Complex numbers

Complex number

Let’s now recall some basic notion on Complex Numbers: let’s consider the set

C ={ (x, y): x, y ∈ R}.

Of each of his couples:

• x is called the real part,
• y is called the imaginary part.

C is called the "set of complex numbers". In set C, the following are defined:

• equality, by setting (x, y) = ( x′, y′) ⟺ x = x′, y = y′;
• addition, by setting (x, y) + (x′, y′) =( x + x ′, y+ y′);
• the multiplication, setting (x, y) (x′, y′) = (xx′ - yy′, xy′ + yx′).

As regards the operations defined in set C, the following properties apply:

• addition enjoys the commutative, associative properties, there is the neutral element that is (0,0) and for each (x, y) there is the opposite (−x, −y);
• multiplication enjoys the commutative, associative, distributive properties respect to the sum, and there is a neutral element which is (1,0);
• for every complex number (x, y) ≠ (0,0) there is an inverse (we will see this shortly).

These properties make the set C with its operations a field.

Consider now the function $\varphi \mathrm{:}R\to C$ such that $\varphi (x)=(x,0)$ with x ∈ R and (x, 0) ∈ C.

The following properties apply:

$\varphi$ is injective, in other words (x ≠ x ′) implies $\varphi (x)\ne \varphi (x\text{'})$ since (x, 0) ≠ (x ′, 0);

$\varphi$ respects the addition and multiplication operations, in the sense that

φ(x) + φ(x′) =φ(x + x′)  and φ(x) φ(x′) = φ(xx′)  ∀x, x ′, ∈ R.

In other words φ is an injective homomorphism of the real field in the complex one, which identifies the Real set R as the subfield of C formed by complex numbers of the form (x,0) with x ∈ R: in this sense C is an expansion of R .To facilitate the treatment of complex numbers, we will identify each pair (x,0) with the real x.

We observe then that, for x, y ∈ C

(x, y) = (x, 0) + (0, y) = (x, 0) + (y, 0) (0,1) = x + y (0,1) = x + iy,

where the complex number i = (0,1) is called the imaginary unit. This representation is

called the "algebraic form of complex numbers". Recall that in z = x + iy ∈ C we have

x, y ∈ R, moreover:

- x is the real part of z and is denoted by Re(z);

- y is the imaginary part of z and is denoted by Im(z)

It results that $i^2=(\mathrm{0,1})(\mathrm{0,1})=(-\mathrm{1,0})=-1$ hence

$i^2=-1$ , and $i=\sqrt{-1}$

The advantage of the algebraic form is that with complex numbers expressed in this way, the usual rules of polynomial calculus can be used, always taking into account that $i^2=-1$

In this way it is easy to identify the inverse of a non-zero complex x+iy. So one

at least between x and y is different from 0, and therefore the real number $x^2+y^2\ne 0$ , indeed positive. Multiplying x + iy by what we will shortly call its conjugate, that is x - iy, we get right

$(x+\mathit{iy})(x-\mathit{iy})=x^2+y^2$

Dividing by $x^2+y^2\ne 0$ we obtain

$(x+\mathit{iy})\frac{(x-\mathit{iy})}{(x^2+y^2)}=1$

which identifies the inverse of $x+\mathit{iy}$ as

$\frac{x}{(x^2+y^2)}-i\frac{y}{(x^2+y^2)}$

where the fraction symbol indicates the quotient in the real field.

The inverse of (x, y) is (x,y)^-1 and we can indicate it, even in the complex field, with the following notation ${(x,y)}^{-1}=\frac{1}{(x,y)}$