Complex Roots of Unity
A root of unity, also known as a de Moivre number, is a complex number z which satisfies z n = 1, for some positive integer n.
Solving for the nth roots of unity
Note that Maple uses the uppercase letter I, rather than the lowercase letter i, to denote the imaginary unit: I 2 = −1 .
Since z n = 1 is a polynomial with complex coefficients and a degree of n, it must have exactly n complex roots according to the Fundamental Theorem of Algebra.
To solve for all the nth roots of unity, we will use de Moivre's Theorem: cosx+I⁢sinxn=cosn⁢x+I⁢sinn⁢x, where x is any complex number and n is any integer (in this particular case x will be any real number and n will be any positive integer).
First, convert the complex number z to its polar form: z=z⁢cos⁡θ+I⁢sin⁡θ, where z is the modulus of z and q is the angle between the positive real axis (Re) and the line segment joining the point z to the origin on the complex plane. Since z n = 1, it must be true that z = 1, and so the previous equation simply becomes z=cosθ+I⁢sinθ.
Also, converting the real number 1 = 1 +0⋅ I to polar form, we get 1=cos⁡2⁢π⁢k+I⁢sin⁡2⁢π⁢k for any integer k.
Now, z n = cosθ+I⁢sinθn = cos2⁢π⁢k+I⁢sin2⁢π⁢k = 1 and so using de Moivre's Theorem, this equation becomes z n = cosn⁢θ+I⁢sinn⁢θ= cos2⁢π⁢k+I⁢sin2⁢π⁢k = 1. From this form of the equation, we can see that n⁢θ = 2⁢π⁢k , or equivalently, ⁢θ = 2 π kn.
Therefore, the n th roots of unity can be expressed using the formula zk= cos2⁢π⁢kn+I⁢sin2⁢π⁢kn, for k = 0, 1, 2, ... , n−1.
Using Euler's formula: ⅇI⁢θ=cosθ+I⁢sinθ, we can write this formula for the n th roots of unity in its most common form: zk = e I 2⁢π⁢kn, for k = 0, 1, 2, ... , n−1.
When the nth roots of unity are plotted on the complex plane (with the real part [Re] on the horizontal axis and the imaginary part [Im] on the vertical axis), we can see that they all lie on the unit circle and form the vertices of a regular polygon with n sides and a circumradius of 1.
Degree of Polynomial, n
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