We all naturally develop a sense of number from the qualitative (in the early stages
of life) to the quantitative around the age of scholarization. Toddlers will often
say that the second row below
\begin{align*}
&\blacksquare\hspace{1cm}\blacksquare\hspace{1cm}\blacksquare\hspace{1cm}\blacksquare\\
&\blacksquare\hspace{1.5cm}\blacksquare\hspace{1.5cm}\blacksquare\hspace{1.5cm}\blacksquare
\end{align*}
contains more objects than the first (try!). Almost no school child, however, will
fail to recognize that the two rows have the same number of elements in them. He or
she will have developed by then a sense of natural number and will mentally verify
that there is a one-to-one correspondence between objects in the two rows.
We shall take the set of ''counting'' numbers, so-called natural numbers to be given
and accepted
$$
\mathbb{N}:=\{ 1,2,3,\dots\}
$$
and use it to count the number of elements in finite sets. A set $S$ will have $n$
elements iff a bijective (one-to-one and onto) map $c$ can be found from the
prototype set $\{1,2,\dots,n\}=:\mathbb{N}_n$ to $S$
$$
c:\mathbb{N} _n\to S\, ,\: k\mapsto c(k)=:c_k\, .
$$
The sets $\mathbb{N} _n$ play the role of a blueprint of all finite sets with exactly
$n$ elements. Analogously $\mathbb{N}$ will be the model of an infinite set $R$ with
a countable number of elements. Countable here means that elements can be enumerated,
labeled in a unique fashion by using natural numbers by means of a bijective map
$$
e:\mathbb{N} \to R\, ,k\mapsto e(k)=:e_k\, .
$$
We shall then use $\mathbb{N}$ to construct two bigger sets, $\mathbb{Z}$ and
$\mathbb{Q}$, in a completely algebraic and intuitive fashion.
Sequences and convergence, the trademark of calculus (and analysis), will naturally
appear in the construction of real numbers. If you think of real, non rational
numbers, such as $\sqrt{2}$, in their decimal form, e.g.
$$
\sqrt{2}=1.41421356237\cdots
$$
you realize rightaway that they cannot be captured by a finite number of digits. As
we are unable to process infinitely many digits at once, we can only grasp $\sqrt{2}$
as a sequence of rationals which better and better approximate it, i.e.
$$
1,1.4,1.41,1.414,1.4142,\dots
$$
Defining limit taking and convergence is thus unavoidable when constructing real
numbers. The ''existence'' of the latter is of course justified by practical issues
such as measuring the length of the diagonal $d$ of a unit square
Complex number are maybe the least ''natural'' numbers (judging from most people's
spontaneous reaction when they are first introduced to them). While they can be
introduced and fully understood in purely algebraic terms, we choose here to
highlight an analytical reason for their relevance. The function
$$
f:\mathbb{R} \to(0,\infty)\, ,\: x\mapsto \frac{1}{1+x^2}
$$
is as ''nice'' as a function can possibly be. It is positive, bounded, and infinitely
many times differentiable with bounded derivatives (check!). In spite of this, its
power series representation
$$
\frac{1}{1+x^2}=1-x^2+x^4-x^6\pm\dots
$$
fails to converge for $|x|\geq 1$. The function experiences absolutely no issue when
its argument passes through $-1$ or $1$ but the corresponding power series ceases to
converge. This phenomenon can really only be understood by working in the larger
field $\mathbb{C}$ of complex numbers where
$$
1+x^2 =0\text{ has solutions }x_{1,2}=\pm i\, ,
$$
thus revealing that the function $f$ does undergo an ''explosion'' at distance $1$ from
$x=0$. While it does so in the complex, its reverberations are felt on the real line
as well!
Natural Numbers
In this course natural numbers are assumed to be given. The standard notation
$$
\mathbb{N}=\{ 1,2,3,\dots\}
$$
is used throughout. The number $0$ is not an element of
$\mathbb{N}$. We also assume that addition and multiplication have
been defined on $\mathbb{N}$ in the usual way. Natural
numbers can be used to define what
finite and countably infinite sets are.
Definitions (finite and countably infinite)
Let $S$ be any set. It is said to be finite iff there is a natural number
$n\in \mathbb{N}$ and a bijection (that is, a one-to-one and onto map)
$$
\varphi:\{1,2,\dots,n\}\to S\, , \: k\mapsto s(k)=s_k\, ,
$$
which counts the elements of $S$ to be $n$ many. The set $S$ is called
infinite iff it is not finite. Next we introduce the concept
of countable infinity, which characterizes sets which have ``as many elements as''
$\mathbb{N}$ has. The set $S$ is said to be countably infinite iff there exists a
bijection
$$
\varphi: \mathbb{N} \to S\, ,\: k\mapsto s_k\, ,
$$
which enumerates all elements of the set.
Notice that somewhat counterintuitive phenomena can occur for infinite sets.
Example
Let $\mathbb{Z}=\{\dots, -2,-1,0,1,2,\dots\}$ be the set of integers
to be defined properly in the next section. Is it countable
(short for countably infinite)?
Yes. Indeed
$$
\varphi(k):=\begin{cases}
k/2\, ,&\text{if }k\text{ is even}\, ,\\
-(k-1)/2\, ,&\text{if }k\text{ is odd}\, ,
\end{cases}
$$
defines a bijection between the two sets $\mathbb{N}$ and $\mathbb{Z}$. Make sure to
verify this for yourself!
Integer Numbers
Starting with the natural numbers, we can construct all other number
systems that we need.
Definition (Whole Numbers)
Consider the set $\mathbb{N}\times \mathbb{N}$ define an equivalence
relation by
\begin{equation*}
(m,n)\sim(m',n')\Leftrightarrow m+n'=m' +n.
\end{equation*}
By the properties of equivalence relation, the set of equivalence
classes of the form
$$
[m,n]=\{ (m',n')\in \mathbb{N}\times \mathbb{N}\, |\, (m,n)\sim(m',n')\}
$$
yields a partition of the set $\mathbb{N}\times \mathbb{N}$. It is
denoted by $\mathbb{Z}$ and is taken to be the set of integer
numbers.
Recall that an equivalence relation on a set $S$ is a
relation $\sim$ satisfying
(er1) $s\sim s$ for all $s\in S$, i.e. the relation is reflexive.
(er2) $s\sim r$ implies that $r\sim s$ for all $r,s\in S$,
i.e. $\sim$ is symmetric.
(er3) $s\sim r$ and $r\sim t$ imply that $s\sim t$ for all $r,s,t\in S$,
i.e. $\sim$ is transitive.
If $\sim$ is an equivalence relation on $S$, then $[s]$ denotes the equivalence
class of the element $s$ given by
$$
[s]=\{ r\in S\, |\, r\sim s\}\, .
$$
An equivalence relation partitions a set into disjoints subsets (the equivalence
classes). Please check this for yourself! The set containing all equivalence classes
is typically denoted by $S/\sim$.
Clearly we think of an equivalence class $[m,n]$ as the number $m-n$. If $m>n$, then
$[m,n]$ corresponds to the natural number $m-n$ but we think of it as
the relationship between $n$ and $m$, i.e. as ''$m$ is $m-n$ numbers after $n$''. On the other
hand, if $m< n$, we think of $[m,n]$ as indicating that ''$m$ is $n-m$ numbers
before $n$''. The fact that a natural number can be before or after another one
yields the idea of positive and negative number. Notice that we think of $[m,m]$ as
the number $0$.
Exercises
Show that the above relation used to define $\mathbb{Z}$ is
indeed an equivalence relation.
How can you think of $\mathbb{N}$ as a subset of $\mathbb{Z}$?
Define the operations of addition and multiplication for integer
numbers and verify that they are well-defined.
Show that addition and multiplication are commutative,
associative, and have an identity element, and that additive
inverses exist.
It clearly holds that $(m,n)\sim(m,n)$ as $m+n=m+n$. Next, if
$(m,n)\sim(m',n')$, then $m+n'=m'+n$, which is obviously the same as
$m'+n=m+n'$ and thus $(m',n')\sim(m,n)$. Finally assume that
$(m,n)\sim(m',n')$ and $(m',n')\sim(m'',n'')$, which amounts to
$$
m+n'=m'+n\text{ and }m'+n''=m''+n'.
$$
Adding the left and right hand sides of these identities yields
$$
m+n'+m'+n''=m'+n+m''+n'\text{ or, equivalently }m+n''=m''+n,
$$
and thus $(m,n)\sim(m'',n'')$. What was used implicitly in this
argument?
We identify $m\in\mathbb{N}$ with $[m+1,1]$.
Operations can be defined by setting
$$ [m,n]+[p,q]=[m+p,n+q]\text{ and }
[m,n]\cdot[p,q]=[mp+nq,mq+np],$$
for any choice of $m,n,p,q\in\mathbb{N}$. To show that these
operations are well defined, we take other representatives $(m',n')$
and $(p',q')$ for the equivalence classes and verify that they lead
to an equivalent representation of the sum and of the
product. To see that, we need
$$
(m'+p',n'+q')\sim(m+p,n+q)\text{ or that }
(m'+p')+(n+q)=(m+p)+(n'+q'),
$$
which follows from
$$
m+n'=m'+n\text{ and } p+q'=p'+q
$$
by addition of the identities. While multiplication can be handled
similarly, we urge you to carry the calculations out.
Commutativity and associativity follow from the corresponding
properties of the operations on $\mathbb{N}$ and the definition of
addition and multiplication for whole numbers. Addition gains an
identity since
$$
[m,n]+[1,1]=[m+1,n+1]=[m,n],
$$
for any choice of $m,n\in\mathbb{N}$, and inverses since
$$
[m,n]+[n,m]=[m+n,n+m]=[1,1]
$$
for any $m,n\in\mathbb{N}$.
Notation
It is more convenient to introduce the following simplified notation for equivalence
classes.
\begin{align*}
m&\qquad \text{ for the equivalence class }[m+1,1]\text{ and } m\in \mathbb{N},\\
0&\qquad \text{ for the equivalence class }[1,1],\\
-m&\qquad \text{ for the equivalence class }[1,m+1]\text{ and } m\in \mathbb{N}.
\end{align*}
Then $\mathbb{Z}=\{ \dots,-2,-1,0,1,2,\dots\}$.
Rational Number
Next we construct the set of rational numbers using only integer numbers. To do so we
first consider pairs $(m,n)$ of integers where $m\in \mathbb{Z}$ and $n\in
\mathbb{N}$. The idea is to obtain a new number by splitting $m$ into $n$ equal
pieces, that is something that will be called $\frac{m}{n}$. We
proceed in a similar way as we did for the introduction of whole
numbers by having multiplication play the role that addition played in
that construction.
Definition (Rational Numbers)
The set of rational numbers is given as
$$
\mathbb{Q}=\mathbb{Z}\times \mathbb{N}/\sim
$$
the set of equivalence classes with respect to the relation defined by
$$
(m,n)\sim (\bar m,\bar n)\text{ iff } m\bar n=\bar mn.
$$
This definition is very intuitive as for instance
$$
\frac{1}{2}=[(1,2)]=[(5,10)]=[(13,26)]=\dots
$$
indicating the one half is the relation between $1$ and $2$ or any
other pair in the equivalence class.
Exercise
Show that integers can canonically be viewed as a subset of the
rationals and show that $\mathbb{Q}$ admits compatible addition and
multiplication. Finally show that any non-zero element has a
multiplicative inverse.
Is there any substatial difference between the way in which
rational numbers are constructed from integers and integer numbers are
constructed from natural numbers?
Is $\mathbb{Q}$ countable? Before answering this question we make two
observations.
Remark
Any infinite subset $R$ of a countably infinite set $S$ is itself countable
A short way of saying that $S$ is countable is
$$
S=\{ s_1,s_2, \dots\}
$$
which follows directly from the definition. Then $R$, as an infinite subset, will
contain some or all of the elements of $S$. By going through the elements of $S$
sequentially, we make a note of which elements belong to $R$ by recording the
corresponding index. In other words
$$
R=\{s_{k_1},s_{k_2},\dots \}\, ,
$$
where $s_{k_1}$ is the first element of $S$ which also lies in $R$ and so on. We
clearly just found a bijection
$$
\varphi:\mathbb{N} \to R\, ,\: j\mapsto s_{k_j}\, ,
$$
proving the countability of $R$.
Remark
The product $S\times T$ is a countable set whenever the sets $S$ and $T$
share the same property.
Since the two sets are countable we can enumerate
their elements and list all elements of the Cartesian product $S\times T$
\begin{align*}
&(s_1,t_1)\, ,\:(s_1,t_2)\, ,\:(s_1,t_3)\, ,\:\dots\, ,\: (s_1,t_m)\, ,\: \dots\\
&(s_2,t_1)\, ,\:(s_2,t_2)\, ,\:(s_2,t_3)\, ,\:\dots\, ,\\
&\dots\\
&(s_n,t_1)\, ,\:(s_n,t_2)\, ,\:(s_n,t_3)\, ,\:\dots
\end{align*}
Next we use a diagonal counting argument by which the elements
of $S\times T$ are numbered sequentially going through all diagonals
$D_k$ of $k=1,2,\dots$ elements
$$
D_k=\{(s_n,t_m)\, |\, n+m=k\}
$$
from top to bottom.
Exercise
Give an explicit formula for the map obtained in the discussion of the
above remark about product sets.
We are now in position to answer the question about the countability
of $\mathbb{Q}$. We only need to observe that $\mathbb{Q}$ is an
infinite set (since it contains the countably infinite set
$\mathbb{Z}$) and that
$$
\mathbb{Q} =\bigl(\mathbb{Z} \times \mathbb{N} /\sim\bigr) \subset
\mathbb{Z} \times \mathbb{N}
$$
to conclude by the above remarks that $\mathbb{Q} $ is indeed
countable.
Exercise
How do you make sense of the inclusion $ \bigl(\mathbb{Z} \times
\mathbb{N} /\sim\bigr) \subset\mathbb{Z} \times \mathbb{N}$ that was
used above?
Real Numbers
There are many reasons why rational numbers are not enough for calculus. One of them
is that natural geometric objects, like the unit square, cannot be fully described by
natural or rational numbers. We indeed know since Pythagoras that the diagonal length
$d$ of a unit square satisfies
$$
1^2+1^2=d^2\, .
$$
Unfortunately
Remark
There is no rational number $d$ for which $d^2=2$.
If $d\in \mathbb{Q}$ then $d=\frac{m}{n}$ for some integers $m$ and $n$ which can
assume do not share any common factors (since otherwise we simply remove them). In
particular they cannot be both even. However,
$$
m^2=2n^2
$$
implies that $m^2$ is even and thus also $m$ is. In this case $m=2k$ for some $k\in
\mathbb{N}$ and, consequently
$$
2n^2=4k^2
$$
so that $n^2$ and thus $n$ must be even as well. This is clearly a contradiction to the
assumption that $d$ is rational and so it isn't.
Consider now the sequence of rational numbers defined recursively by
$$
\begin{cases}
x_0\in \mathbb{Q}_{>0} \, ,&\\
x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\, ,&n\geq 0\, ,
\end{cases}
$$
where the notation $\mathbb{Q}_{>0}$ is self-explanatory. Can you find any pattern to
the behavior of the sequence $(x_n)_{n\in\mathbb{N}}$?
Definition (Sequence)
A sequence $(x_n)_{n\in\mathbb{N}}$ in a set S is a map
$$
x:\mathbb{N}\to S,\: n\mapsto x(n)=x_n,
$$
and is determined by its values $x_n=x(n)\, ,\: n\in\mathbb{N}$, hence the notation.
We collect properties of this sequence.
Remarks
(a) $x_n\in \mathbb{Q}$ for all $n\in \mathbb{N} $. This follows from the fact that
$\mathbb{Q} $ is a field with respect to addition and multiplication and mathematical
induction on $n$.
(b) The sequence $(y_n)_{n\in\mathbb{N}}$ of squares given by
$$\begin{cases}
y_0=x_0^2\in \mathbb{Q}\, ,&\\
y_{n+1}=x^2_{n+1}=(x_n/2+1/x_n)^2=y_n/4+1+1/y_n\, ,& n\geq 0
\end{cases}
$$
converges to $y_\infty=2$.
Notice that a sequence of rationals
$(x_n)_{n\in\mathbb{N}}$ is said to converge to a rational number
$x_\infty$ if it comes closer and closer to it with increasing index,
i.e. if $|x_n-x_\infty|\to 0$ as $n\to\infty$, i.e. iff
$$
\text{ for all }\epsilon>0\text{ there is }N_\epsilon\in \mathbb{N}\text{ such that
} |x_n-x_\infty|\leq\epsilon\text{ for }n\geq N_\epsilon\, .
$$
(b) Observe that
$$
y/4+1+1/y>2\:\Leftrightarrow\: (y-2)^2>0
$$
implies that $y_1>2$ so long as $y_0\neq 2$. If $y_0=2$, the sequence becomes
stationary, i.e. $y_n=2$ for each $n\in \mathbb{N}$. In particular, whatever $2\neq y_0>0$
is, $y_1>2$ and thus by induction
$y_n>2$ for all $n\in \mathbb{N} $. Now if $y_n=2+\epsilon$ for some $\epsilon>0$,
then
$$
y_{n+1}-2=\frac{(4+\epsilon)^2}{4(2+\epsilon)}-2=\frac{\epsilon ^2}{4(2+\epsilon)}\, ,
$$
and thus
$$
|y_{n+1}-2|=\frac{\epsilon}{4}\frac{\epsilon}{(2+\epsilon)}\leq \frac{|y_n-2|}{4}\, .
$$
This clearly shows that $y_n\to 2$ as $n\to \infty$.
Since we also have that $x_n^2=y_n$, it appears as if $x_n=\sqrt{y_n}$ should
converge to "$\sqrt{2}$''. Now, while we do not have $\sqrt{2}$, we do have that
$(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence (the definition is coming up
shortly) since
\begin{equation*}
0\leq |x_n-x_m|\leq(x_n+x_m)|x_n-x_m|=|x_n^2-x_m^2|=|y_n-y_m|\to 0\text{ as }
n,m\to\infty\, .
\end{equation*}
What does it mean for $(x_n)_{n\in\mathbb{N}}$ to be a Cauchy
sequence? It means that more and more digits of $x_n$ stop changing as
$n$ grows. Indeed, take for instance $x_0=3$,
\begin{equation*}
x_1=1.8\bar 3\, ,\: x_2=\underline{1}.46\overline{21}\, ,\:
x_3=\underline{1.4}14998\cdots\, ,\: x_4=\underline{1.41}42137\cdots\, ,\:
x_5=\underline{1.414213}56\, ,\: \dots
\end{equation*}
We can therefore think of the Cauchy sequence $(x_n)_{n\in\mathbb{N}}$
as a representation of a number with infinitely many digits and call
it $\sqrt{2}$.
To extend this idea and construct all of $\mathbb{R}$ we need the
concept of Cauchy sequence.
Definition (Cauchy Sequence)
A sequence $(x_n)_{n\in\mathbb{N}}$ of rational numbers is called a
Cauchy sequence iff
$$
\text{ for all }\epsilon>0\text{ there is }N_\epsilon\in
\mathbb{N}\text{ such that } |x_n-x_m|\leq\epsilon\text{ for }m,n\geq
N_\epsilon\, .
$$
A sequence is Cauchy iff all but finitely many of its members are at most distance
$\varepsilon>0$ from each other no matter how small $\varepsilon$
is. This condition is in principle easier to verify than convergence
to a limit since it does not require knowledge of the limit. In
fact, in general, Cauchy sequences do not even need to possess a limit
in the first place just like the sequence in $\mathbb{Q}$ used above
to "detect" $\sqrt{2}$.
Denote by
$$
\text{CS}(\mathbb{Q})=\{(x_n)_{n\in\mathbb{N}}\, |\, x_n\in
\mathbb{Q}\text{ and }(x_n)_{n\in\mathbb{N}}\text{ is Cauchy}\}
$$
the set of all Cauchy sequences of rational numbers and introduce the
following equivalence relation on it
$$
(x_n)_{n\in\mathbb{N}}\sim(y_n)_{n\in\mathbb{N}}\text{ iff }
\lim_{n\to\infty}(x_n-y_n)=0\, .
$$
Exercise
Prove that the above is an equivalence relation on the set of Cauchy
sequences of rationals.
The set of equivalence classes $\text{CS}(\mathbb{Q})/\sim$ contains sets of
sequences. The difference of any two elements of the same equivalence class is a null
sequence (i.e. a sequence with limit zero). Thus if one element of an equivalence
class converges so do all others (and to the same limit). If an equivalence class
does not contain any convergent sequence, we take the sequence (better the
equivalence class itself) to "be'' the limit.
Notice that by identifying $r\in \mathbb{Q} $ with
\begin{equation*}
[(r,r,r,\dots)]=[(r,r+1/2,r+1/3,\dots)]=[(r,r-1/2,r+1/4,\dots))]=\dots
\end{equation*}
we recover $\mathbb{Q} $ as a subset of
$\text{CS}(\mathbb{Q})/\sim$. We define
Definition (Real Numbers)
$$
\mathbb{R}=\text{CS}(\mathbb{Q})/\sim
$$
Exercise
Show that this identification of $\mathbb{Q}$ with a subset of
$\mathbb{R}$ is unambiguous. Define operations and an order relation
on $\text{CS}(\mathbb{Q})/\sim$ which extend the operations and order
relation of the rationals. Show that $\mathbb{R}$ is a field.
For our example, we now have that
$$
[(x_n)_{n\in\mathbb{N}}]^2=[(x_n^2)_{n\in\mathbb{N}}]=[(y_n)_{n\in\mathbb{N}}]=2
$$
and thus can find a solution to the equation $d^2=2$ in the new field
of numbers.
We shall see later that a set with a distance function defined on it (what is it in
the case of rationals or reals?) is complete iff all Cauchy sequences in it
converge (i.e. there are no "holes''). The new set $\mathbb{R}$ is complete (by
construction). It is, in fact, the completion of $\mathbb{Q}$ with respect to
its distance function.
Question
Do you think that $\mathbb{R}$ is countable? Try and figure out a way
to prove your conjecture.
We also mention the following important structure theorem on real
numbers which you prove using the construction of the reals sketched
above.
Theorem
Every $x \in \mathbb{R}$ can be represented as
$$
x = m+ \sum_{j=1}^\infty\frac{d_j}{10^j}\text{ where }d_j \in \{0,1,...,9\}\, ,
$$
and for some $m\in \mathbb{Z}$. In particular $1 =0.9999....=1.0$.
Remark
We spent some time describing one possible way to construct real
numbers because it is exemplary of the way of analysis. You will have
noticed that convergence and limiting procedures play a central
role. These concepts will reappear in different contexts time and
again and a deep understanding can only be beneficial.
Complex Numbers
We conclude this lecture with a brief introduction of complex
numbers. While the inability of solving the equation $x^2=-1$ can be
used as motivation for looking for a larger number field in which it
possesses a solution, there are farther reaching analytical (and
other) reasons to introduce complex numbers. They can, for instance,
shed light on the reason why the power series expansion
$$
\frac{1}{1+x^2}=\sum _{k=0}^\infty (-1)^kx^{2k}
$$
does not converge for $|x|\geq 1$ in spite of the fact that the
function it represents has no apparent issue with arguments larger
than one in absolute value!
Definition (Complex Numbers)
Complex numbers can be introduced following a purely algebraic
procedure. Consider $\mathbb{R}^2$ and define the following operations
on it
\begin{gather*}
x+y=(x_1+y_1,x_2+y_2)\, ,\\
xy=(x_1y_1-x_2y_2,x_1y_2+x_2y_1) \, ,
\end{gather*}
for $x=(x_1,x_2)\, ,\: y=(y_1,y_2)\in \mathbb{R} ^2$. Then
$\mathbb{C}=(\mathbb{R}^2,+,\cdot)$.
Remarks
(a) Observe that $(0,1)\cdot(0,1)=(-1,0)$.
(b) It is easily checked that $\mathbb{R}$ can be identified
with
$$
\{(r,0)\, |\, r\in\mathbb{R} \}=\mathbb{R}\times\{ 0\}\subset
\mathbb{R} \times\mathbb{R}
$$
and that the operations defined above extend the standard ones on
$\mathbb{R}$. Indeed
\begin{gather*}
(r,0)+(s,0)=(r+s,0)\, ,\\
(r,0)\cdot(s,0)=(rs,0)\, ,
\end{gather*}
for any $r,s\in \mathbb{R}$. It can be checked that $\mathbb{R} ^2$
with these operations is a field. It is denoted by $\mathbb{C}$ to
distinguish it from the vector space $\mathbb{R} ^2$.
(c) Elements $z=(x,y)$ of $\mathbb{C}$ are simply referred to
as $z=x+iy$. This is justified by
$$
z=(x,y)=x(1,0)+y(0,1)\, ,
$$
by observing that $(1,0)=1\in \mathbb{R}$, and setting
$(0,1)=i$. Notice that $i^2=-1$.
(d) It is often
convenient to describe complex numbers by their polar coordinate representation
$$
z=x+iy=\sqrt{x^2+y^2}e^{i\arctan{y/x}}=re^{i\theta}=\bigl(
r\cos(\theta),r\sin(\theta)\bigr)\, ,
$$
where we used the exponential function
$$
e^z=\sum _{k=0}^\infty \frac{z^k}{k!}\, .
$$
We refer to Lecture 3 for the convergence analysis of power series required to verify
that this definition is indeed justified.
Definitions (Modulus and Argument)
(i) We use the notation
$$
r=\sqrt{x^2+y^2}=|z|
$$
for the modulus (length) of the complex number $z$ and
$\theta\in[0,2\pi)$ for its so-called argument.
(ii) Given $z=x+iy$, the complex number
$$
\bar z=x-iy
$$
is referred to as the complex conjugate of $z$. It holds
that
$$
|z|^2=z\bar z=(x+iy)(x-iy)\, .
$$
What is $\bar z$ if $z=re^{i\theta}$?
Exercise
Prove that $\cos(\theta)+i\sin(\theta)=e^{i\theta}\, ,\: \theta\in
\mathbb{R}$, by using the power series definition of the trigonometric
functions and not worrying about convergence issues for now.
Definitions (Distance and Convergence)
Based on the modulus, the following distance is defined on
$\mathbb{C}\times\mathbb{C}$
$$
d(z,w)=|z-w|=\sqrt{(x-u)^2+(y-v)^2}\, ,\: z=x+iy\, ,\: w=u+iv\, .
$$
Given a sequence $(z_n)_{n\in\mathbb{N}}$ of complex numbers and
$z_\infty\in \mathbb{C} $ we say that
$$
z_n\to z_\infty\text{ as }n\to\infty\text{ iff }d(z_n,z_\infty)\to 0\text{ as
}n\to\infty\, .
$$
Exercise
Do Cauchy sequences of complex numbers converge? Give a proof or a
counterexample.
Remark
We conclude this section by recalling the important fact that the
field $\mathbb{C}$ is algebraically closed, that is, a
polynomial of any degree $n\in\mathbb{N}$ with complex coefficients
has exactly $n$ roots (counting multiplicity). It follows that
$\mathbb{C}$ puts an end to the quest for solutions of polynomial
equations.
Exercise
Let $z,w\in \mathbb{C}^n$ be arbitrary. Prove the following Cauchy-Schwartz
inequality
$$
|\sum _{j=1}^nz_j\overline{w}_j|^2\leq\bigl(\sum _{j=1}^n|z_j|^2\bigr)\bigl(\sum
_{j=1}^n|w_j|^2\bigr)\, .
$$