Looking for a good solution to system of complex equations

Suppose that z1,z2,z3z_1, z_2, z_3 are complex numbers which satisfy

{z1=z2=z3,z1+z2+z3=0. \begin{cases} |z_1| = |z_2| = |z_3|, \\ z_1 + z_2 + z_3 = 0. \\ \end{cases}

Can we conclude that z1,z2,z3 z_1, z_2, z_3 form the vertices of an equilateral triangle in the complex plane?

Note by Calvin Lin
5 years, 6 months ago

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1st 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 - 5 years, 6 months ago

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Simple and sweet!!

Vishal Sharma - 5 years, 6 months ago

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Ah, nice interpretation!

Calvin Lin Staff - 5 years, 6 months ago

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Fantastic!

Eloy Machado - 5 years, 6 months ago

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Nice one.

jatin yadav - 5 years, 5 months ago

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Following @Micheal Mendrin, we just have to prove that GOG\equiv O implies that ABC ABC is an equilateral triangle. This is a one-shot-one-kill proof through trilinear coordinates, or through the following observation - if M M is the midpoint of ABAB, then GG lies on CMCM while OO lies on the perpendicular bisector through MM, so GOG\equiv O implies that CMABCM\perp AB, or just AC=AB AC=AB. In the same way we have AB=BCAB=BC.

Jack D'Aurizio - 5 years, 6 months ago

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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 z1z_1 and z2z_2 be α\alpha. Let the angle between z2z_2 and z3z_3 be β\beta and the angle between z3z_3 and z1z_1 be γ\gamma.

Since, z1+z2=z3z_1+z_2=-z_3; z3-z_3 lies on the angle bisector of α\alpha. So, z3z_3 also lies on the angle bisector of α\alpha. That means β=γ\beta=\gamma. Similarly α=β\alpha=\beta and that means they are all equal to 120120^\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 not conclude 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 - 5 years, 6 months ago

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One solution that I have:

WLOG, we can rotate the system such that z1=1z_1 = 1 . Let z2=a+bi z_2 = a + bi , then the second condition gives us z3=1abi z_3 = -1-a-bi .

The first condition gives us a2+b2=1,(1+a)2+b2=1 a^2 + b^2 = 1, (1+a)^2 + b^2 = 1 and hence a2=(1+a)2a=12 a^2 = (1+a)^2 \Rightarrow a = - \frac{1}{2} , which gives us that z1,z2,z3z_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 - 5 years, 6 months ago

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

Finn Hulse - 5 years, 6 months ago

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Magnitude, or distance from the origin. For example, 5+12i=52+122=13|5+12i| = \sqrt{5^2+12^2} = 13.

Justin Wong - 5 years, 6 months ago

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The bars on the sides are the "modulus" sign, for example the modulus of a+bia+bi is a2+b2\sqrt {a^{2}+b^{2}}.

Tristan Shin - 5 years, 6 months ago

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@Tristan Shin Cool beans! Thank you! I used them in my proof! :D

Finn Hulse - 5 years, 6 months ago

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I think treating complex numbers as vectors can help us

Harsh Depal - 5 years, 6 months ago

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Wait really?!?!?!

Finn Hulse - 5 years, 6 months ago

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If we consider z1,z2,z3z_{ 1 },z_{2},z_{3} as three vectors of equal magnitude,say k\boxed{k} (already given) then each of these vectors will be at 120° from each other to satisfy z1+z2+z3z_{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 z2z1|z_{2}-z_{1}| subtends 120° at the center so it'll subtend 60° on the circumference i.e.z3 z_{3} Similarly we can prove for the other two as well. Hence proved that it'll be an equilateral triangle

Vishal Sharma - 5 years, 6 months ago

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The important part is justifying why those vectors must be at 120 degrees to each other.

For example, the conclusion no longer holds true for 4 or more variables.

Calvin Lin Staff - 5 years, 6 months ago

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As you are aware I treated the complex numbers as vectors, and since they are having equal magnitude, as well as their sum is zero. Therefore they have to be at 120° w.r.t each other. For any other angle between the vectors z1+z2+z3z_{1}+z_{2}+z_{3}=0 won't be true

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 - 5 years, 6 months ago

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@Vishal Sharma The fact that they must be 120 120 ^\circ with respect to each other is the crux that you have to show.

It is not true that for 4 complex numbers, they (the vectors) must be inclined at 90 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 90^\circ .

Calvin Lin Staff - 5 years, 6 months ago

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@Calvin Lin 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.. Thanks

Vishal Sharma - 5 years, 6 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 - 5 years, 6 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 - 5 years, 6 months ago

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Let the angle between z1z_{1} and z2z_{2} be α\alpha, i.e. z2=z1eiαz_{2} = z_{1} e^{i \alpha}

Clearly, z1+z2=z3|z_{1}+ z_{2}| = |-z_{3}|

Square both sides to get:

2+2cosα=12 + 2 \cos \alpha = 1

cosα=12\Rightarrow \cos \alpha = - \dfrac{1}{2}

Hence, α=2π3\alpha = \dfrac{2 \pi}{3}

Thus, z2=z1ei2π3=z1ωz_{2} = z_{1} e^{i \dfrac{2\pi}{3}} = z_{1} \omega

z3=(z1+z2)=z1ω2z_{3} = -(z_{1}+z_{2}) = z_{1} \omega^2

Hence, the numbers are z1,z1ω,z1ω2z_{1} , z_{1} \omega , z_{1} \omega^2, which are vertices of equilateral triangle, as the vectors from origin to these are rotated by 120120^{\circ} to get the next vector.

jatin yadav - 5 years, 5 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 (θ\theta) must be equal to 120120^\circ, which is just 3603\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 z1=biz_1=bi thus z1z_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 (θ=120\theta=120^\circ), then z2z_2 and z3z_3 must be symmetric via. the complex axis with one in the third quadrant and one in the fourth. Thus, they form a 3030^\circ angle with the real axis and their coordinates are  ±3z12z1i/2\ \pm \frac{\sqrt{3}|z_1|}{2}-|z_1|i/2 by the laws of 30609030-60-90 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 2z122\frac{-|z_1|}{2} which is just z1-z_1. Adding that to the first leg (z1z_1) is zero and thus equation two is satisfied! Looking back at the coordinates for z2z_2 and z3z_3, we can use the Pythagorean theorem to verify that they have the same length as z1z_1. Plugging their coordinates in to the Distance Formula produces (z124)+(3z124(z1)2(\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 120120^\circ angles and all vectors have the same length, so the rest is just angle chasing! Letting AA, BB, and CC be the endpoints of each vector, and letting OO be the origin, we can see that ABCABC is an equilateral triangle by connecting ABAB, BCBC, and ACAC. From there, we know that the angle AOCAOC is 120120^\circ, and it form an isosceles triangle (OAOA and OCOC are equal, remember?). Thus, the angles OCAOCA and OACOAC are both 3030^\circ. This can be proven for the other two equal isosceles triangles which means that each angle ABCABC, ACBACB, and BACBAC are 6060^\circ. And we're done! Yipee! *REMINDER TO ALL READERS, I AM A STUDENT IN ALGEBRA 1! :D

Finn Hulse - 5 years, 6 months ago

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You need to be careful with why "the three values must be equally spaced apart". There is a lot more going on here, and in fact the crux is argubaly showing that the angle is 120 120^\circ .

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 - 5 years, 6 months ago

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Yes, but I proved that with the other part. I see what you're saying though. :D

Finn Hulse - 5 years, 6 months ago

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