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.

Question

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?

Answer 1

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.

Answer 2

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.