Online Course Discussion Forum

Roots of Unity

 
 
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Roots of Unity
by James Shi - Monday, 29 June 2020, 9:39 PM
 

While I was doing my homework for Math Challenge II-B, on Chapter 4, problems 4.26 and 4.29, I realized that it would be much easier if I could say that the roots of any equation z^n = m are evenly distributed on the unit circle in the complex plane. This means, that, when connecting these roots, a regular n-sided polygon would be formed on the complex plane. I was wondering if this was true and if it is, would the explanation for why it is true be the same as the explanation for the roots of unity? I was also wondering if all n of these roots would have the same modulus. Thank you.

 
Picture of David Reynoso
Re: Roots of Unity
by David Reynoso - Wednesday, 1 July 2020, 5:45 PM
 
If we write a complex number in polar form, that is, $z = r (\cos(\theta) + i \sin(\theta))$, de Moivre's Theorem says that $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$. Thus, the solutions to the equation $z^n = m$ ($z = m (\cos(2\pi) + i \sin(2\pi))$) are precisely $$z = m^{\frac{1}{n}} \left(\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right),$$ for $k = 0, \dots, n-1$.


This is exactly why the roots of unity are evenly spaced around the unit circle, and the roots of the equation $z^n = m$ are evenly spaced around a circle with radius $m^{\frac{1}{n}}$.