Let $z_1,z_2,z_3\in\mathbb{C}$ such that $|z_1|=|z_2|=|z_3|=1$.

If $z_1+z_2+z_3\ne0$ and $z_1^2+z_2^2+z_3^2=0$ then prove $|z_1+z_2+z_3|=2$.

What I did


$\Rightarrow z_1^3+z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^3+z_2^2z_3+z_3^2z_1+z_3^2z_2+z_3^3=0$


(HERE IS THE MISTAKE) $z_1^3+z_2^3+z_3^3=(z_1+z_2+z_3)(\underbrace{z_1^2+z_2^2+z_3^2}_0-z_1z_2-z_1z_3-z_2z_3)=$



Substituting $(z_1^3+z_2^3+z_3^3)$ in $(*)~(z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2)$ reduces $\Rightarrow$

$-3z_1z_2z_3=0\Rightarrow z_1z_2z_3=0$. That means one of them must be $0$. Let's prove it.

$z_1=a_1+b_1i,~z_2=a_2+b_2i~(a_1,b_1,a_2,b_2\in\mathbb{R})$. We want to see when $z_1z_2=0$.


$ \begin{cases} a_1a_2-b_1b_2=0\Rightarrow a_1a_2=b_1b_2&(1) \\ a_1b_2+a_2b_1=0&(2) \end{cases} $

Let's consider $a_1,b_2\ne0$ ,so we can divide by $a_1b_2$ in $(1)$.

$\frac{a_2}{b_2}=\frac{b_1}{a_1}\Rightarrow a_2=\frac{b_1}{a_1}\cdot b_2$

Substitute in $(2)$ and $a_1b_2+\frac{b_1}{a_1}\cdot b_2b_1=0\Rightarrow b_2\cdot\frac{a_1^2+b_1^2}{a_1}=0$. $b_1\ne0\Rightarrow a_1^2+b_1^2=0$, and since $a_1,b_1\in\mathbb{R}$ the only possibility is that $a_1=b_1=0\Rightarrow z_1=0$

So when $z_2$ is complex ($b_2\ne0$) $z_1=0$ and by symmetry when $z_1$ is complex $z_2=0$.

Conclusion: $z_1z_2=0\Leftrightarrow z_1=0$ or $z_2=0$. The relation applies for multiple variables also.

Thus, one of $z_1,z_2,z_3$ must be $0$. But in the hypothesis $|z_1|=|z_2|=|z_3|=1$.

How is that possible?

  • 1
    $\begingroup$ If $\omega$ is a sixth root of unity then $z_1=1$, $z_2= \omega$, $z_3=\omega^5$ satisfy the conditions, but none of them is zero. You can use that example to check where your calculation is wrong. $\endgroup$
    – Martin R
    Commented Jun 24, 2021 at 9:05
  • 3
    $\begingroup$ You've made an algebra error on line 7 of your post. You claim that (I'm changing variables to make it easier to write) $(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=a^3+b^3+c^3$. But this is not true. That actually expands to $a^3+b^3+c^3-3abc$. $\endgroup$
    – Tom Sharpe
    Commented Jun 24, 2021 at 9:05
  • $\begingroup$ The identity is $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 -ab -bc- ca)$ actually. You missed out $-3abc$. $\endgroup$
    – bigbang
    Commented Jun 24, 2021 at 9:05
  • 2
    $\begingroup$ Also, you don't need to prove that $abc=0$ implies one of $a,b$ or $c$ is $0$. That follows from the fact that $\mathbb{C}$ is a field. $\endgroup$
    – Tom Sharpe
    Commented Jun 24, 2021 at 9:07
  • $\begingroup$ You are right! Problem solved. Thank you guys! $\endgroup$
    – Neox
    Commented Jun 24, 2021 at 9:08


You must log in to answer this question.

Browse other questions tagged .