Suppose that $z_1, z_2, z_3, z_4, z_5$ are complex numbers which satisfy

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

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).

How does adding in the third condition help us?

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## Comments

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TopNewestRearranging the third equation $z_2(z_1+z_3)=-z_5(z_4+z_1)$ and taking modulus on both the sides, one obtains: $|z_1+z_3|=|z_1+z_4|$. This means that the distance of $z_1$ from $z_3$ and $z_4$ is same. Join the centre with $z_1$. Call this line OA. Now, OA divides the circle into two semicircles. $z_3$ and $z_4$ obviously cannot lie on the same semicircle. They lie symmetrically about OA on different semicircles. Further divide the semicircle in four quadrants. Here is an image to go with my explanation:

Notice that $z_3$ and $z_4$ cannot lie in 1st and 4th quadrant. If they did, $z_3+z_4$ would be along $z_1$ and $|z_1|+|z_3+z_4|>1$ but since $|z_2+z_5|<1$, it would be impossible to satisfy the second condition. So, $z_3$ and $z_4$ must lie in 3rd and 2nd quadrants.

By a similar argument, $z_2$ and $z_5$ must lie in 3rd and 2nd quadrant. They too must lie symmetrically about OA and on different sections of semicircles.

Hence, it is a pentagon but not necessarily a regular one.

I am not sure if this is correct, though.

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This is a great piece of analysis!

I just realized that I screwed up the 3rd equation, and missed out the $z_3 z_4$ term. Let me post a new note to reflect that.

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Ah!

Thanks Calvin! :)

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They may be 5 th roots of unity

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They may be vertices of regular Pentagon inscribed the in a unit circle

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