Roots of Unity

The square roots of unity	1 and −1
The cube roots of unity	1 and
The quartic roots of unity	1, −1, i and −i

Page Contents

1. Roots_of_a_Complex_Number:_Theorem
2. Roots_of_Unity
3. Square Roots
4. Cube Roots
5. Quartic Roots of Unity

Polynomials have exactly as many roots as their degree, so, for example, there are exactly 2 square roots of a number, although they might be the same value (multiplicities). Naturally, equations do not need to have to be a natural number; they could be fractions. However, negative numbers raised to fractional powers and rational numbers might be unsolvable.

Other Pages:

Complex Number Arithmetic (Notes)

Roots of a Complex Number: Theorem

We are mainly interested in the roots of unity; however, we begin by being general and presenting a theorem.

The cyclotomic equation is:

[1.01]

Where [1.02], n=2,3...  and in the complex plane.

And because it is not surprisingly, cyclotomic means 'related to the roots of unity' and often refers to the equation:

[1.03]

The theorem is that the roots of are:

[1.04]

Where z and a are complex numbers, l=0,1,2, ... ,n-1, and −π<φ<π. We are interested, here, with z=1

Roots of Unity

So when z=1, φ=0 in [1.02] So from [1.04], we find that the roots of zn=1, where |z|=1, are:

[1.05]

Where n=2,3... and l=0,1,2, ... ,n-1

 

Comment

Because one of the roots of , and because 1 does not have a complex part, all the roots must make −1 when multiplied together (Viete's Theorem) so each root  must be the conjugate of another (or equal to 1); that is when multiplied the complex part must disappear. Again, from Viete's Theorem, the sum of the roots must be zero, because the coefficient of xn−1 in [1.02] is zero.

We note:

If a+b·i is a root, then so is a−b·i. 

The number of roots is n. So, in the complex plain, the circle, of radius 1 (modulus 1), is divided into n equal parts.

Square Roots of Unity

The square roots of unity are 1 and −1. We use this fact to discover square roots. For instance, √(4) produces a principle root of 2. We find the the two roots by multiplying 2 by the square roots of unity. The roots are therefore:
2·(1), and
2 ·(−1)
Hence √(4)=2 and −2
[Incidentally, the factors of 4 are 2·2 and (−2)·(−2). ]
Another way to discover the square roots of unity is from this equation:
x²=1, or
x²−1=0, and factorising gives us:
(x−1)(x+1)=0
Another way is to use de Moivre's Theorem. That is, we use the theorem 1.05

We wish to find the 2 roots of (1)^1/2
|a|=1
φ=0

n=2
The 2 roots of 1 are from the theorem 1.05:
_________________________

Letter l=0:
φsub>0=0+2 ·π /2·0=0
Root₀=1 ·cos(0)+1 · i sin(0)
Root₀=1
(As a check) Root₀²=1
_________________________________

Letter l=1:
φsub>1=0+2 ·π/2·1=π
Root₁=1 ·cos(π)+1 · i sin(π)
Root₁= – 1

• (As a check) Root₁²=1
• Also the sum of the roots is zero, as required.
• Also, the number of roots is 2, so the circle is divided into 2 angles, 0 and -π
  

Cube Roots of Unity

There are three cube roots for a number. We can discover the cube root of a number by multiplying its principle value by the cube roots of unity. The cube roots of unity are (one real and two imaginary):

1. 1,
2. −0.5+√(3)/2i
3. −0.5−√(3)/2i

So in seeking the ∛(8), we multiply the principle root, 2 by the three cubic roots of unity:

1. 2 ·1=2,
2. 2 ·(−0.5+√(3)i/2)=−1+√(3)
3. 2 ·(−0.5−√(3)i/2)=−1−√(3)

By solving:

x³=1, we find:

x³−1=0, and by taking out the factor (x−1):

(x−1)(x²+x+1)=0, one root is 1, and the others are the roots of x²+x+1=0, which we can find using the quadratic equation formula:

Another way is to use de Moivre's Theorem, that is using the theorem 1.05

While de Moivre's Theorem is harder to use than solving the equation, as we progress up the degrees of root, we find de Moivre to be the easier option.

We wish to find the 3 roots of (1)^1/3
a=1

|a|=1

phi=0

n=3

The 3 roots of 1 are:
_________________________

Letter l=0

φsub>0=0+2 ·π/3·0=0

Root₀=1 ·cos(0)+1 · i sin(0)

Root₀=1

(As a check) Root₀³=1
_________________________________

Letter l=1
φsub>1=0+2 ·π/3 ·1=2π/3
Root₁=1 ·cos(2π/3)+1 · i sin(2π/3)
Root₁= – 1/2+√(3)i /2

(As a check) Root₁³=1

_________________________________

Letter l=2

φspan class="style5">2=0+2 ·π/3 ·2=4π/3
Root₂=1 ·cos(4π/3)+1 · i sin(4π/3)
Root₂= – 1/2– √(3)i /2

• (As a check) Root₂³=1
• Sum of the roots=0, as required
• There are 3 roots, so the circle is divided into three equal angles of 2π//3
  

Quartic Roots of Unity

While we can still use the equation:
x⁴=1
x⁴−1=0
(x−1)(x³+x²+x+1)=0

We would have to solve the cubic equation. In this case, because de Moivre is easier, we will use de Moivre, that is, 1.05:

We wish to find the 4 roots of (1)^1/4

a=1
|a|=1
φ=atan=0
n=4
The 4 roots of 1 are:
________________________

Letter l=0.
φ₀=0+2 · π/4·0=0
Root₀=1 ·cos(0)+1 · i sin(0)
Root₀=1
(As a check) Root₀⁴=1
________________________________

Letter l=1
φ₁=0+2 ·π/4 ·1=π/2
Root₁=1 ·cos(π/2)+1 · i sin(π/2)
Root₁=i
(As a check) Root₁⁴=1
________________________________

Letter l=2
φ₂=0+2 ·π/4 ·2=π
Root₂=1 ·cos(π)+1 · i sin(π)
Root₂= – 1
(As a check) Root₂⁴=1
________________________________

Letter l=3
φ₃=0+2 ·π/4 ·3=3π/4
Root₃=1 ·cos(3π/4)+1 · i sin(3π/4)
Root₃= – i
(As a check) Root₃⁴=1
If we seek ∜(16), we need to multiply the principle root, 2, by the quartic roots of unity:

(−1) ·2=−2

1 ·2    =2

i ·2     =2i

(− ·2)=−2i

Showing the 4th Roots of 16 are −2, 2, 2i, −2i

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Faster Arithmetic - by Ken Ward