Introduction

In this first section we recall some basic concepts about complex numbers. A deep knowledge of these notions is essential for the comprehension of the following topics of the course.

The need of extending the real field ${\displaystyle \mathbb {R} }$ has its origins in the impossibility of finding a solution of particular equations, such as:

${\displaystyle x^{2}+1=0\ ,}$

for which it is clear that ${\displaystyle \nexists \ x\in \mathbb {R\colon } \ x^{2}+1=0}$.

We define the complex field ${\displaystyle \mathbb {C} }$, so that we can solve equations such as the previous one.

${\displaystyle \mathbb {C} }$ is defined through the extension of the well-known properties of ${\displaystyle \mathbb {R} }$.

We remember that ${\displaystyle \mathbb {R} }$ is a vector field if it is equipped with the following operations:

${\displaystyle \forall x,y\in \mathbb {R} {\mbox{ we have that }}x+y\in \mathbb {R} \ ,}$

1. Multiplication

${\displaystyle \forall x,y\in \mathbb {R} {\mbox{ we have that }}xy\in \mathbb {R} \ ,}$

defined in an axyomatic way, in order to satisfy the following properties:

• Commutative property: ${\displaystyle \forall x,y\in \mathbb {R} }$ we have that:

${\displaystyle x+y=y+x\ ,}$

${\displaystyle xy=yx\ .}$

• Associative property: ${\displaystyle \forall x,y,z\in \mathbb {R} }$ we have that:

${\displaystyle x+(y+z)=(x+y)+z\ ,}$

${\displaystyle x(yz)=(xy)z\ .}$

• Distributive property: ${\displaystyle \forall x,y,z\in \mathbb {R} }$ we have that:

${\displaystyle x(y+z)=xy+xz\ .}$

• Identity element: ${\displaystyle \forall x\in \mathbb {R} }$ we have that:

${\displaystyle x+0=0+x=x\ ,}$

${\displaystyle 1x=x1=x\ .}$

${\displaystyle \forall x\in \mathbb {R} \,\exists x\in \mathbb {R} \,\colon \,x+(-x)=(-x)+x=0\ .}$

• Inverse element (multiplication):

${\displaystyle \forall x\in \mathbb {R} \setminus \{0\}\,\exists {\frac {1}{x}}\in \mathbb {R} \,\colon \,x\left({\frac {1}{x}}\right)=\left({\frac {1}{x}}\right)x=1\ .}$

We define the complex field ${\displaystyle \mathbb {C} }$ as the set of the pairs ${\displaystyle (a,b)}$ with ${\displaystyle a,b\in \mathbb {R} }$ equipped with the properties:

1. Equality:

${\displaystyle \forall (a,b)\,\&\,(c,d)\,\in \mathbb {C} {\mbox{ we have that }}(a,b)=(c,d)\,\Longleftrightarrow {\begin{cases}a=c\\b=d\end{cases}}\ ,}$

${\displaystyle \forall (a,b)\,\&\,(c,d)\,\in \mathbb {C} {\mbox{ we have that }}(a,b)+(c,d)=(a+c,b+d)\ ,}$

1. Multiplication:

${\displaystyle \forall (a,b)\,\&\,(c,d)\,\in \mathbb {C} {\mbox{ we have that }}(a,b)(c,d)=(ac-bd,ad+bc)\ .}$

we see that if we define ${\displaystyle \mathbb {C} _{0}\equiv \{(a,b)\in \mathbb {C} \colon b=0\}}$ we have that:

• ${\displaystyle \mathbb {C} _{0}\subset \mathbb {C} \ ;}$
• ${\displaystyle \mathbb {C} _{0}}$ is a field, with theoperations and properties inherited by ${\displaystyle \mathbb {C} \ ;}$
• ${\displaystyle \mathbb {C} _{0}}$ is isomorphic to ${\displaystyle \mathbb {R} \ ,}$ that is:

${\displaystyle \exists f:\mathbb {R} \rightarrow \mathbb {C} \;x\mapsto f(x)=(x,0){\mbox{ is an isomorphism.}}}$

Moreover, it can be proved that the function:

${\displaystyle g:\mathbb {R} ^{2}\rightarrow \mathbb {C} {\mbox{ definita come: }}\ g(x,y)=(x,y)\,\forall x,y\in \mathbb {R} }$

is an isomorphism, that is ${\displaystyle \mathbb {C} {\mbox{ is isomorphic to }}\mathbb {R} ^{2}}$.
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