In a previous blog, I showed how de Moivre came up with his famous formula based on the work of Francois Viete. In today's blog, I will show how he provided a proof for the existence of the roots of unity.
In 1739, de Moivre used his formula to find the nth root for any complex number (a + bi).
Theorem 1: For any complex number, there are n distinct nth roots.
if a,b ≠ 0, then:

where:
0 ≤ k ≤ n-1
Proof:
(1) Let a,b be real numbers such that either a ≠ 0 or b ≠ 0
(2) We note first that:


(3) Since:

It is clear that:

And further that:

(4) Therefore, there exists φ such that [see here for review of cosine if needed]:

(5) Since cos2(φ) + sin2(φ) = 1 (see Corollary 2, here), we can see that:
sin2(φ) = 1 - cos2(φ) =

(6) So that we also have now:

(7) We can now conclude:

since:



(8) Using De Moivre's formula (see Theorem 1, here), we know that:


(9) From the basic properties of sin and cosine, we know that:
cos(φ + 2kπ) + isin(φ 2kπ) = cos(φ) + isin(φ)
(10) Multiplying √a2 + b2 to both sides gives us:



(11) Combining step #7 with step #10 gives us:

(12) Now, we know that each of the values are distinct for 0 ≤ k ≤ n-1 since none of the angles vary by a multiple of 2π.
QED
Corollary 1.1: Roots of Unity
There are n roots of unity such that:

where:
0 ≤ k ≤ n-1
Proof:
(1) Let a=1 and b=0
(2) Then a+bi = 1+0*i = 1
(3) Using Theorem 1 above, we have:

(4) Using step #4 and step #6 in Theorem 1 above, we have:

and

(5) So, φ = 0
(6) Since:

we are left with:

where:
0 ≤ k ≤ n-1
QED
Corollary 1.2: Roots of -1
There are n roots of unity such that:

where:
0 ≤ k ≤ n-1
Proof:
(1) Let a=-1 and b=0
(2) Then a+bi = -1+0*i = -1
(3) Using Theorem 1 above, we have:

(4) Using step #4 and step #6 in Theorem 1 above, we have:

and

(5) So, φ = π
(6) Since:

we are left with:


where:
0 ≤ k ≤ n-1
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001