1$ and $a_2^2+b_2^2=1$. –  May 19 '19 at 19:58
  • Often easier just means "with less detail." For example, why can we take $z_3 = 1$ without loss of generality? It's intuitively clear and, depending on context, might not require justification, but if you had to justify it, could you? Could you do it in fewer steps than MathOverview's solution takes? – Charles Hudgins May 19 '19 at 20:06
  • Probably not, sorry. :-( –  May 19 '19 at 20:13
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As $|z_j|=1$ then $\overline z_j=z_j^{-1}$. Then as $\overline{z_1}+\overline{z_2}+\overline{z_3}=0$, $$z_2z_3+z_3z_1+z_1z_2=z_1z_2z_3(z_1^{-1}+z_2^{-1}+z_3^{-1})=0$$ and then the $z_j$ are the roots of $$w^3-z_1z_2z_3=0.$$ As the three cube roots of a complex number, they form the vertices of an equilateral triangle centred at the origin.

Angina Seng
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