Suppose that \(z_1, z_2, z_3\) are complex numbers which satisfy

\[ \begin{cases} |z_1| = |z_2| = |z_3|, \\ z_1 + z_2 + z_3 = 0. \\ \end{cases} \]

Can we conclude that \( z_1, z_2, z_3 \) form the vertices of an equilateral triangle in the complex plane?

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TopNewest1st condition: The circumcenter of {z1, z2, z3} is {0,0}.

2nd condition: The centroid oif {z1, z2, z3} is also {0,0}.

Without going into the details, only when {z1, z2, z3} are vertices of an equilateral triangle both 1) and 2) are true. This is probably just saying that one can use complex numbers to prove certain things in geometry, and vice-versa. This is what makes math so fun. – Michael Mendrin · 2 years, 11 months ago

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– Vishal Sharma · 2 years, 11 months ago

Simple and sweet!!Log in to reply

– Eloy Machado · 2 years, 11 months ago

Fantastic!Log in to reply

– Calvin Lin Staff · 2 years, 11 months ago

Ah, nice interpretation!Log in to reply

– Jatin Yadav · 2 years, 10 months ago

Nice one.Log in to reply

I saw this note right before I went to sleep last night and came up with two solutions. However, I see that @Michael Mendrin has beat me to it with the first one.

Here's the second one. We're going to use the fact that the resultant of two equal (in magnitude) vectors lies on the bisector of the angle between the two vectors. Its proof is pretty simple [and the fact is pretty intuitive as well].

Let the angle between \(z_1\) and \(z_2\) be \(\alpha\). Let the angle between \(z_2\) and \(z_3\) be \(\beta\) and the angle between \(z_3\) and \(z_1\) be \(\gamma\).

Since, \(z_1+z_2=-z_3\); \(-z_3\) lies on the angle bisector of \(\alpha\). So, \(z_3\) also lies on the angle bisector of \(\alpha\). That means \(\beta=\gamma\). Similarly \(\alpha=\beta\) and that means they are all equal to \(120^\circ\).

I'm sorry if this is too short but it's almost midnight here. I'll add a diagram and do the vector proof tomorrow.

EDIT: Oh, by the way, you can

notconclude that they are the vertices of an equilateral triangle. They don't form the vertices of an equilateral triangle when the numbers are all equal to zero :) – Mursalin Habib · 2 years, 11 months agoLog in to reply

Following @Micheal Mendrin, we just have to prove that \(G\equiv O\) implies that \( ABC \) is an equilateral triangle. This is a one-shot-one-kill proof through trilinear coordinates, or through the following observation - if \( M\) is the midpoint of \(AB\), then \(G\) lies on \(CM\) while \(O\) lies on the perpendicular bisector through \(M\), so \(G\equiv O\) implies that \(CM\perp AB\), or just \( AC=AB\). In the same way we have \(AB=BC\). – Jack D'Aurizio · 2 years, 11 months ago

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If we consider \(z_{ 1 },z_{2},z_{3}\) as three vectors of equal magnitude,say \(\boxed{k}\) (already given) then each of these vectors will be at 120° from each other to satisfy \(z_{1}+z_{2}+z_{3}\) =0 Now we assume a circle of radius 'k' with center at the origin and obviously all three complex numbers will lie on the circle. We have seen that any two complex numbers subtend 120° at the origin which is indeed the center of circle. Now We can definitely say that the chord \(|z_{2}-z_{1}|\) subtends 120° at the center so it'll subtend 60° on the circumference i.e.\( z_{3}\) Similarly we can prove for the other two as well. Hence proved that it'll be an equilateral triangle – Vishal Sharma · 2 years, 11 months ago

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For example, the conclusion no longer holds true for 4 or more variables. – Calvin Lin Staff · 2 years, 11 months ago

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Also I agree it won't be true for 4 or more complex numbers. But those cases would be different from the one above. For e.g if I have four complex numbers then for the vector sum to be zero. They must be inclined at 90° to each other. In other words angle subtended at origin differs for different numbers.

(This is an explanation for what I have understood from your ques. , if this not what you meant , pls let me know.)

By the way thanks for posting such notes which inspire new thinking and we can learn all sorts of different approaches from people around the world. Pls keep posting in future as well... – Vishal Sharma · 2 years, 11 months ago

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It is not true that for 4 complex numbers, they (the vectors) must be inclined at \( 90^\circ \) to each other. In fact, any 4 complex numbers who form the vertices of a rectangle suffice. Note that the angles formed by the vectors (at the origin) need not be \( 90^\circ \). – Calvin Lin Staff · 2 years, 11 months ago

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– Vishal Sharma · 2 years, 11 months ago

OK, I get it. You're are right that for 4 numbers sum can be zero even when they aren't at 90° to each other. Now I will try to prove that they'll be at 120° by some different way.. ThanksLog in to reply

I think treating complex numbers as vectors can help us – Harsh Depal · 2 years, 11 months ago

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– Finn Hulse · 2 years, 11 months ago

Wait really?!?!?!Log in to reply

One solution that I have:

WLOG, we can rotate the system such that \(z_1 = 1 \). Let \( z_2 = a + bi \), then the second condition gives us \( z_3 = -1-a-bi \).

The first condition gives us \( a^2 + b^2 = 1, (1+a)^2 + b^2 = 1 \) and hence \( a^2 = (1+a)^2 \Rightarrow a = - \frac{1}{2} \), which gives us that \(z_1, z_2, z_3 \) are the vertices of an equilateral triangle.

I would prefer a solution that isn't so algebraic / brute force. – Calvin Lin Staff · 2 years, 11 months ago

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– Finn Hulse · 2 years, 11 months ago

Remind me what the little absolute value thingies mean in terms of complex numbers. :DLog in to reply

– Justin Wong · 2 years, 11 months ago

Magnitude, or distance from the origin. For example, \(|5+12i| = \sqrt{5^2+12^2} = 13\).Log in to reply

– Tristan Shin · 2 years, 11 months ago

The bars on the sides are the "modulus" sign, for example the modulus of \(a+bi\) is \(\sqrt {a^{2}+b^{2}}\).Log in to reply

– Finn Hulse · 2 years, 11 months ago

Cool beans! Thank you! I used them in my proof! :DLog in to reply

Let the angle between \(z_{1}\) and \(z_{2}\) be \(\alpha\), i.e. \(z_{2} = z_{1} e^{i \alpha}\)

Clearly, \(|z_{1}+ z_{2}| = |-z_{3}|\)

Square both sides to get:

\(2 + 2 \cos \alpha = 1\)

\(\Rightarrow \cos \alpha = - \dfrac{1}{2}\)

Hence, \(\alpha = \dfrac{2 \pi}{3}\)

Thus, \(z_{2} = z_{1} e^{i \dfrac{2\pi}{3}} = z_{1} \omega\)

\(z_{3} = -(z_{1}+z_{2}) = z_{1} \omega^2\)

Hence, the numbers are \(z_{1} , z_{1} \omega , z_{1} \omega^2\), which are vertices of equilateral triangle, as the vectors from origin to these are rotated by \(120^{\circ}\) to get the next vector. – Jatin Yadav · 2 years, 10 months ago

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If one triangle has the same centroid and circumcenter then it'e equilateral. It's easy to prove, because it means that every median is also perpendicular on each side. – Adrian Neacșu · 2 years, 11 months ago

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We can treat them as vectors and hence we get that they are making an angle 120○with each other ....or else we can consider the system as BF3 molecule where all bonds are equal and arranged in a way that they are as far away as possible.... – Rajsuryan Singh · 2 years, 11 months ago

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From the second equation, by vector addition, the three values must be equally spaced apart (assuming this is referring to the Argand diagram). Thus the angle between each vector (\(\theta\)) must be equal to \(120^\circ\), which is just \(\frac{360}{3}\)). From the first equation, their lengths are equal, thus they form the center of an equilateral triangle. A more rigorous proof:

Let \(z_1=bi\) thus \(z_1\) is pure imaginary, and when plotted is simply a vertical line lying on the imaginary axis. Without loss of generality, we will assume that the triangle is centered at the origin. Assuming the above is true (\(\theta=120^\circ\)), then \(z_2\) and \(z_3\) must be symmetric via. the complex axis with one in the third quadrant and one in the fourth. Thus, they form a \(30^\circ\) angle with the real axis and their coordinates are \(\ \pm \frac{\sqrt{3}|z_1|}{2}-|z_1|i/2\) by the laws of \(30-60-90\) triangles. Now, let's add up all of their coordinates to see if it's zero. Obviously the \(\pm\) causes the real portion to cancel out, leaving \(2\frac{-|z_1|}{2}\) which is just \(-z_1\). Adding that to the first leg (\(z_1\)) is zero and thus equation two is satisfied! Looking back at the coordinates for \(z_2\) and \(z_3\), we can use the Pythagorean theorem to verify that they have the same length as \(z_1\). Plugging their coordinates in to the Distance Formula produces \((\frac{|z_1|^{2}}{4})+(\frac{3|z_1|^{2}}{4} \Longrightarrow (|z_1|)^{2}\) which shows that they all posses the same length and we're done! We've proved that the vectors form \(120^\circ\) angles and all vectors have the same length, so the rest is just angle chasing! Letting \(A\), \(B\), and \(C\) be the endpoints of each vector, and letting \(O\) be the origin, we can see that \(ABC\) is an equilateral triangle by connecting \(AB\), \(BC\), and \(AC\). From there, we know that the angle \(AOC\) is \(120^\circ\), and it form an isosceles triangle (\(OA\) and \(OC\) are equal, remember?). Thus, the angles \(OCA\) and \(OAC\) are both \(30^\circ\). This can be proven for the other two equal isosceles triangles which means that each angle \(ABC\), \(ACB\), and \(BAC\) are \(60^\circ\). And we're done! Yipee! *REMINDER TO ALL READERS, I AM A STUDENT IN ALGEBRA 1! :D – Finn Hulse · 2 years, 11 months ago

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For example, if we had 4 (or more) vectors whose sum is 0, then it need not be true that they "must be equally spaced apart". – Calvin Lin Staff · 2 years, 11 months ago

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– Finn Hulse · 2 years, 11 months ago

Yes, but I proved that with the other part. I see what you're saying though. :DLog in to reply