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System Of Complex Equations - Add A Constraint Corrected

Update: Abbreviated solution added.


Suppose that \(z_1, z_2, z_3, z_4, z_5 \) are complex numbers which satisfy

\[ \begin{cases} |z_1| = |z_2| = |z_3| = |z_4| = |z_5| =1, \\ z_1 + z_2 + z_3 + z_4 + z_5 = 0, \\ z_1 z_2 + z_2 z_3 +z_3z_4 + z_4 z_5 + z_5 z_1 = 0 \\ \end{cases} \]

What can we conclude about \( z_1, z_2, z_3, z_4, z_5 \)?


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
2 years, 8 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 · 2 years, 8 months ago

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@Calvin Lin @Calvin Lin , would you kindly post a solution? Mursalin Habib · 2 years, 8 months ago

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

\[ \begin{cases} z_2 &+ & z_5 &= -1 - z_3 - z_4 \\ (1 +z_3) z_2 &+ & ( 1 + z_4) z_5 &= - z_3z_4 \\ \end{cases} \]

Show that if \( z_3 = z_4 \), then we don't get any solutions. Otherwise, we get the solutions

\[ z_2 = \frac{ ( 1 + z_4)^2 + z_3} { z_3 - z_4} , z_5 = \frac{ ( 1 + z_3)^2 + z_4 } { z_4 - z_3} \]

Use the fact that \( |z_2 | = |z_5| = 1 \) to conclude that \( z_3 z_4 = 1 \) and also that \( 1 + z_3 + z_3^2 + z_3 ^3 + z_3 ^4 = 1 \). Hence, \(z_3 \) must be a fifth root of unity. Substitute this back in, and we get that the \(z_i = \omega^{i-1} \) for some root of unity. Calvin Lin Staff · 2 years, 7 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 \(z_1=1\) (we may as well take \(z_1\) as a random complex number on \(|z|=1\)). Consider the roots of \(z^5=1\). They are: \(1,\alpha,\alpha^2,\alpha^3,\alpha^4\) where \(\alpha=e^{i\pi/5}\). Notice that sum of the roots is zero so we may say that \(z_2=\alpha\), \(z_3=\alpha^2\), \(z_4=\alpha^3\) and \(z_5=\alpha^4\). 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 · 2 years, 8 months ago

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I think these should be 5th roots of unity Saurabh Jain · 2 years, 7 months ago

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the complex numbers are the fifth roots of unity Sharath Aravind · 2 years, 8 months ago

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