# System Of Complex Equations - Add A Constraint Corrected

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
6 years, 12 months ago

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I think these should be 5th roots of unity

- 6 years, 10 months ago

the complex numbers are the fifth roots of unity

- 6 years, 11 months ago

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?

Staff - 6 years, 12 months ago

@Calvin Lin , would you kindly post a solution?

- 6 years, 11 months ago

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.

Staff - 6 years, 11 months ago

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.

- 6 years, 12 months ago