Find the nth roots of unity.

Let me rephrase this question slightly:
 

"Find all the roots of the equation x^n - 1 = 0."

We know by the fundamental theorem of algebra that an nth degree polynomial has exactly n roots. So the excersice has now been reduced to something as simple as: can you find n different numbers  (call them x) such that  x^n = 1.

Well we know 1 works. Let us call it eoiπ from now on. We still need to find the (n-1) other roots. The key to this is using the fact that 1 = eoiπ, e2, e4, ... So long as our number when raised by n goes to any one of these numbers, we are done. Well, we can see e2iπ/n does the job, and we can also see e4/n does the job, so more generally e2riπ/n does the job for all positive integer r. Now we just need to find n of these numbers that are actually distinct (recall that there are infinitely many different ways of writing a number depending on how you write its argument, so while two numbers may be written differently they will actually be the same).
But fear not! If we look at e2riπ/n for 0 <= r < n, and r an integer, these are all distinct! (If they were not distinct then we could prove that eix = 1 for an x such that 0 < x < 2π, which is false).

ER
Answered by Edison R. Further Mathematics tutor

8245 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Show that the sum from 1 to n of 1/(2n+1)(2n-1) is equal to n/(2n+1) by Induction


Given that α= 1+3i is a root of the equation z^3 - pz^2 + 18z - q = 0 where p and q are real, find the other roots, then p and q.


I don't understand how proof by mathematical induction works, can you help?


Find the general solution to the differential equation: d^2y/dx^2 - 8 dy/dx +16y = 2x


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning