# Paradoxical homework

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

$$z_1^2+z_2^2+z_3^2=0~~|\cdot(z_1+z_2+z_3)$$

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

$$z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2+(z_1^3+z_2^3+z_3^3)=0~~~(*)$$

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

$$=-(z_1+z_2+z_3)(z_1z_2+z_1z_3+z_2z_3)=$$

$$=-(3z_1z_2z_3+z_1^2z_2+z_1^3z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2)$$

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 $$z_1z_2=0$$.

$$(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i=0$$

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

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?

• 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. Commented Jun 24, 2021 at 9:05
• 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 $a^3+b^3+c^3-3abc$. Commented Jun 24, 2021 at 9:05
• 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 $-3abc$. Commented Jun 24, 2021 at 9:05
• 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. Commented Jun 24, 2021 at 9:07
• You are right! Problem solved. Thank you guys!
– Neox
Commented Jun 24, 2021 at 9:08