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## Homework Statement

Hey, I am attempting to fully factorize z[itex]^{n}[/itex]-1=0 for all integers of n where n does not equal zero, and where z is a complex number in the form a+bi. The question asks to first factorize the equation when n=3,4,5. I know how to factorize when n=3 and 4, but I get stuck at n=5. I don't know where to begin.

(Below is just an explanation of what the question is referring to)

This question is from a set of finding patterns in complex numbers. We are asked to plot all solutions of z[itex]^{n}[/itex]-1=0 using DE MOIVRE'S THEOREM on an Argand diagram or a unit circle. This step was easy; it was noticed that for any integer value of n, there are n roots of unity such that when all roots are plotted on an Argand diagram and I connect all adjacent roots, I form a regular n-gon (for example, if n=5, then there are 5 roots of unity. When adjacent roots are connected, a regular pentagon is formed). (The connecting of the adjacent roots was NOT asked in the question. I just told you guys this to help you hopefully visualize what I am talking about).

The question then asks to choose ONE root and form line segments from that one root to ALL other roots (including roots that are adjacent to that one root). Then, we had to find the distances of the line segments that were just formed.

To cut to the chase of what I found in the end, I basically noticed that when all distances of line segments (for any value of n) are multiplied together, the result is exactly n.

**However, I do not know how to prove this conjecture, so any help would be appreciated.**

## Homework Equations

de Moivre's theorem, binomial expansion (not sure about this)

## The Attempt at a Solution

I do not know where to begin. Apparently, the factorization of z[itex]^{n}[/itex]-1=0 has to do with the distances of the line segments that I talked about up there^ (I am not sure of this though, the two may even have no relation at all).

My first reaction as to how to factorize the equation is to substitute z for a+bi. Therefore I would have (a+bi) [itex]^{n}[/itex]-1=0, and I could use binomial expansion from there, but I don't see how that would help me in any way.

**TL;DR How do I factorize z[itex]^{n}[/itex]-1=0 where z is a complex number and n is any integer other than zero, without the use of de Moivre's theorem?**