# Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$.

Let $$z_1$$, $$z_2$$, $$z_3$$ be distinct complex numbers such that $$|z_1| = |z_2| = |z_3| > 0$$. If $$z_1 + z_2z_3$$, $$z_2 + z_3z_1$$, and $$z_3 + z_1z_2$$ are real numbers, prove that $$z_1z_2z_3 = 1$$.

I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).

I tried several approaches: The first was to substitute $$a + ib$$, $$c + id$$, and $$f + ig$$ for $$z_1$$, $$z_2$$, and $$z_3$$, respectively, and bash. I used $$(1+z_1)(1+z_2)(1+z_3)$$, which didn't really get me anywhere.

After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

• Aren't you missing some type of constraint? Like they need to be magnitude 1. You could have $z_1 = 1/2$, $z_2 = 1/4$ and $z_3 = 1/3$.
– IanJ
Commented Dec 28, 2020 at 21:15
• @IanJ Those numbers don't have the same magnitude :-) Commented Dec 28, 2020 at 21:16
• @JyrkiLahtonen In my head those $=$ signs were commas.
– IanJ
Commented Dec 28, 2020 at 21:30

Let $$r = |z_1| = |z_2| = |z_3| > 0$$ be the common modulus and $$P = z_1z_2z_3$$. Then $$z_1 + z_2 z_3 = \overline{z_1} + \overline{z_2} \cdot \overline {z_3} = \frac{r^2}{z_1} + \frac{r^4}{z_2 z_3} \, .$$ Multiplication with $$z_1$$ gives $$z_1^2 + P = r^2 + \frac{r^4 z_1^2}{P} \implies z_1^2 \left( 1 - \frac{r^4}{P}\right) = r^2 - P \, .$$ The last identity holds for $$z_2$$ and $$z_3$$ as well. It follows that $$P = r^4$$ because otherwise $$z_1^2=z_2^2=z_3^2$$, which is not possible for distinct $$z_1, z_2, z_3$$.

So we have $$z_1z_2z_3 = r^4$$. Taking the absolute value implies that $$r=1$$.

• Oh nice. In fact, not only $P = r^4$, but also $r^2 - P = 0$, which gives $r = 1$. Commented Dec 29, 2020 at 6:14
• @CalvinLin: You are completely right. I could not decide whether to use that all $z_i$ have modulus $r$, or that $r^2-P=0$ from the last equation, so I did the former. Commented Dec 29, 2020 at 7:03
• This is elegant. Commented Dec 29, 2020 at 7:32

My solution relies heavily on the fact that the two sides of a parallelogram are equal if and only if the diagonals of the said parallelogram are orthogonal (when the parallelogram is a rhombus). On the complex plane this can be reformulated as follows: two complex numbers share the same magnitude if and only if their sum and difference vectors are orthogonal to each other.

I will first prove that the common length of the three complex numbers is $$=1$$. I will show that $$|z_3|=1$$, the others are done in the same way.

Consider the known real numbers $$u=z_1+z_2z_3$$ and $$v=z_2+z_1z_3$$. We have that $$u+v=(z_1+z_2)(1+z_3)$$ and $$u-v=(z_1-z_2)(1-z_3)$$ are also both real. Let us denote $$\phi_k=\arg z_k, k=1,2,3$$. Because $$|z_1|=|z_2|$$, the parallelogram with sides $$z_1$$ and $$z_2$$ is a rhombus. From this (or by a number of other means) it follows that modulo an integer multiple of $$\pi$$ we have $$\arg(z_1+z_2)\equiv\frac{\phi_1+\phi_2}2$$ and $$\arg(z_1-z_2)\equiv\frac{\pi+\phi_1+\phi_2}2.$$ In other words, the arguments of these two numbers differ from each other by $$\pm\pi/2$$.

As $$\arg(u+v)\equiv \arg(u-v)\equiv0\pmod\pi$$, the arguments of $$1-z_3$$ and $$1+z_3$$ must also differ from each other by $$\pm\pi/2$$. Therefore the diagonals $$1\pm z_3$$ of the parallelogram with sides $$1$$ and $$z_3$$ are orthogonal, making the said parallelogram a rhombus. Therefore $$|z_3|=1$$ as claimed.

Another consequence of $$|z_3|=1$$ is that $$\arg(1-z_3)=\frac{\pi+\phi_3}2.$$ Putting these pieces together we get (modulo $$\pi$$) $$0\equiv\arg(u-v)=\arg(z_1-z_2)+\arg(1-z_3)=\frac{2\pi+\phi_1+\phi_2+\phi_3}2\equiv\frac{\phi_1+\phi_2+\phi_3}2.$$ So $$\phi_1+\phi_2+\phi_3$$ is an integer multiple of $$2\pi$$ and the claim follows.

Alternative approach:
$$x$$ and $$y$$ say hello from kludgiville.

Given that $$(z_1\overline{z_1}) = (z_2\overline{z_2}) = (z_3\overline{z_3}). \tag1$$

Let $$(x_k + iy_k)$$ denote $$z_k : k \in \{1,2,3\}.$$

From the constraints:
$$(x_1 + iy_1) + (x_2 + iy_2)(x_3 + iy_3) = a \in \mathbb{R}.$$
$$(x_2 + iy_2) + (x_1 + iy_1)(x_3 + iy_3) = b \in \mathbb{R}.$$
$$(x_3 + iy_3) + (x_1 + iy_1)(x_2 + iy_2) = c \in \mathbb{R}.$$

Let $$s_1 = (a-z_1), s_2 = (b-z_2), s_3 = (c-z_3).$$
Then $$(z_1 z_2 z_3) = (s_1 z_1) = (s_2 z_2) = (s_3 z_3).\tag2$$

Therefore:
$$s_1 = (a - z_1) = a - (x_1 + iy_1) = a + (x_1 - iy_1) - 2x_1 = [a + \overline{z} - 2x_1] \implies$$
$$(s_1 z_1) = (az_1) + (z_1\overline{z_1}) - 2x_1z_1 = (z_1\overline{z_1}) + z_1(a - 2x_1).\tag3$$

Similarly:
$$(s_2 z_2) = (z_2\overline{z_2}) + z_2(b - 2x_2).$$ $$(s_3 z_3) = (z_3\overline{z_3}) + z_3(c - 2x_3).$$

Using equations (1) and (2) above, this implies that
$$z_1(a - 2x_1) = z_2(b - 2x_2) = z_3(c - 2x_3).$$

Suppose that $$(a - 2x_1) \neq 0.$$

$$\text{Then} ~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k$$

$$\text{and} ~z_1 = kz_2 ~\text{and} |z_1| = |z_2|.$$

The only way that this is possible is if $$z_1 = z_2$$.
This is a contradiction, since it violates the constraint that $$z_1, z_2, z_3$$ are all distinct.

Edit
The above analysis is definitely the right approach, but contains a flaw. See the comment/analysis of JyrkiLahtonen immediately following this answer. Applying his comment/analysis, the explicit remedy is as follows:

Exploring the hypothesis that $$(a - 2x_1) \neq 0$$, and keeping in mind that none of $$z_1,z_2,z_3$$ can $$= 0$$ [else, by equation (1) above, they all $$= 0$$ and are therefore not distinct], you have that

$$~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_1$$

$$\text{and} ~\frac{c - 2x_3}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_2.$$

This implies that $$~z_1 = k_1z_2 = k_2z_3,~~$$ with $$~~|z_1| = |z_2| = |z_3|.$$

This implies that $$~|k_1| = 1 = |k_2|.$$

Since, in this hypothetical $$~~k_1, k_2 \in \mathbb{R},$$
this implies that $$~k_1,k_2~$$ are both $$~\in \{+1, -1\}.$$

In order for $$z_2$$ and $$z_3$$ to both be distinct from $$z_1$$, it is necessary for $$k_1 = -1 = k_2.$$ However, in that case, you have that $$z_2 = z_3$$.

Therefore, the assumption that $$(a-2x_1) \neq 0$$ implies that $$z_1, z_2, z_3$$ can not all be distinct. Therefore, one can conclude that $$(a-2x_1) = 0.$$

Therefore, from equation (3) above, $$(z_1 z_2 z_3) = (s_1 z_1) = z_1 \overline{z_1} \implies (z_2 z_3) = \overline{z_1}. \tag4$$

Using equation (1) above, this implies that $$|z_1|^2 = |z_2 z_3| = |\overline{z_1}| = |z_1| \implies |z_1| = 1.$$

Therefore, by equation (4) above, $$(z_1 z_2 z_3) = 1.$$

• I guess that in $z_1=kz_2$ it is possible to have $k=-1$ without violating the assumption on distinctness. But then you can bring in the other pairs. Three complex numbers, pairwise linearly dependent over $\Bbb{R}$ cannot share the same length without one pair of them being equal? Commented Dec 29, 2020 at 15:44
• @JyrkiLahtonen +1 good catch, my oversight. I agree with your comment. I will edit my answer accordingly. Commented Dec 29, 2020 at 21:12
• A nice fix.+1 was there already. Commented Dec 29, 2020 at 23:11

We have $$\dfrac{z}{|z|}+\dfrac{|z|}{z} \in \mathbb{R}$$ when $$z \ne 0$$

So we have $$\dfrac{z_1}{|z_1|}+\dfrac{|z_1|}{z_1} \in \mathbb{R}$$ and $$z_1+z_2z_3 \in \mathbb{R} \Rightarrow \dfrac{z_1}{|z_1|}+\dfrac{z_2z_3}{|z_1|} \in \mathbb{R}$$

Subtracting we get $$\dfrac{z_1z_2z_3-|z_1|^2}{z|z_1|} \in \mathbb{R} \Rightarrow z_1z_2z_3=|z_1|^2$$

Hence $$|z_1z_2z_3|=|z_1|^3=|z_1|^2 \Rightarrow |z_1|=1$$ and the conclusion follows

Gosh... well, I just looked at the three-way symmetry, noted all points must lie on a circle around the origin, and deduced $$z_1 = 1$$, $$z_2 = e^{2 i \pi/3}$$ and $$z_3 = e^{-2 i \pi/3}$$, and confirmed these worked.

I suppose one could work through the algebra more rigorously, but on an exam... I think this is fine!

• On an olympiad such as RMO, this would probably not be accepted, at least not full credit. You need to show that either all configurations are similar to this or this is the only possible configuration. Commented Dec 28, 2020 at 20:54
• Hmm... I think this works for anything in the form of $z_1 = 1$, and $z_2 = z_3^{-1}$. The issue is that the it should be proven for all configurations, like the above comment also said. Commented Dec 28, 2020 at 20:55
• By symmetry.... Commented Dec 28, 2020 at 21:23
• Given that $z_1=1$, $z_{2,3}=e^{\pm i\phi}$ is a solution for all $\phi$ I am a little bit skeptical about an answer claiming that there is a unique solution. And, I agree, this would score very few points at an olympiad. Commented Dec 28, 2020 at 21:30
• @JyrkiLahtonen all that is necessary is to demonstrate that $|z_1| = 1$ and that $\text{Arg}(z_1) + \text{Arg}(z_2) + \text{Arg}(z_3) = 0,~$ within a modulus of $(2\pi).$ Commented Dec 28, 2020 at 21:33