# Find the value of $z_1^2 + z_2^2 + z_3^2$ given that $z_1, z_2, z_3 \in \mathbb{C}$, $z_1 + z_2 + z_3 = 0$ and $|z_1| = |z_2| = |z_3| = 1$.

So I'm given the task of finding the value of $$z_1^2 + z_2^2 + z_3^2$$ For $z_1, z_2, z_3 \in \mathbb{C}$ such that $z_1 + z_2 + z_3 = 0$ and $|z_1| = |z_2| = |z_3| = 1$.

(Edit: of course $|z|$ represents the modulus of $z$)

I have done some manipulations but have arrived at nothing which I consider to be useful. If anyone can guide me, I'd be very thankful.

[I'm aware there is a question which is very similar and speaks of the fact that $z_1, z_2$ and $z_3$ are the points of an equilateral triangle although I do not see a correlation with this problem, so as a side request, if this has to do with that, I'd also appreciate to understand why, although this is more out of curiosity]

• Start from $(z_1 + z_2 +z_3)^2=0$ – FormerMath Nov 28 '16 at 1:27
• I had tried that previously, reaching the following equality: $z_1^2 + z_2^2 + z_3^2 = -2(z_1z_2 + z_1z_3 + z_2z_3)$ – D. Brito Nov 28 '16 at 1:37
• Is $Re [z_1 z_2] = |z_1| |z_2| \cos \theta$? – David G. Stork Nov 28 '16 at 1:53
• @DavidG.Stork Do you mean $\text{Re}\,z_1z_2$? – A.Γ. Nov 28 '16 at 1:54
• Think of complex numbers as vectors in the plane. Complex number addition corresponds to the usual "arrow-to-tail" parallelogram rule of vector addition. The condition $z_{1}+z_{2}+z_{3}=0$ simply means that, placing the arrows end-to-end, the head of the last meets the tail of the first; hence, we have a "triangle of forces" (to borrow a phrase from physics). The condition on the moduli of the numbers means exactly that the arrows are all the same length. Hence we have an equilateral triangle, whose vertices are the points $z_{1},$ $z_{2}$ and $z_{3}$ on the unit circle. – Will R Nov 28 '16 at 2:52

$$0 = \bar{z_1}\bar{z_2}\bar{z_3}(z_1+z_2+z_3) = \bar{z_2}\bar{z_3}+\bar{z_3}\bar{z_1}+\bar{z_1}\bar{z_2}$$ and hence $$z_1z_2 + z_2z_3 + z_3z_1 = 0$$ Now $$0 = (z_1 +z_2+z_3)^2 = z_1^2 + z_2^2 +z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)$$ and thus $$z_1^2 + z_2^2 + z_3^2 = 0$$

• Nice, and purely algebraical. +1 – dxiv Nov 28 '16 at 3:41
• @dxiv Thanks for the comment and +1 – user348749 Nov 28 '16 at 3:55
• This is very clever! I don't think I would have spotted this. – Will R Nov 28 '16 at 4:09
• This is extremeley efficient, great thinking. First step never would have occured to me. – D. Brito Nov 28 '16 at 4:46

(Not an answer proper, but too long for a comment.)

I'm aware there is a question which is very similar and speaks of the fact that $z_1 ,z_2$ and $z_3$ are the points of an equilateral triangle although I do not see a correlation with this problem

The respective question is Vertices of equilateral triangle inscribed in the unit circle which states:

Prove that if $z_{1}+z_{2}+z_{3}=0$ and $|z_{1}|=|z_{2}|=|z_{3}|=1$ then the points $z_{1},z_{2},z_{3}$ are the vertices of an equilateral triangle inscribed in the unit circle $|z|=1$.

Your question is directly equivalent to the other one, and both are equivalent to proving that $z_1 z_2 + z_2 z_3 + z_3 z_1 = 0$ so that $z_k$ are the roots of a cubic $z^3-w=0$ with $|w|=1$.

I don't see a direct proof to your problem which wouldn't at some point derive and use the fact that $\triangle Z_1Z_2Z_3$ is equilateral, or $z_k = z_1 \omega_k$ where $\omega_k$ are the cube roots of unity.

Think geometrically about how $3$ numbers in the unit circle could sum $0$. Assuming some familiarity with the complex plane one should soon get the easy solution $$S_0=\{1,e^{i\frac{2\pi}{3}},e^{-i\frac{2\pi}{3}}\}.$$ After this one rapidly generalizes to $$S_\theta=e^{i\theta}S_0=\{e^{i\theta},e^{i(\theta+\frac{2\pi}{3})},e^{i(\theta-\frac{2\pi}{3})}\}.$$ Graphically:

(The three blue dots are the elements of $S_0$, and the three red points the elements of $S_\theta$).

So we have a parametric family of solutions of $z_1+z_2+z_3=0$ in the unit circle. Are there other solution $S$ to this problem that is not of this form? No: If $S$ is not of this form then there are two elements $e^{i\alpha}, e^{i(\alpha+\beta)}\in S$ with $\beta\neq \pm\frac{2\pi}{3}$. Call $w$ the other element of $S$. As the sum of the elements of $S$ must be $0$ we must have $$-w=e^{i\alpha}+e^{i(\alpha+\beta)},$$ so the RHS must have modulus equal to one $1$. Calculating \begin{align} |e^{i\alpha}+e^{i(\alpha+\beta)}|^2=&(\cos\alpha+\cos(\alpha+\beta))^2+(\sin\alpha+\sin(\alpha+\beta))^2\\=&\underbrace{\cos^2\alpha+\cos^2\beta}_{1}+\underbrace{\cos^2(\alpha+\beta)+\sin^2(\alpha+\beta)}_{1}+2\underbrace{(\cos\alpha\cos(\alpha+\beta)+\sin\alpha\sin(\alpha+\beta))}_{\cos\beta}. \end{align}

we arrive at $1^2=2+2\cos\beta\hspace{.1cm}$; i.e. $\cos\beta=-\frac{1}{2}$ which only holds if $\beta=\pm\frac{2\pi}{3}$.

So every solution to $z_1+z_2+z_3=0$ in the unit circle is $S_\theta$ for some $0\leq\theta<2\pi$. This reduces your question to $$\text{"if S_\theta=\{w_1,w_2,w_3\}, then what is the value of w_1^2+w_2^2+w_3^2 ?",}$$ which becomes obvious with the observation that $$(z\mapsto z^2)(S_\theta)=S_{2\theta}.$$ I.e. $S_{2\theta}=\{w_1^2,w_2^2,w_3^2\}$, which is a solution of $z_1+z_2+z_3=0$. !!