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 
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.
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Complex Number Arithmetic (Notes)
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, ... ,n1, and −π<φ<π. We are interested, here, with z=1
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, ... ,n1
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.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:
x^{2}=1, or
x^{2}−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
Root_{0}=1 ·cos(0)+1 · i sin(0)
Root_{0}=1
(As a check) Root_{0}^{2}=1
_________________________________
Letter l=1:
φsub>1=0+2 ·π/2·1=π
Root_{1}=1 ·cos(π)+1 · i sin(π)
Root_{1}= – 1
So in seeking the ∛(8), we multiply the principle root, 2 by the three cubic roots of unity:
By solving:
x^{3}=1, we find:x^{3}−1=0, and by taking out the factor (x−1):
(x−1)(x^{2}+x+1)=0, one root is 1, and the others are the roots of x^{2}+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
Root_{0}=1 ·cos(0)+1 · i sin(0)
Root_{0}=1
(As a check) Root_{0}^{3}=1
_________________________________
Letter l=1
φsub>1=0+2 ·π/3 ·1=2π/3
Root_{1}=1 ·cos(2π/3)+1 · i sin(2π/3)
Root_{1}= –
1/2+√(3)i /2
(As a check) Root_{1}^{3}=1
_________________________________
Letter l=2
φspan class="style5">2=0+2 ·π/3
·2=4π/3
Root_{2}=1 ·cos(4π/3)+1 · i sin(4π/3)
Root_{2}= –
1/2– √(3)i /2
While we can still use the equation:
x^{4}=1
x^{4}−1=0
(x−1)(x^{3}+x^{2}+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.
φ_{0}=0+2 · π/4·0=0
Root_{0}=1 ·cos(0)+1 · i sin(0)
Root_{0}=1
(As a check) Root_{0}^{4}=1
________________________________
Letter l=1
φ_{1}=0+2 ·π/4 ·1=π/2
Root_{1}=1 ·cos(π/2)+1 · i sin(π/2)
Root_{1}=i
(As a check) Root_{1}^{4}=1
________________________________
Letter l=2
φ_{2}=0+2 ·π/4 ·2=π
Root_{2}=1 ·cos(π)+1 · i sin(π)
Root_{2}= – 1
(As a check) Root_{2}^{4}=1
________________________________
Letter l=3
φ_{3}=0+2 ·π/4 ·3=3π/4
Root_{3}=1 ·cos(3π/4)+1 · i sin(3π/4)
Root_{3}= – i
(As a check) Root_{3}^{4}=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
