Complex Numbers

Over the set of ordered pairs of reals numbers

we define the following two operations

This set is a field: the pair (0,0) is the neuter element for addition, (1,0) is the neuter element for multiplication. If (a,b) ≠ (0,0), then

Thus each element posses a multiplicative inverse as well.

The set of pairs (a,0), a ∈ ℝ, is the set of reals numbers, because the mapping

is an injection, such that f(r + r') = f(r) + f(r') and f(r ⋅ r') = f(r) ⋅ f(r') ∀r,r' ∈ ℝ.

Definition 4.13.1 A complex number is an ordered pair of real numbers, i.e. an element of ℂ = ℝ x ℝ. With respect to the operations defined above, ℂ is a field, containing the real field.

If we indicate the pair (0,1) with the symbol i, we can write

The element i ∈ ℂ is known as imaginary unit. Using this notation we have

To add two complex numbers together we just add up their real and imaginary parts.

We multiply two complex numbers together as we multiply real numbers together, and we make use of the fact that i² = −1.
We can thus think of ℂ much like we think of ℝ² (i.e., we think of the complex number a + bi ∈ ℂ as the vector (a, b) ∈ ℝ²), but with a multiplication operation that we do not have on ℝ².

Given z = x + iy, we call complex conjugate of z and we denote it by z̄ the number

for example,

Complex conjugates give us another way to interpret reciprocals. You can easily verify that a complex number z = x + yi times its conjugate x – yi is the square of its absolute value |z|².

Therefore, 1/z is the conjugate of z divided by the square of its absolute value |z|².

Division of complex numbers

Consider 11 + 6i divided by 2 + 3i: 11 + 6i / 2 + 3i; this ratio can be written as a complex number in the form a + bi if we multiply numnerator and denominator by 2 − 3i

Given z, z₁ and z₂ the following identities hold

The real number zz̄ is known as norm of the complex number z and its square root $\sqrt{z\overline{z}}=\sqrt{x^2+y^2}$ is knwon as modulus of z and indicated by |z|.

Definition 4.13.2 For any complex number z₁ = a + bi and z₂ = c + di

Proof. Suppose z₁ = a + bi and z₂ = c + di are complex numbers. Then

The following properties hold

The last identity is known as Triangle Inequality. If we identify the complex number z = x + iy as the point of the plane P(x,y) the modulus |z| represents the lenght of the segment OP.

Geometrically, the right-hand part of the triangle inequality states that the sum of the lengths of any two sides of a triangle is greater than the length of the remaining side.

Complex plane; Trigonometric and polar forms of complex numbers

Much like we think of ℝ as a line, we can think of ℂ as a plane, called the complex plane. The set of real numbers takes the place of the x-axis (which we call the real axis) and the set of imaginary numbers takes the place of the y-axis (which we called the imaginary axis), so that the number a + bi has coordinates (a, b) on that plane, as in the following Figure

The number $r=\sqrt{x^2+y^2}$ corresponds to the modulus or absolute value of x + yi and the angle θ is called the argument or amplitude of x + yi.
With the similarities with ℝ² in mind, the absolute value is the norm or magnitude of a complex number a + bi, denoted as

which is simply the length of the associated vector (x, y) (i.e., it is the distance between x + yi and the origin in the complex plane).

Let θ be the angle subtended between the positive real axis and the line segment OP, and let r² = x² + y², we have the following relations

Thus we can represent the complex number x + iy in the following trigonometric form

The angle θ is known as the argument of the complex number z, the number r corresponds to the modulus of z: r = |z| = (x² + y²)^1/2. The modulus, r, is unique, but the angle θ is not unique since there are many angles in standard position with P on their terminal side.

Euler's identity provides yet another convenient way to represent complex numbers. Indeed, if z = r(cos θ + i sin θ) is a complex number, then since e^iθ = cos θ + i sin θ), we obtain the exponential representation or polar form

When we multiply two complex numbers, geometrically what happens is that we multiply their distances to their origin and you add thei angles.

De Moivre’s formula

If z and z' have the following representation z = r[cosθ + i sinθ] and z' = r'(cosθ' + i sinθ') then as it can be easily verified

the product of two complex numbers in trigonometric form is equal to a complex number having as argument the sum of the arguments and as modulus the product of moduli. The trigonometric form of a complex number is particularly useful to calculate the powers; for every integer n > 0, we have the following relation known as De Moivre Formula, valid for every integer n and every θ ∈ ℝ

it results also

n-th roots

Proposition 4.13.3. Let n a positive integer greater than 1, and let u and ζ two complex numbers. If

there are precisely n complex roots of equation 4.13.2, and ζ is known as an nth root of u.

Proof. Let

since ζⁿ = u, we can deduce that r = ρⁿ, i.e. ρ = r^1/n, thus ρ is the unique real positive number such that the n-th power is r. Moreover from zⁿ = u we can deduce as well that: cos nφ = cos θ and sin nφ = sin θ, then nφ = θ + 2kπ, i.e.

For an arbitrary integer k ∈ ℤ, we obtain the complex number

For k = 0, 1, n−1 the ζ_k are all distinct and represent the n-th roots of u.  □

For u = 0 there is a unique root which is zero. For u = 1, we have the n-th roots of unity.

Definition 4.13.4 The roots in ℂ of the polynomial xⁿ − 1 are called the complex nth roots of unity.

If ζ is an nth root of unity, and if n is the smallest positive integer for which ζⁿ = 1, we say that ζ is a primitive nth root of unity. Thus, i is an 8th root of unity, but it is not a primitive 8th root of unity; however, i is a primitive 4th root of unity.

For u = 1, we have the nth roots of unity: 1 = cos(nθ) + i sin(nθ), and hence cos(nθ) = 1 and sin(nθ) = 0. The only way this can be verified is when nθ is an integer multiple of 2π, or equivalently θ = 2kπ/n for some integer k. So the nth roots of unity are cos(2kπ/n) + i sin(2kπ/n), for k = 0,...,n − 1. Let

Then every n-th root of the unit is equal by De Moivre formula, to the k-th power of ζ, i.e.

All nth roots of a complex number u may be obtained by multiplying one of these values by all the nth roots of unity. Indeed let z₀ be one of the values of the n-th root of the number u, i.e., z₀ⁿ = u and let ζ be an arbitrary value of the n-th root of unity, that is ζⁿ = 1. Then

In the complex plane, the n-th roots of unity correspond to the n vertices of a regular n-sided polygon inscribed inside the unit circle. We view the vertices of the pentagon as the five 5th roots of unity, that is, the five roots of ζ⁵ − 1 = 0.

Example 4.13.5. Find the roots of x⁴ + 1.

Solution. We start by finding a particular solution: −1 = e^iπ and taking the 4th root (−1)^1/4 = e^{i π/4} = √2/2 + i √2/2. We know multiply this particular solution ω = √2/2 + i √2/2 = 1/√2 + i /√2, by the roots of unity. Notice that the primitive 4th root of unity” is just “i” or “−i”. We take i. Thus the four roots are

Thus the roots are (1 ± i)/√2 and (−1 ± i)/√2.  ■

Proposition 4.13.3 states that the equation zⁿ = a (with a complex) has exactly n complex solutions. However in the real field the equation xⁿ +a can have one, two or no solutions (e.g. x² − 1 = 0, x³ − 1 = 0, x² + 1 = 0); Now we know that there are always n roots in ℂ, and eventually one or two of those roots are also in ℝ.
This result can be generalized to an arbitrary equation with degree n as follows:

Theorem 4.13.5. (Fundamental Theorem of algebra) A polynomial

with a_i ∈ ℂ has exactly n roots in ℂ (counting multiplicity).

Exercises

1. Find the roots of x³ − 1.

2. Calculate the module of the following complex numbers

   1. 1 + i + i² + i³ + i⁴ + i⁵

   2. (1 + i)²

   3. (1 + i)(1 − i)

3. Calculate $\sqrt[5]{1+i}$

4. Write in algebraic form (1 + i)⁷.

5. Calculate

   1. The 4-th roots of i.

   2. The primitive 6th roots of the unity

Solutions

1. Let a primitive root

   ω = cos(2π/3) + i sin(2π/3) 2

   Then the roots are

   $${\omega }^0=1,\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2},{\omega }^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$

   Both ω and ω are primitive cube roots of unity while 1 is obviously not.  ■

2. 1. |1 + i −1 + − i + 1 + i| = |1 + i| = √2

   2. (1 + i)² = (1 + 2i −1) = 2i, thus the module is (0 + 2²)^1/2 = 2.

   3. |(1 + i)(1 − i)| = |1 −i + i + 1| = |2| = 2.

3. The modulus of (1 + i) is √2 and the argument π/4. Then in exponential form

   √2 [cos (π/4 + 2kπ) + i sin (π/4 + 2kπ)]  for k = 0,1,2,3,4

   The 5th roots of unity are

   (2)^1/10 [cos (π/20 + 2kπ/5) + i sin (π/20 + 2kπ/5)]  for k = 0,1,2,3,4

4. We calculate the modulus and argument and then apply De Moivre law

   |1 + i| = √2    arg(1 + i) = π/4

   Thus 1 + i = √2e^iπ/4

   (1 + i)⁷ = (√2e^iπ/4)⁷ = (√2)⁷ e^{i 7π/4} = 8√2e^{i 7π/4}

   This can be rewritten as (1 + i)⁷ = (√2)⁷ e^{i 7π/4} = 8√2[e^{i8π/4 − π/4}] = 8√2[e^−π/4] = 8√2 [cos π/4 − i sin π/4] = 8√2 (1/√2 − i/√2) = 8 − 8i. ■

5. 1. i = r (cos θ + i sin θ) con θ = π/2, r = 1. The four roots are z_k = cos (θ + 2kπ/4) + i sin (θ + 2kπ/4), with k = 0,1,2,3. Replacing θ = π/2: z_k = cos (π/2 + kπ/2) + i sin (π/2 + kπ/2), with k = 0,1,2,3

   2. z_k = cos (2kπ/6) + i sin (2kπ/6), with k = 0,1,2,3,4,5. The primitive roots are those satisfying z^k ≠ 1, namely those for k = 1, k = 5.