System Of Complex Equations

1) Suppose that z1,z2,z3,z4z_1, z_2, z_3, z_4 are complex numbers which satisfy

{z1=z2=z3=z4=1,z1+z2+z3+z4=0. \begin{cases} |z_1| = |z_2| = |z_3| = |z_4| = 1 , \\ z_1 + z_2 + z_3 + z_4 = 0. \\ \end{cases}

What can we conclude about z1,z2,z3,z4 z_1, z_2, z_3, z_4 ? Must they lie on the corners of a square?


2) Suppose that z1,z2,z3,z4,z5z_1, z_2, z_3, z_4, z_5 are complex numbers which satisfy

{z1=z2=z3=z4=z5=1,z1+z2+z3+z4+z5=0. \begin{cases} |z_1| = |z_2| = |z_3| = |z_4| = |z_5| =1, \\ z_1 + z_2 + z_3 + z_4 + z_5 = 0. \\ \end{cases}

What can we conclude about z1,z2,z3,z4,z5 z_1, z_2, z_3, z_4, z_5 ? Must they lie on the corners of a regular pentagon?

You can refer to this similar note, which deals with the 3 variable case.

Note by Calvin Lin
5 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

For 1, I'm going to use the same argument as before.

Without loss of generality, let z1z_1, z2z_2, z3z_3 and z4z_4 be arranged counter-clockwise around the origin.

The resultant of two equal [in magnitude] vectors lies on the bisector of the angle between them.

So, z1+z2=a1z_1+z_2=a_1 must lie on the bisector of the angle between z1z_1 and z2z_2.

Similarly, z3+z4=a2z_3+z_4=a_2 must lie on the bisector of the angle between z3z_3 and z4z_4.

Since a1+a2=0a_1+a_2=0, we can conclude that they lie on the same line.

That means [you can see this visually, I don't know if it's rigorous] z1z_1 and z4z_4 lies on the same line and z3z_3 and z2z_2 lies on the same line.

And z1+z4|z_1|+|z_4| and z2+z3|z_2|+|z_3| are the diagonals of a quadrilateral and since they are equal and bisect one another, they are the diagonals of a rectangle.


You should add that the complex numbers are non-zero. I will work on 2 now.

Mursalin Habib - 5 years, 5 months ago

Log in to reply

For 2)

Must they lie on the corners of a regular pentagon?

Not necessarily. Here's a counter-example. If z1z_1, z3z_3 and z5z_5 are the vertices of an equilateral triangle and z2=z4z_2=-z_4 and the angle between z1z_1 and z2z_2 is not equal to 7272 degrees, then the complex numbers are not the vertices of a regular pentagon.

Notice that you can rotate the line that contains z2z_2 and z4z_4 along the origin. I'll add a picture later.

What can we conclude about z1z_1, z2z_2, z3z_3, z4z_4, z5z_5?

Well, we can say that the complex numbers are the vertices of a erm... a pentagon! :)


We can generalize this. Given any nn complex numbers that satisfy the following conditions, it is not necessary that the numbers are the vertices of a regular nn gon.

EDIT: Adding a picture.

Imgur Imgur

BBB1BB'B'_1 is an equilateral triangle and the complex numbers ABAB, ACAC, ABAB', AB1AB'_1 and ADAD are five complex numbers that satisfy the conditions but are not the vertices of a regular pentagon.

Mursalin Habib - 5 years, 5 months ago

Log in to reply

For 5 points, since we're looking to classify these sets, can we conclude that the set "must either be an equilateral triangle and a diameter, or a regular pentagon"? Or is there another set that is more 'random'?

Calvin Lin Staff - 5 years, 5 months ago

Log in to reply

@Calvin Lin I think there exist various other sets. For example,

z1=a+i0z_1 = a + i0

z2=0.1a+icz_2 = -0.1a + ic

z3=0.1aicz_3 = -0.1a - ic

z4=0.4a+idz_4 = -0.4a + id

z5=0.4aidz_5 = -0.4a - id

A rough figure of the shape formed by these points can be seen here.

These complex numbers doesn't form a nice geometrical shape. We can just call it a pentagon.

Prakhar Jaiswal - 5 years, 5 months ago

Log in to reply

I think we cannot conclude about both (1) and (2) . If we take z1,z3 as diametrical end points of unit circle and z2 , z4 as diametrical end points of same circle but a different diameter. It will satisfy both the conditions. And it is not necessary that both diameters will be perpendicular to each other. hence the 2 conditions are not sufficient to conclude that z1,z2,z3,z4 will be vertices of square

Harsh Depal - 5 years, 5 months ago

Log in to reply

THE ARE THE ROOTS OF THE EQUATION Z^5=1

Sai Krishna Chary - 5 years, 5 months ago

Log in to reply

1) If you take any two numbers as diametrical points on a circle of radius 1. You get two diameters and if you rotate these two diameters about the origin independently you will get all the possible values of z1 z2 z3 z4 and they will not necessarily form a square 2) Similarly, If you take any three points as vertices of an equilateral triangle circumscribed by the circle of radius 1 and the remaining 2 points as diametrical points of the same circle you can rotate these two figures independently about the origin to get all the 5 points.

In general, for a system with n points, you can form subsets of regular polygon(s) and diameter(s) and rotate them to get the values.

Ajinkya Jadhav - 5 years, 5 months ago

Log in to reply

(Z1Z2)/(Z3Z4)=1(Z_1*Z_2) / (Z_3*Z_4) =1 ////no

El Kattan - 5 years, 5 months ago

Log in to reply

Hm, is that a sufficient and necessary condition? What does that tell us about the complex numbers? Is there any possible description?

Calvin Lin Staff - 5 years, 5 months ago

Log in to reply

According to me I think it the four complex numbers can lie on any rectangle

Milun Moghe - 5 years, 5 months ago

Log in to reply

I mean the vertices of a rectangle

Milun Moghe - 5 years, 5 months ago

Log in to reply

As with the previous note, the challenge is to prove your claim.

Calvin Lin Staff - 5 years, 5 months ago

Log in to reply

1) Let z1=a+biz_1 = a+bi. Then, z2=±c±a2+b2c2iz_2 = \pm c \pm \sqrt{a^2+b^2-c^2}i, z3=abiz_3 = -a-bi, z4=ca2+b2c2iz_4 = \mp c \mp \sqrt{a^2+b^2-c^2}i, cRc \in \mathbb{R} (or some permutation of this; WLOG z1z_1, z2z_2, z3z_3 and z4z_4 are arranged clockwise or counter-clockwise around the origin), i.e. we have that z1z_1 and z3z_3 are reflections of each other over the origin, as are z2z_2 and z4z_4. Geometrically, the object z1z2z3z4z_1 z_2 z_3 z_4 is a rectangle centered around the origin. I'll come back to you on 2)!

Michael Lee - 5 years, 5 months ago

Log in to reply

Can you substantiate why ziz_i must have that form? I see that z1,z2 z_1, z_2 are still unrestricted, but why must z3,z4 z_3, z_4 have that particular form?

Calvin Lin Staff - 5 years, 5 months ago

Log in to reply

@Finn Hulse @Rajsuryan Singh @Adrian Neacșu @Harsh Depal @Vishal Sharma @Jack D'Aurizio @Mursalin Habib @Michael Mendrin Thanks for working on the 3 variable case. Care to tackle the 4 variable and 5 variable case?

Calvin Lin Staff - 5 years, 5 months ago

Log in to reply

I'd love to! I'll get right to work... tomorrow. :D

Finn Hulse - 5 years, 5 months ago

Log in to reply

that z1=z3=-1 and z2=z4=1

Nabil Ahmed - 5 years, 5 months ago

Log in to reply

No, It is not necessary that they should lie on corners of square. Second condition gives points are collinear and first gives they lie on a circle with radius 1. example (1,0) (-1,0) (1/2,root3/2) (-1/2,-root3/2)

Manish Mehta - 5 years, 5 months ago

Log in to reply

THEY ARE THE ROOTS OF THE EQUATION Z^5=0

Sai Krishna Chary - 5 years, 5 months ago

Log in to reply

well, since z1=z2=z3=z4=1\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =\left| { z }_{ 4 } \right| =1 is true, z1,z2,z3,z4{ z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 },{ z }_{ 4 } all lie on the unit circle with radius one and center as origin which is one obvious conclusion you make...so they make a cyclic quadrilateral when joined..and one of the properties of a cyclic quadrilateral states that the sum of the opposite angles is always π \quad \pi radians... from the statement 2 given z1+z2+z3+z4=0{ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 }=0 using zi=cosθi+isinθi{ z }_{ i }=\cos { { \theta }_{ i } } +i\sin { { \theta }_{ i } } i get cosθ1+cosθ2+cosθ3+cosθ4=0\cos { { \theta }_{ 1 } } +\cos { { \theta }_{ 2 } } +\cos { { \theta }_{ 3 } } +\cos { { \theta }_{ 4 } } =0 and similarly sinθ1+sinθ2+sinθ3+sinθ4=0\sin { { \theta }_{ 1 } } +\sin { { \theta }_{ 2 } } +\sin { { \theta }_{ 3 } } +\sin { { \theta }_{ 4 } } =0 and i know θ1+θ2+θ3+θ4=2π{ \theta }_{ 1 }+{ \theta }_{ 2 }+{ \theta }_{ 3 }+{ \theta }_{ 4 }=2\pi if that cyclic quadrilateral were to be a square,,then i would have the sum of any two angles as π\quad \pi ..but from the two conditions we just got.. 2cosθ1+θ22cosθ1θ22=(2)cosθ3+θ42cosθ3θ422\cos { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 1 }-{ \theta }_{ 2 } }{ 2 } } =(-2)\cos { \frac { { \theta }_{ 3 }+{ \theta }_{ 4 } }{ 2 } } \cos { \frac { { \theta }_{ 3 }-{ \theta }_{ 4 } }{ 2 } }

2sinθ1+θ22cosθ1θ22=(2)sinθ1+θ22cosθ3θ422\sin { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 1 }-{ \theta }_{ 2 } }{ 2 } } =(-2)\sin { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 3 }-{ \theta }_{ 4 } }{ 2 } } we get tanθ1+θ222=1{ \tan { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } }^{ 2 }=1 if it were a square i should get the sum of any two angles as π\pi but that doesn't satisfy the condition we just ended on..so..i conclude the points don't have to make a square...

this is my thinking....i might be wrong..if so,then i would be glad to know the flaw in my approach...

shriya mandarapu - 5 years, 5 months ago

Log in to reply

z1,z2,z3,z_{1}, z_{2}, z_{3}, and z4z_{4} are vertices of a rectangle ABCD. Here is the proof(rigorous):

In general, we can prove it for any z=a(>0)|z| = a(>0) rather than 1.

First, you should know that the value of p+q2=2p2(1+cosθ)|p+q|^2 = 2|p^2|(1 + \cos \theta), where θ\theta is theta is the angle that complex numbers pp and qq lying on a circle centered about origin subtend at origin.

Let α\alpha be the angle between lines OA and OB, and β\beta be the angle between lines OC and OD,

z1+z2=(z3+z4)z_{1}+ z_{2} = -(z_{3}+z_{4})

Taking modulus and squaring, we get:

2z2(1+cosα)=2z2(1+cosβ)2|z^2|(1+ \cos \alpha|) = 2|z^2| (1 + \cos \beta)

cosα=cosβ\cos \alpha = \cos \beta , or α=2πβ\alpha = 2 \pi - \beta (not possible)

Similarly, if γ\gamma and δ\delta are angles between OAOA and ODOD , and OBOB and OCOC respectively, we can prove that

γ=δ\gamma = \delta

Now, as α,β\alpha, \beta, and γ,δ\gamma, \delta are pairs of vertically opposite angles, we get that

Lines AOC , BOD are straight, i.e. the diagonals intersect at O.

Thus, all angles of the quadrilateral are angle in semicircle, i.e. 9090^{\circ} , and the quadrilateral is rectangle.

Also, it is not necessiraly a square. As a counter example, z,zˉ,z,zˉz , \bar{z},-z , -\bar{z} always satisfy the given relations, but don't necessarily form a square.

jatin yadav - 5 years, 5 months ago

Log in to reply

i think it's not possible .if condition 1 is true ,then 2 must be false and vice-verse

Mamun Uddin - 5 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...