System Of Complex Equations - Add A Constraint Corrected

Update: Abbreviated solution added.


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,z1z2+z2z3+z3z4+z4z5+z5z1=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, \\ z_1 z_2 + z_2 z_3 +z_3z_4 + z_4 z_5 + z_5 z_1 = 0 \\ \end{cases}

What can we conclude about z1,z2,z3,z4,z5 z_1, z_2, z_3, z_4, z_5 ?


You may refer to System Of Complex Equations, in which Mursalin and Prakhar showed that if only the first 2 conditions are true, then we can't really conclude too much. Other than a regular pentagon, the 5 points could be that of an equilateral triangle along with a diameter, or even 5 'somewhat random' points on the circle (no easy description).

In Add A Constraint, Pranav demonstrated a way of analyzing these conditions using the Argand diagram.

How does adding in the third condition help us?

Note by Calvin Lin
5 years, 6 months ago

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Hm, I see that people are liking and resharing this, but not making a comment. Can you vote up this comment if you are trying this problem, but haven't made any progress?

Calvin Lin Staff - 5 years, 6 months ago

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@Calvin Lin , would you kindly post a solution?

Mursalin Habib - 5 years, 5 months ago

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WLOG, we may assume that z1=1 z_1 = 1 (otherwise divide throughout by z1z_1 ). We want to show that zi z_i must be a fifth root of unity. We solve the system of equations

{z2+z5=1z3z4(1+z3)z2+(1+z4)z5=z3z4 \begin{cases} z_2 &+ & z_5 &= -1 - z_3 - z_4 \\ (1 +z_3) z_2 &+ & ( 1 + z_4) z_5 &= - z_3z_4 \\ \end{cases}

Show that if z3=z4 z_3 = z_4 , then we don't get any solutions. Otherwise, we get the solutions

z2=(1+z4)2+z3z3z4,z5=(1+z3)2+z4z4z3 z_2 = \frac{ ( 1 + z_4)^2 + z_3} { z_3 - z_4} , z_5 = \frac{ ( 1 + z_3)^2 + z_4 } { z_4 - z_3}

Use the fact that z2=z5=1 |z_2 | = |z_5| = 1 to conclude that z3z4=1 z_3 z_4 = 1 and also that 1+z3+z32+z33+z34=1 1 + z_3 + z_3^2 + z_3 ^3 + z_3 ^4 = 1 . Hence, z3z_3 must be a fifth root of unity. Substitute this back in, and we get that the zi=ωi1z_i = \omega^{i-1} for some root of unity.

Calvin Lin Staff - 5 years, 5 months ago

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I found a solution set but I don't see how to prove that this is the only possible one.

For simplicity sake, let z1=1z_1=1 (we may as well take z1z_1 as a random complex number on z=1|z|=1). Consider the roots of z5=1z^5=1. They are: 1,α,α2,α3,α41,\alpha,\alpha^2,\alpha^3,\alpha^4 where α=eiπ/5\alpha=e^{i\pi/5}. Notice that sum of the roots is zero so we may say that z2=αz_2=\alpha, z3=α2z_3=\alpha^2, z4=α3z_4=\alpha^3 and z5=α4z_5=\alpha^4. See that this set also satisfies the third equation. It can be concluded that the complex numbers lie on the vertices of a regular pentagon.

Pranav Arora - 5 years, 6 months ago

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the complex numbers are the fifth roots of unity

Sharath Aravind - 5 years, 5 months ago

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I think these should be 5th roots of unity

Saurabh Jain - 5 years, 4 months ago

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