The complex number
Prerequisiti di algebra lineare
Introduction
Addition and Multiplication Operations on \Bbb{R}^2
Let’s consider the set \Bbb{R}^2 of all ordered pairs of real numbers:
\Bbb{R}^2 = \{(a, b) \mid a, b \in \Bbb{R}\}
Two ordered pairs (a, b) and (c, d) are equal if and only if:
(a, b) = (c, d) \quad \Leftrightarrow \quad a = c \text{ and } b = d
We define two operations on \Bbb{R}^2:
Addition: Given two elements (a, b) and (c, d) of \Bbb{R}^2, their sum is the element (a+c, b+d) of \Bbb{R}^2 defined by: (a, b) + (c, d) = (a+c, b+d) \tag{1}
Multiplication: Given two elements (a, b) and (c, d) of \Bbb{R}^2, their product is the element (ac-bd, ad+bc) of \Bbb{R}^2 defined by: (a, b) \cdot (c, d) = (ac-bd, ad+bc) \tag{2}
Remarks:
The operations of addition and multiplication defined on \Bbb{R}^2 extend the usual operations of addition and multiplication between real numbers, by identifying a real number a with the element (a, 0) of \Bbb{R}^2.
With these operations, the set \Bbb{R}^2 forms a field, known as the field of complex numbers.
where the addition and multiplication operations in the right-hand sides of equations Equazione 1 and Equazione 2 are the standard operations defined in the field of real numbers \Bbb{R}.
The Field of Complex Numbers
Since the set
\Bbb{R}^\prime = \{ (a,0) \quad\vert\quad a \in \Bbb{R} \}
is contained in \Bbb{R}^2, and is easily identifiable (via a one-to-one correspondence) with \Bbb{R}, it is reasonable and useful to simply write a instead of (a,0) for each pair in \Bbb{R}^\prime. We also define the imaginary unit {\color{blue}\imath} as:
{\color{blue}\imath}\equiv (0,1),
and we can thus express every pair (a,b) \in \Bbb{R}^2 in the form:
(a,b) = (a,0) + (0,b) = a + (0,b) \cdot (1,0) = a + (b,0) \cdot (0,1) = a + {\color{blue}\imath}b.
With this notation, we observe that {\color{blue}\imath} satisfies the relation:
{\color{blue}\imath}^2 = -1.
We can now define the set of complex numbers.
Definition (Complex Numbers)
The set of complex numbers, denoted by \Bbb{C}, consists of pairs of real numbers on which the operations of addition and multiplication, defined in Equazione 1 and Equazione 2, have been introduced:
\Bbb{C} = \left( \Bbb{R}^2, +, \cdot \right).
The commonly used notation for a complex number z \in \Bbb{C} is:
z = a + {\color{blue}\imath}b, \qquad \text{with} \qquad a, b \in \Bbb{R},
where
a = \text{Re}\{z\} \quad \text{is called the real part of } z,
and
b = \text{Im}\{z\} \quad \text{is called the imaginary part of } z.
We note that \Bbb{R} \subset \Bbb{C}, since every real number a can be written as a = a + {\color{blue}\imath}\cdot 0. Moreover, the product of two complex numbers z_1 = a + {\color{blue}\imath}b and z_2 = c + {\color{blue}\imath}d follows the usual rules of polynomial multiplication. Indeed, the product z_1 \cdot z_2 can be expanded both algebraically:
(a + {\color{blue}\imath}b) \cdot (c + {\color{blue}\imath}d) = ac + {\color{blue}\imath}ad + {\color{blue}\imath}bc + {\color{blue}\imath}^2 bd = ac - bd + {\color{blue}\imath}(ad + bc),
and using the definition of multiplication in \Bbb{R}^2, yielding the same result.
The properties of the addition and multiplication operations introduced in \Bbb{C} show that \Bbb{C} is a commutative ring with a multiplicative identity.
We now aim to demonstrate that \Bbb{C} is also a field, meaning that every non-zero complex number has a multiplicative inverse. To this end, we introduce the notions of conjugate and modulus of a complex number.
Definition
Let z = a + {\color{blue}\imath}b. We define the conjugate of z as the complex number \overline{z} = a - {\color{blue}\imath}b.
Properties of the Conjugation Operation
Let z = a + {\color{blue}\imath}b \in \Bbb{C}. The following properties hold:
z + \overline{z} = 2a, from which it follows that \mathop{\mathrm{Re}}\left\{z\right\} = \dfrac{z + \overline{z}}{2}
z - \overline{z} = 2{\color{blue}\imath}b, from which it follows that \mathop{\mathrm{Im}}\left\{z\right\} = \dfrac{z - \overline{z}}{2{\color{blue}\imath}}
\overline{\left(\overline{z}\right)} = z
\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}
\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}
z = \overline{z} \quad \Longleftrightarrow \quad z \in \Bbb{R}
z \cdot \overline{z} = a^2 + b^2 is a non-negative real number.
For property 7, we introduce the following definition:
Definition of Modulus
The modulus of a complex number z is defined as:
|z| = \sqrt{z \cdot \overline{z}}.
Properties of the Modulus
Let z = a + {\color{blue}\imath}b \in \Bbb{C}. The following properties hold:
Modulus of the conjugate: |z| = |\overline{z}|.
Non-negativity: |z| \text{ is a non-negative real number and} |z| = 0 \quad \Longleftrightarrow \quad a^2 + b^2 = 0 \quad \Longleftrightarrow \quad a = b = 0 \quad \Longleftrightarrow \quad z = 0.
Multiplicative property: |z_1 \cdot z_2| = |z_1| \cdot |z_2|.
Triangle inequality: |z_1 + z_2| \leq |z_1| + |z_2|.
Geometric Interpretation of the Triangle Inequality
Complex numbers can be graphically represented as points in the Cartesian plane. For example, the number z = a + ib is represented by the point with coordinates (a, b). The origin (0, 0) represents the complex number 0, the point (1, 0) represents the complex number 1 = 1 + 0i, and the point (0, 1) represents the complex number i = 0 + 1i.
In the complex plane:
- Points on the x-axis correspond to real numbers (x, 0) \equiv x + 0i, and this axis is called the real axis.
- Points on the y-axis correspond to purely imaginary numbers (0, y) \equiv 0 + yi, and this axis is called the imaginary axis.
This graphical representation helps visualize the triangle inequality in the context of the complex plane.
Geometrically, the triangle inequality reflects the fact that in a triangle, the length of any side is always less than or equal to the sum of the lengths of the other two sides (see figure below).
From property 2 of the modulus, it follows that for any complex number z \neq 0, the number 1/|z| is well defined. Moreover, it is easy to verify that the inverse of z is given by:
z^{-1} = \frac{\overline{z}}{|z|^2}.
Example (Calculating the Inverse of a Complex Number)
Let’s consider a few examples to calculate the inverse of a complex number:
For z = 3 - 2i:
|z|^2 = 3^2 + (-2)^2 = 13, \quad \text{thus} \quad z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{3 + 2i}{13}.
For z = i:
|z|^2 = 1^2 = 1, \quad \text{thus} \quad z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{-i}{1} = -i.
For z = a, where a \in \mathbb{R}:
|z|^2 = a^2, \quad \text{thus} \quad z^{-1} = \frac{1}{a} \quad \text{for any} \quad a \neq 0.
These examples show how to calculate the inverse of complex numbers in various cases.
Referring to the last example, we note that the inverse of a real number, considered as an element of \mathbb{C}, coincides with that calculated as usual in \mathbb{R}. Therefore, the definition of the inverse in \mathbb{C} naturally extends to that of \mathbb{R}.
Problem
Now we aim to solve the following problems:
- Calculate 1 + {\color{blue}\imath}^{2000}.
- Find the solutions in \Bbb{C} of the equation z^n = w, with w \in \Bbb{C} fixed.
To tackle these problems, it is useful to introduce the trigonometric representation of a complex number.
Polar form of complex numbers
Let z be a complex number written in algebraic form z = a + {\color{blue}\imath}b with a, b \in \Bbb{R}. We define \rho as the modulus of z and \theta as the angle that the segment from the origin to the point (a, b) forms with the x-axis. These values \rho and \theta are called polar coordinates:
\rho = |z| \quad \text{(modulus of } z)
\theta = \mathop{\mathrm{arg}}(z) \quad \text{(argument of } z)
Using the properties of right triangles, we obtain:
a = \rho \cos \theta \quad \text{and} \quad b = \rho \sin \theta
or, equivalently:
\rho = \sqrt{a^2 + b^2}, \quad \theta = \arctan\left(\frac{b}{a}\right)
From this follows the polar representation of z:
z = a + {\color{blue}\imath}b = \rho \cos \theta + {\color{blue}\imath}\rho \sin \theta = \rho \left(\cos \theta + {\color{blue}\imath}\sin \theta\right)
This representation is illustrated in figure Figura 3.
Observation
It is important to note that the argument of z is defined up to integer multiples of 2\pi. In other words, varying the angle \theta by 2\pi does not change the argument of z. Keeping \rho constant and increasing \theta by 2\pi, you obtain the same complex number, as you complete a full rotation around the circle centered at (0,0) with radius \rho, thus returning to the starting point.
Moreover, since the functions \sin\theta and \cos\theta are 2\pi-periodic, we have that:
z = \rho\left(\cos\theta + {\color{blue}\imath}\sin\theta\right) = \rho\left(\cos(\theta + 2k\pi) + {\color{blue}\imath}\sin(\theta + 2k\pi)\right), \quad k \in \Bbb{Z}.
Observation
Let us consider the Taylor series expansions for the functions \mathrm{e}^{{\color{blue}\imath}\theta}, \cos \theta, and \sin \theta. We have:
\begin{aligned} \mathrm{e}^{{\color{blue}\imath}\theta} &= 1 + {\color{blue}\imath}\theta + \frac{({\color{blue}\imath}\theta)^2}{2!} + \frac{({\color{blue}\imath}\theta)^3}{3!} + \frac{({\color{blue}\imath}\theta)^4}{4!} + \frac{({\color{blue}\imath}\theta)^5}{5!} + \cdots \\ &= 1 + {\color{blue}\imath}\theta - \frac{\theta^2}{2!} - {\color{blue}\imath}\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + {\color{blue}\imath}\frac{\theta^5}{5!} + \cdots \\ &\quad + (-1)^k \frac{\theta^{2k}}{(2k)!} + {\color{blue}\imath}(-1)^k \frac{\theta^{2k+1}}{(2k+1)!} + \cdots \\ &= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots + (-1)^k \frac{\theta^{2k}}{(2k)!} + \cdots \right) \\ &\quad + {\color{blue}\imath}\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots + (-1)^k \frac{\theta^{2k+1}}{(2k+1)!} + \cdots \right) \\ &= \cos \theta + {\color{blue}\imath}\sin \theta. \end{aligned}
Therefore, the complex number \rho \left(\cos \theta + {\color{blue}\imath}\sin \theta\right) can be written as:
\rho \left(\cos \theta + {\color{blue}\imath}\sin \theta \right) = \rho \mathrm{e}^{{\color{blue}\imath}\theta}.
Product of Complex Numbers in Polar Form
For the product of complex numbers expressed in polar form, the following proposition is particularly useful and simplifies the calculation of powers.
Proposition
The modulus of the product of two complex numbers is equal to the product of their moduli, while the argument of the product is equal to the sum of the arguments of the two complex numbers.
Let z_1 and z_2 be two complex numbers. Then we have:
|z_1 \cdot z_2| = |z_1| \cdot |z_2|
and
\mathop{\mathrm{arg}}(z_1 \cdot z_2) = \mathop{\mathrm{arg}}(z_1) + \mathop{\mathrm{arg}}(z_2).
Proof
Consider the complex numbers z_1 and z_2 expressed in their trigonometric form:
z_1 = \rho (\cos \theta + {\color{blue}\imath}\sin \theta), \qquad z_2 = \sigma (\cos \alpha + {\color{blue}\imath}\sin \alpha).
The product z_1 \cdot z_2 is calculated as follows:
\begin{aligned} z_1 \cdot z_2 &= \rho \sigma (\cos \theta + {\color{blue}\imath}\sin \theta) (\cos \alpha + {\color{blue}\imath}\sin \alpha) \\ &= \rho \sigma \left[ (\cos \theta \cos \alpha - \sin \theta \sin \alpha) + {\color{blue}\imath}(\cos \alpha \sin \theta + \sin \alpha \cos \theta) \right] \\ &= \rho \sigma \left[ \cos (\theta + \alpha) + {\color{blue}\imath}\sin (\theta + \alpha) \right]. \end{aligned}
In the final step, we have used the trigonometric identities that express the sine and cosine of the sum of two angles in terms of the sines and cosines of the individual angles.
Observation
Assuming the properties of the exponential function are valid in the complex field as well, we can derive the proposition more directly. Consider the complex numbers in exponential form:
z_1 = \rho e^{{\color{blue}\imath}\theta}, \qquad z_2 = \sigma e^{{\color{blue}\imath}\alpha}.
The product of z_1 and z_2 is:
z_1 \cdot z_2 = \rho e^{{\color{blue}\imath}\theta} \cdot \sigma e^{{\color{blue}\imath}\alpha} = \rho \sigma e^{{\color{blue}\imath}(\theta + \alpha)}.
Thus, the modulus of the product is the product of the moduli, and the argument is the sum of the arguments, as stated in the previous proposition.
This proposition can also be extended to the calculation of the ratio of two complex numbers, as described in the following corollary.
Corollary
Let w = \frac{z_1}{z_2} \in \Bbb{C}. Then:
|w| = \frac{|z_1|}{|z_2|}
and
\mathop{\mathrm{arg}}(w) = \mathop{\mathrm{arg}}(z_1) - \mathop{\mathrm{arg}}(z_2).
Proof
We can write w as the product of z_1 and the inverse of z_2:
w = z_1 \cdot \frac{1}{z_2} = z_1 \cdot \frac{\overline{z_2}}{|z_2|^2}.
Applying the previous proposition, we obtain:
\begin{aligned} |w| &= \left| z_1 \cdot \frac{\overline{z_2}}{|z_2|^2} \right| \\ &= |z_1| \cdot \left| \frac{\overline{z_2}}{|z_2|^2} \right| \\ &= |z_1| \cdot \frac{|\overline{z_2}|}{|z_2|^2} \\ &= |z_1| \cdot \frac{|z_2|}{|z_2|^2} \\ &= \frac{|z_1|}{|z_2|}. \end{aligned}
As for the argument, we have:
\mathop{\mathrm{arg}}(w) = \mathop{\mathrm{arg}}\left(z_1 \cdot \frac{\overline{z_2}}{|z_2|^2} \right) = \mathop{\mathrm{arg}}(z_1) + \mathop{\mathrm{arg}}\left(\frac{\overline{z_2}}{|z_2|^2}\right).
Note that the argument of a complex number multiplied by a real number does not change, so:
\mathop{\mathrm{arg}}\left(\frac{\overline{z_2}}{|z_2|^2}\right) = \mathop{\mathrm{arg}}(\overline{z_2}).
Since the argument of a complex number and its conjugate are opposites:
\mathop{\mathrm{arg}}(\overline{z_2}) = -\mathop{\mathrm{arg}}(z_2),
it follows that:
\mathop{\mathrm{arg}}(w) = \mathop{\mathrm{arg}}(z_1) - \mathop{\mathrm{arg}}(z_2).
In the last steps, we took into account the fact that the argument of a complex number does not vary if that complex number is multiplied by any real number, that is:
\mathop{\mathrm{arg}}\left(\alpha z\right)=\mathop{\mathrm{arg}}\left(z\right)\qquad\forall z\in\Bbb{C}\quad\textrm{and}\quad \forall \alpha\in\Bbb{R}
and the fact that
\mathop{\mathrm{arg}}\left(\overline{z}\right) =-\mathop{\mathrm{arg}}\left(z\right)\qquad\forall z\in\Bbb{C}.
Such verifications are left as an exercise.
De Moivre’s Formula
Proposition
Let \theta \in \Bbb{R} and n \in \Bbb{N}. Then:
\left(\cos\theta + {\color{blue}\imath}\sin\theta\right)^n = \cos(n\theta) + {\color{blue}\imath}\sin(n\theta).
Proof
Consider the complex number z = \cos\theta + {\color{blue}\imath}\sin\theta. We have |z| = 1. Using the rules of multiplication in polar form, we can calculate the powers of z.
For n = 2:
\begin{aligned} z^2 &= (\cos\theta + {\color{blue}\imath}\sin\theta) \cdot (\cos\theta + {\color{blue}\imath}\sin\theta) \\ &= \cos^2\theta - \sin^2\theta + {\color{blue}\imath}(2 \cos\theta \sin\theta) \\ &= \cos(2\theta) + {\color{blue}\imath}\sin(2\theta). \end{aligned}
Assuming the formula is true for n, that is:
z^n = \cos(n\theta) + {\color{blue}\imath}\sin(n\theta),
we need to show that it is also true for n+1. Consider:
z^{n+1} = z^n \cdot z = \left(\cos(n\theta) + {\color{blue}\imath}\sin(n\theta)\right) \cdot \left(\cos\theta + {\color{blue}\imath}\sin\theta\right).
Using the formula for the product of complex numbers in trigonometric form, we obtain:
z^{n+1} = \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta + {\color{blue}\imath}(\cos(n\theta)\sin\theta + \sin(n\theta)\cos\theta).
Simplifying, we get:
z^{n+1} = \cos((n+1)\theta) + {\color{blue}\imath}\sin((n+1)\theta).
Thus, by induction, the formula holds for every n \in \Bbb{N}.
Observation
Assuming the properties of the exponential function for exponentiation are valid in the complex field as well, the proposition can be obtained more easily. If z = e^{{\color{blue}\imath}\theta}, then:
z^n = (e^{{\color{blue}\imath}\theta})^n = e^{{\color{blue}\imath}n\theta}.
Example of Calculation
Let’s calculate (1 + {\color{blue}\imath})^{2000} using De Moivre’s formula. First, we express 1 + {\color{blue}\imath} in its trigonometric form:
1 + {\color{blue}\imath}= \sqrt{2} \left( \cos \frac{\pi}{4} + {\color{blue}\imath}\sin \frac{\pi}{4} \right).
Applying De Moivre’s formula:
\begin{aligned} (1 + {\color{blue}\imath})^{2000} &= \left[ \sqrt{2} \left( \cos \frac{\pi}{4} + {\color{blue}\imath}\sin \frac{\pi}{4} \right) \right]^{2000} \\ &= \left( \sqrt{2} \right)^{2000} \left( \cos \left( 2000 \cdot \frac{\pi}{4} \right) + {\color{blue}\imath}\sin \left( 2000 \cdot \frac{\pi}{4} \right) \right) \\ &= \left( \sqrt{2} \right)^{2000} \left( \cos \left( 500 \pi \right) + {\color{blue}\imath}\sin \left( 500 \pi \right) \right) \\ &= 2^{1000} \left( \cos 0 + {\color{blue}\imath}\sin 0 \right) \\ &= 2^{1000}. \end{aligned}