# Thread: [SOLVED] Complex number solution

1. ## [SOLVED] Complex number solution

Four points $\displaystyle z_1, z_2, z_3, z_4$ satisfy the conditions
$\displaystyle z_1 + z_2 + z_3 + z_4 = 0$

$\displaystyle |z_1| = |z_2| = |z_3| = |z_4| = 1$

Show that the points lie either at the vertices of a square inscribed in the unit circle or else coincide in pairs.
(I've been taking the z's to be points in the Argand plain, but I suppose you could simply think of the z's as (x, y) points.)

I was able to solve a similar problem involving 3 points and an inscribed equilateral triangle, but I've only been able to go so far with the 4 point problem.

Here's what I've got so far, and you all can redirect me or give me clue how to continue as you will.

Since
$\displaystyle |z_1| = |z_2| = |z_3| = |z_4| = 1$
all points lie on the unit circle, and thus are of the form $\displaystyle z_n = e^{i \theta_n}$.

Without loss of generality I will take $\displaystyle z_1 = 1$.

Thus the remaining condition becomes:
$\displaystyle e^{i \theta _2} + e^{i \theta _3} + e^{i \theta _4} = -1$

I'm rather stuck at this point. I used a similar approach to the three point problem and it worked nicely. Obviously it isn't working here because I've got that extra variable playing around. Is there, perhaps, a different approach that would work for both problems, or is there another simplifying step I'm missing for the 4 point problem?

-Dan

2. Four points $\displaystyle z_1, z_2, z_3, z_4$ satisfy the conditions
$\displaystyle z_1 + z_2 + z_3 + z_4 = 0$

$\displaystyle |z_1| = |z_2| = |z_3| = |z_4| = 1$

Show that the points lie either at the vertices of a square inscribed in the unit circle or else coincide in pairs.
The problem relates to four points on the unit circle with centroid coincident with the centre of the circle.

As such, what you are asked to show is false. Let the points be the vertices of any rectangle, but not a square, inscribed in the circle. Then the specified properties hold for this configuration, but it is not a square nor do any of the points coincide.

RonL

3. Originally Posted by CaptainBlack
The problem relates to four points on the unit circle with centroid coincident with the centre of the circle.

As such, what you are asked to show is false. Let the points be the vertices of any rectangle, but not a square, inscribed in the circle. Then the specified properties hold for this configuration, but it is not a square nor do any of the points coincide.

RonL
You are correct. Apparently my problem was that I was trying to "force a square to fit into a circle." (If you don't mind the pun, the comment is actually literal.) It would seem that my choice of $\displaystyle z_1 = 1$ was obscuring a more logical way to look at the problem. Thank you!

-Dan