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

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?

## marked as duplicate by Martin R, Lord Shark the Unknown, Xander Henderson, Leucippus, max_zornMay 20 at 5:35

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

• Sorry, It must be $|z_i|=1$ – Analysis May 19 at 18:46
• And how? How did you find $a,b$? – Analysis May 19 at 18:47
• I got it. :-) Thanks – Analysis May 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 $a_2^2+b_2^2=1$. – Analysis May 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 at 20:06

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.

• Thank you very much. Is my attempt wrong? – Analysis May 19 at 19:02