Category Archives: Numbers

Description of the numbers

Numbers, Real numbers, Infinitesimal calculus

Irrational numbers

Irrational numbers

Irrational numbers are numbers that cannot be expressed, such as a fraction \frac{m}{n}, where m, n \in\mathbb{Z} and n\not = 0

This property is fulfilled by the real numbers that are not rational

An infinite decimal place (i.e. with infinite figures) aperiodic, as \sqrt{7} = 2,645751311064591 can not be represented as a rational number

Such numbers are called irrational numbers

This designation means the impossibility of represent that number as a ratio of two whole numbers

Are denoted as \mathbb{I}

Irrational numbers most well-known

  • Pi number:
    reason between the length of a circumference and its diameter, \pi\approx 3,14159\cdots
  • Euler Number:
    e=\lim\limits_{n\to\infty}\left(1+{\frac {1}{n}}\right)^{n}\approx 2,7182\cdots
  • Aureum number:
    \Phi={\frac {1+{\sqrt {5}}}{2}}\approx 1,6180\cdots

Demonstration: Root of 2 is irrational

The value that verifies d^2=2 it is denoted by \sqrt{2} and it is a real number

Consider the set: S=\{x\in\mathbb{R}|x\geq 0, x^2 \leq 2\}

Set S is non empty (1 \in S) and is superiorly bound, as x\in S, x^2\leq 2 < 4 = 2^2, then x < 2

As a non empty set dimensioned above, we will have for the axiom of completeness, which S possesses supreme, that supreme we will denote by v. It cannot happen that v^2 > 2 or v^2 < 2 and, therefore, you have to v^2 = 2; that is, it will be the value we have denoted as \sqrt{2}

Suppose that v^2 > 2, then taking h=min\{v,\frac{(v^2 - 2)}{2\cdot v}\} it would have h > 0, v-h \geq 0 and (v-h)^2=v^2+2\cdot h \cdot v + h^2 \leq v^2 +2\cdot h \cdot v +h \cdot v = v^2 + 3 \cdot h \cdot v \leq v^2 + (2-v^2)=2, that is, v+h\in S

But this is not possible, because v+h > v and change \forall x \in S you have to x \leq v. So, \sqrt{2} it is an irrational number

Gauss's motto

Gauss's motto was published in article 42 of the Disquisitiones Arithmeticae (1801) and says:

Each real root of a polynomial monico with integral coefficients is integer or irrational

Demonstration of Gauss's motto

Be r a real root of the monican polynomial: P(x)=x^n+c_{n-1}\cdot x^{n-1}+\cdots + c_0 where n is a positive integer, and c_0,\cdots,c_{n-1} are integers and suppose that r is rational but not integer

Then there exists a unique integer q such that q < r < q+1. Since r is rational, so will r^2, \cdots, r^{n-1}

Therefore, the set: M=\{m>0| m, m\cdot r, m\cdot r^2, \cdots, m\cdot r^{n-1}\text{ they are integers}\} possesses an element and is not empty

Considering that r is the root of P(x), the identity will be fulfilled: r^n=-(c_{n-1}\cdot r^{n-1}+\cdots +c_0) and in addition \forall m \in M, m(c_{n-1}\cdot r^{n-1}+\cdots +c_0) is an integer, and therefore, m\cdot r^n it is also an integer

Let's look for a contradiction with the principle of good sorting of natural numbers, thus proving that set M has no minimum element; I mean: \forall m \in M, \exists m'\in M \Rightarrow 0 < m' < m

We consider m\in M and we take m'=(r-q)\cdot m. So we have to for every i=0,1,2,\cdots, n-1 we have to m'\cdot r^i=m\cdot r^{i+1}-q\cdot m^i, then it's an integer and it's true that m'\in M and 0 < m' < m because 0 < r-q < 1

Then M can not have an item minimum, and the root r must be integer or irrational

Numbers, Natural numbers, Integer numbers, Rational numbers, Real numbers, Infinitesimal calculus

Inequalities

Inequalities

Below you are going to list some inequalities that are useful or that have not been mentioned

Up to now we have shown some properties that are verified in the various sets of numbers

Considering that each numerical set we have defined above contains the previous one and the set of real ones contains them all, all of them (except the principle of good sorting of natural numbers) are checked for the actual numbers

For example, using some of the inequalities mentioned above, it is possible to prove that:

  • 0\leq a \leq b \Rightarrow a^2 \leq b^2
  • 0 < a \leq b \Rightarrow \frac{1}{b} \leq \frac{1}{a}

Absolute value of a real number

It is defined for a real number in the same way that it has already been made for integers, but it also meets some inequalities that deserve to be pointed out

  • -\|a\| \leq a \leq \|a\|
  • \|a\| \leq b \Leftrightarrow -b \leq a \leq b
  • \|a\| \geq b \Leftrightarrow \begin{cases} a \geq b \\ a \leq -b \end{cases}
  • \|a\cdot b\| = \|a\|\cdot \|b\|
  • a^2 \leq b^2 = \|a\| \leq \|b\|

Keep in mind that in the equality \sqrt{a^2}=\|a\|, will only be true \sqrt{a^2}=a if a \geq 0

Distance

Dice a, b \in \mathbb{R}, is called distance between a and b to the actual non negative number \|a-b\|

This notation is fundamental to interpret inequalities of the form \|x-a\| \leq b, such as the distance of x to a is less than or equal to b

Triangular inequality

We've seen it before, but because it's an important inequality we remember it for real numbers

Dice a, b \in \mathbb{R} is true that \|a+b\| \leq \|a\|+\|b\|

Demonstration: triangular inequality

We take \begin{cases} -\|a\|\leq a \leq \|a\| \\ -\|b\|\leq b \leq \|b\| \end{cases}

We add both inequalities and we obtain -(\|a\|+\|b\|) \leq a+b \leq \|a\|+\|b\|

And therefore \|a+b\| \leq \|a\|+\|b\|

Inequality triangular reverse

Dice a, b \in \mathbb{R} is true that \|a\|-\|b\| \leq \|a-b\|

Demonstration: reverse triangular inequality

The inequality triangular inverse is equivalent to prove that -(a-b) \leq \|a\|-\|b\| \leq \|a-b\|

By the inequality triangle has to

\|a\|=\|a-b+b\| \leq \|a-b\|+\|b\|
\|a\|-\|b\| \leq \|a-b\|

with what the right-hand side of the inequality is proved

By the inequality triangle has to

\|b\|=\|b-a+a\| \leq \|b-a\|+\|a\|
\|b\|-\|a\| \leq \|b-a\|
-(a-b) \leq \|a\|-\|b\|

with what the left-hand side of the inequality is proved

Inequality between arithmetic and geometric mean

One of the most useful and popular inequalities is the inequality between the arithmetic and geometric mean (sometimes referred to as AM – GM). Which is defined as follows:

Dice a_1, a_2, \cdots, a_n \in \mathbb{R^+}

We define the arithmetic mean as M_{n, 1}=\frac{a_1, a_2, \cdots, a_n}{n}

It defines the average geomética as M_{n, 0}=\sqrt[n]{a_1, a_2, \cdots, a_n}

And inequality is defined as M_{n , 0} \leq M_{n, 1}

Demonstration: inequality between arithmetic and geometric mean

This demonstration was published in the magazine Mathematical Intelligencer in 2007, vol. 29, number 4 by M. D. Hirschhorn. It is simple to understand and is based on an induction on n

If n=1 then M_{1, 0}=M_{1, 1}

Suppose that it is true for n

Let's use the following observation, seemingly unrelated, to achieve the objective pursued:

x^{n+1}-(n+1)\cdot x + n \geq 0\text{, if }x > 0

The demonstration of this fact is evident by using the identity

x^{n+1}-(n+1)\cdot x +n=(x-1)^2\cdot(x^{n-1}+2\cdot x^{n-2}+\cdots +(n-1)\cdot x+ n)

It can also be tested by induction. If n=1 is inferred from the identity x^2-2\cdot x+1=(x-1)^2

Suppose that it is true for n

(x-1)^2\cdot (x^n+2\cdot x^{n-1}+\cdots + n\cdot x + n +1)= =(x-1)^2\cdot [x\cdot (x^{n-1}+2\cdot x^{n-2}+\cdots + (n-1)\cdot x + n) + n +1]= =x\cdot [(x-1)^2\cdot (x^{n-1}+2\cdot x^{n-2}+\cdots + (n-1)\cdot x + n)] + (x-1)^2\cdot (n +1)= =x\cdot (x^{n+1}-(n+1)\cdot x + n) + (x^2-2\cdot x + 1)\cdot (n +1)=

=x^{n+2}-(n+2)\cdot x + n+1

Now we take \begin{cases}a=\frac{a_1+a_2+\cdots+a_n+a_{n+1}}{n+1} \\ b=\frac{a_1+a_2+\cdots+a_n}{n}\end{cases}, using in the chosen identity x=\frac{a}{b} we will have to \begin{cases}(\frac{a}{b})^{n+1}-(n+1)\cdot\frac{a}{b}+n\geq 0 \\ a^{n+1}\geq ((n+1)\cdot a - n \cdot b)\cdot b^n \end{cases}

Which can be rewritten as

\begin{cases}(\frac{a_1+a_2+\cdots+a_n+a_{n+1}}{n+1})^{n+1}\geq a_{n+1}\cdot (\frac{a_1+a_2+\cdots +a_n}{n})^n \\ (M_{n+1, 1})^{n+1} \geq a_{n+1}\cdot (M_{n,1})^n \end{cases}

since M_{n,0}\geq M_{n,1}

you have to (M_{n+1,1})^{n+1})\leq a_{n+1}\cdot (M_{n,0})^n=a_{n+1}\cdot a_n \cdots a_1

that is equivalent to M_{n+1, 0} \leq M_{n+1,1}

So at the end of the demonstration met the argument of induction

Notes

It is interesting to observe that the equality M_{n,0}=M_{n,1} is true only if and only if a_1=a_2=\cdots=a_n. This fact follows in view of the fact that equality x^{n+1}-(n+1)\cdot x+n=0, for x > 0, is only true if x=1 and has been chosen as the argument of induction suitable

The arithmetic mean and the geometric mean are only two particular cases of a much wider class of means. \forall s \in \mathbb{R}, the mean order s of the positive actual values is defined
a_1,a_2,\cdots,a_n\text{ as }M_{n,s}=(\frac{a_{n}^{s}+\cdots+a_{1}^{s}}{n})^\frac{1}{s}\text{, }s\not=0
M_{n,0} as has already been done. Limit cases can also be considered \begin{cases}M_{n, -\infty}=min\{a_1,a_2\cdots,a_n\} \\ M_{n, +\infty}=max\{a_1,a_2\cdots,a_n\} \end{cases}

Inequality between arithmetic and geometric mean is, in turn, a particular case of a more general chain of inequalities: M_{n, s} \leq M_{n, r}\text{, if }s < r

The average M_{n, -1} is called the harmonic mean and can be inferred from elementary inequality M_{n, 0}\leq M_{n,1} that M_{n, -1}\leq M_{n,0}

Numbers, Complex numbers, Infinitesimal calculus

Complex numbers

Definition

The complex numbers give a solution to the equation x^2+1=0, how x^2=-1 has as a solution x=\pm{\sqrt{-1}}, which has no real solution

We will use as a solution x=\pm{i}. Where i is the imaginary unit of the complex number

We will call complex number expression z=a+b\cdot{i} (binomic form) where a, b\in{\mathbb{R}}. Being to the real part of the number and b its imaginary part

Are an ordered pair of real numbers (a, b) \in{\mathbb{R} \times \mathbb{R}}

In the event that b=0 then we can consider the number as real, since the actual numbers are a subset of the complex

The set of complex numbers is denoted as \mathbb{C}

Two basic types of operations are defined: addition and multiplication

Numbers, Complex numbers, Infinitesimal calculus

Graph of a complex

Graph of a complex

We will draw the graph of a complex

Given a complex number z=a+b\cdot{i}=(a, b) \in \mathbb{R} x \mathbb{R} is an ordered pair of real numbers, represented by the point (a, b) on the XY plane, called complex plane

The dot (a, b) is called the z-complex number

Example of graph of a complex

Given the following complex we want to represent:

z=(-3)+4\cdot{i}

We get the following graph:

Graphical example of complex number

Numbers, Complex numbers, Infinitesimal calculus

Argument

Argument

Argument:

The value of the angle \alpha receives the name of argument

For a given complex number, the argument supports an infinite set of values, which differ from each other in 2\cdot{k}\cdot{\pi}; k\in{\mathbb{Z}}

It's called principal value of the argument the one that meets 0\leq\alpha\leq{2}\cdot{\pi}

It can be calculated by:

\alpha=Arg(z)=atan2(b, a)=\begin{cases} \arctan(\frac{b}{a}) & \text{if }a > 0 \\ \arctan(\frac{b}{a}) + \pi & \text{if }b \geq 0, a < 0 \\ \arctan(\frac{b}{a}) - \pi & \text{if }b < 0, a < 0 \\ \frac{\pi}{2} & \text{if }b > 0, a = 0 \\ \frac{-\pi}{2} & \text{if }b < 0, a = 0 \\ \text{Undefined} & \text{if }b = 0, a = 0 \end{cases}

This result is obtained in radians and sometimes it will be useful to convert it to degrees:

\alpha=\frac{atan2(b, a)\cdot{360}}{2\cdot\pi}

You can also use the following table that expresses trigonometric reasons:

>rad >\sin \alpha >\cos \alpha \tan \alpha = \frac{\sin \alpha}{\cos \alpha}
>0^{\circ} >0 >0 >1 >0
>30^{\circ} >\frac{\pi}{6} >\frac{1}{2} >\frac{\sqrt{3}}{2} >\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}
>45^{\circ} >\frac{\pi}{4} >\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} >\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} >1
>60^{\circ} >\frac{\pi}{3} >\frac{\sqrt{3}}{2} >\frac{1}{2} >\sqrt{3}
>90^{\circ} >\frac{\pi}{2} >1 >0 >\text{Undefined}
>180^{\circ} >\pi >0 >-1 >0
>270^{\circ} >\frac{3\cdot\pi}{2} >-1 >0 >\text{Undefined}

Example of argument

\begin{cases}z=(-3)+4\cdot{i} \\ \alpha=atan2(4, -3)=2.2143\text{ radians}\end{cases}

\alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3}

So we have to:

\alpha\approx\frac{2\cdot\pi}{3}
Numbers, Complex numbers, Infinitesimal calculus

Module and its properties

Module

Module and its properties

Given a complex number z=a+b\cdot{i}, is defined as module, or absolute value to the expression:

r=\|a+b\cdot{i}\|=\sqrt{a^2+b^2}

Given z_1, z_2, \cdots, z_n \in{\mathbb{C}} it is fulfilled that:

  1. \|z_1\cdot{z_2}\cdot\text{ }\cdots\text{ }\cdot{z_n}\|=\|z_1\|\cdot\|z_2\|\cdot\text{ }\cdots\text{ }\cdot\|z_n\|
  2. \|\frac{z_1}{z_2}\|=\frac{\|z_1\|}{\|z_2\|}\text{ con }\|z_2\|\not{=}0
  3. \|z_1+z_2\|\leq\|z_1\|+\|z_2\|
  4. \|z_1+z_2+\text{ }\cdots\text{ }+z_n\|\leq\|z_1\|+\|z_2\|+\text{ }\cdots\text{ }+\|z_n\|

Example of module

z=(-3)+4\cdot{i}

\|z\|=\|(-3)+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

So we have to:

\|z\|=5
Numbers, Complex numbers, Infinitesimal calculus

Forms of a complex number

Forms of a complex number

Form of a complex number:

  • polar
  • trigonometric
  • exponential

Polar

Given the point (a, b) I sharpen the complex number z=a+b\cdot{i} whose module is r and its argument is \alpha, its representation in polar shape it is z=r_\alpha

Example of a polar

z=(-3)+4\cdot{i}

\|z\|=\|(-3)+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

\alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3}

So we have to:

z=5_{\frac{2\cdot\pi}{3}}

Trigonometric

It can also be represented in trigonometric shape where

\begin{cases}a=r\cdot\cos{\alpha} \\ b=r\cdot\sin{\alpha} \end{cases}

with what we have

z=a+b\cdot{i}=r\cdot(\cos{\alpha}+\sin{\alpha}\cdot{i})

Example of a trigonometric

z=(-3)+4\cdot{i}

\|z\|=\|(-3)+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

\alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3}

So we have to:

z=a+b\cdot{i}=5\cdot(\cos(\frac{2\cdot\pi}{3})+\sin(\frac{2\cdot\pi}{3})\cdot{i})

Exponential

It can also be represented in exponentially shape where

\begin{cases}\sin\alpha=\frac{e^{\alpha\cdot{i}}-e^{(-\alpha)\cdot{i}}}{2\cdot{i}} \\ \cos\alpha=\frac{e^{\alpha\cdot{i}}+e^{(-\alpha)\cdot{i}}}{2}\end{cases}

with what we have to

z=a+b\cdot{i}=r\cdot(\cos{\alpha}+\sin{\alpha}\cdot{i})=r\cdot{e^{\alpha\cdot{i}}}

Example of exponential

z=(-3)+4\cdot{i}

\|z\|=\|(-3)+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

\alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3}

So we have to:

z=a+b\cdot{i}=5\cdot{e^{\frac{2\cdot\pi}{3}\cdot{i}}}
Numbers, Complex numbers, Infinitesimal calculus

Moivre formula

Moivre formula

Moivre Formula:

The power n-th entry of a complex number r_\alpha it is another complex module r^n and argument n times the argument of the first

In consequence, we have that

z^n={(r_\alpha)}^n=\overbrace{r_\alpha \cdots r_\alpha}^{n\;\rm times}={(r^n)}_{n\cdot\alpha}

Example of Moivre's formula

\tiny\begin{cases}z^4={((-3)+4\cdot{i})}^4 \\ \|z\|=\|-3+4\cdot{i}\|=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5 \\ \alpha=\frac{atan2(4, -3)\cdot{360}}{2\cdot\pi}=\frac{2.2143\cdot{360}}{2\cdot\pi}=\frac{797.148}{2\cdot\pi}=126.8701^{\circ}\approx\frac{2\cdot\pi}{3} \end{cases}

z^4={(5_{\frac{2\cdot\pi}{3}})}^4=(5^4)_{4\cdot{\frac{2\cdot\pi}{3}}}=625_{\frac{8\cdot\pi}{3}}

So we have to:

z^4={(r_\alpha)}^n=625_{\frac{8\cdot\pi}{3}}

Roots n-ésimas of a complex number

Given a complex number z=r_{\alpha}\text{ si }w=s_{\beta} it is a root n-th we have that

\begin{cases} z={(s_\beta)}^n={(r^n)}_{n\cdot\beta}=r_{\alpha} \\ s^n=r \\ n\cdot{\beta}=\alpha+2\cdot{k}\cdot\pi \\ s=\sqrt[n]{r} \\ \beta=\frac{(\alpha+2\cdot{k}\cdot\pi)}{n} \\ z_k=\sqrt[n]{r}\cdot{e}^{{\frac{(\alpha+2\cdot{k}\cdot\pi)}{n}}\cdot{i}} \end{cases}

with k=0,1,2,\cdots,(n-1) since for k=n gives the same value as for k=0

There are thus n roots of the n-ésimas different if z\not=0

Example of cube roots

\begin{cases} z^3+2=0\rightarrow{z^3=-2}\rightarrow{z=(-2)^{1/3}} \\ \|z\|=\|(-2)+0\cdot{i}\|=\sqrt{(-2)^2+0^2}=\sqrt{4+0}=\sqrt{4}=2 \\ \alpha=\frac{atan2(0, -2)\cdot{360}}{2\cdot\pi}=\frac{\pi\cdot{360}}{2\cdot\pi}=\frac{360}{2}=180^{\circ}\approx\pi \\ \beta=\frac{\pi+2\cdot{k}\cdot{\pi}}{3} \end{cases}

\begin{cases} z_1={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{0}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi}{3}}\cdot{i}} \\ z_2={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{1}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi+2\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{3\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{\pi\cdot{i}} \\ z_3={\sqrt[3]{2}} \cdot{e}^{{\frac{\pi+2\cdot{2}\cdot{\pi}}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi+4\cdot\pi}{3}}\cdot{i}}={\sqrt[3]{2}}\cdot{e}^{{\frac{5\cdot\pi}{3}}\cdot{i}} \end{cases}

So we have to:

z^3+2=0\rightarrow\begin{cases} z_1={\sqrt[3]{2}}\cdot{e}^{{\frac{\pi}{3}}\cdot{i}} \\ z_2={\sqrt[3]{2}}\cdot{e}^{\pi\cdot{i}} \\ z_3={\sqrt[3]{2}}\cdot{e}^{{\frac{5\cdot\pi}{3}}\cdot{i}} \end{cases}
Numbers, Complex numbers, Infinitesimal calculus

Operations of the complex

Operations of complex numbers

The operations of the complex z_1, z_2, z_3 \in \mathbb{C}

\begin{cases}z_1=a+b\cdot{i} \\ z_2=c+d\cdot{i} \\ z_3=e+f\cdot{i} \end{cases}

Operation sum

To define the sum you have to comply with the following properties

Properties of the sum

  1. z_1+z_2 \in{\mathbb{C}}
  2. z_1+(z_2+z_3)=(z_1+z_2)+z_3
  3. \exists\text{ }0 \in \mathbb{C} | z_1+0=0+z_1=z_1
  4. \exists\text{ }(-z_1) \in \mathbb{C} | z_1+(-z_1)=(-z_1)+z_1=0
  5. z_1+z_2=z_2+z_1

As meets the above properties, the pair (C, +) has structure of grupo abeliano

Example of sum

\begin{cases}z_1=(-3)+4\cdot{i} \\ z_2=5-2\cdot{i} \\ z=z_1+z_2 \end{cases}

z=((-3)+4\cdot{i})+(5-2\cdot{i})=((-3)+5)+(4-2)\cdot{i}=2+2\cdot{i}

So we have to:

z=z_1+z_2=2+2\cdot{i}

Operation subtraction

Subtraction is a special case of the sum, to realize it you only have to use the sum of the opposite of its real parts and the sum of the opposites of the imaginary parts

Example of subtraction

\begin{cases}z_1=(-3)+4\cdot{i} \\ z_2=5-2\cdot{i} \\ z=z_1-z_2 \end{cases}

z=((-3)+4\cdot{i})-(5-2\cdot{i})=((-3)-5)+(4+2)\cdot{i}=(-8)+6\cdot{i}

So we have to:

z=z_1-z_2=(-8)+6\cdot{i}

Operation multiplication

To define multiplication we have to comply with the following properties

Properties of multiplication

  1. z_1\cdot{z_2} \in{\mathbb{C}}
  2. z_1\cdot(z_2\cdot{z_3})=(z_1\cdot{z_2})\cdot{z_3}
  3. \exists\text{ }1 \in \mathbb{C} | z_1\cdot{1}=1\cdot{z_1}=z_1
  4. \forall\text{ }(z_1) \in \mathbb{C} (z_1\ne{0}), \exists\text{ }{z_1}^-1 \in \mathbb{C} | z_1\cdot{z_1}^-1={z_1}^-1\cdot{z_1}=1
  5. z_1\cdot{z_2}=z_2\cdot{z_1}

As meets the above properties, the pair (\mathbb{C}-\{0\}, \cdot) has structure of grupo abeliano

Example of multiplication

\begin{cases}z_1=(-3)+4\cdot{i} \\ z_2=5-2\cdot{i} \\ z=z_1\cdot z_2 \end{cases}

z=((-3)+4\cdot{i})\cdot(5-2\cdot{i})=((-3)\cdot{5})+((-3)\cdot(-2)\cdot{i}))+(4\cdot{i}\cdot{5})+(4\cdot{i}+(-2)\cdot{i})=(-15)+6\cdot{i}+20\cdot{i}-8\cdot{i^2}=(-15)+26\cdot{i}-8\cdot{i^2}=(-15)+8+26\cdot{i}=(-7)+26\cdot{i}

In the case of obtaining a value of an imaginary square (i^2), we will take that value as your real opposite. In the example we appeared (-8)\cdot{i}, we'll take 8 as the actual number that will add to your real part

So we have to:

z=z_1\cdot{z_2}=(-7)+26\cdot{i}

Operation division

Division is a special case of multiplication, to make it only multiply the numerator and denominator by the conjugate of the denominator

The conjugated is obtained by changing the sign of the imaginary part of the denominator

Example of division

\begin{cases}z_1=(-3)+4\cdot{i} \\ z_2=5-2\cdot{i} \\ z={z_1 \over z_2} \end{cases}

z={((-3)+4\cdot{i})\over(5-2\cdot{i})}={((-3)+4\cdot{i})\over(5-2\cdot{i})}\cdot{(5+2\cdot{i})\over(5+2\cdot{i})}=
{(((-3)\cdot{5})+((-3)\cdot{2}\cdot{i})+(4\cdot{i}\cdot{5})+(4\cdot{i}\cdot{2}\cdot{i}))\over((5\cdot{5})+(5\cdot{2}\cdot{i})+((-2)\cdot{i}\cdot{5})+((-2)\cdot{i}\cdot{2}\cdot{i}))}=
{(-15)+(-6)\cdot{i}+20\cdot{i}+(8\cdot{i^2})\over{25}+10\cdot{i}+(-10)\cdot{i}+(-4)\cdot{i^2}}={(-15)+14\cdot{i}+(8\cdot{i^2})\over{25}+(-4)\cdot{i^2}}=
{(-15)-8+14\cdot{i}\over{25}+4}={(-23)+14\cdot{i}\over{29}}={(-23)\over{29}}+{14\cdot{i}\over{29}}

So we have to:

z={z_1 \over z_2}={(-23)\over{29}}+{14\cdot{i}\over{29}}

Operation sum and product

In addition to sum and product share the following property

Property of the sum and the product

  1. z_1\cdot(z_2+z_3)=z_1\cdot{z_2}+z_1\cdot{z_3}

As the previous property complies with, the terna (\mathbb{C}, +, \cdot) has structure of body-commutative

Example of sum and multiplication

\begin{cases}z_1=(-3)+4\cdot{i} \\ z_2=5-2\cdot{i} \\ z=z_1\cdot(z_2+z_3) \end{cases}

z=((-3)+4\cdot{i})\cdot((5-2\cdot{i})+(8+3\cdot{i}))=((-3)+4\cdot{i})\cdot(5-2\cdot{i}))+((-3)+4\cdot{i})\cdot(8+3\cdot{i}))=((-3)\cdot{5})+((-3)\cdot(-2)\cdot{i})+(4\cdot{i}\cdot{5})+(4\cdot{i}+(-2)\cdot{i})+((-3)\cdot{8})+((-3)\cdot{3}\cdot{i})+(4\cdot{i}\cdot{8})+(4\cdot{i}\cdot{3}\cdot{i})=((-15)+6\cdot{i}+20\cdot{i}-8\cdot{i^2})+((-24)-9\cdot{i}+32\cdot{i}+12\cdot{i^2})=(-15)-24+6\cdot{i}+20\cdot{i}-9\cdot{i}+32\cdot{i}-8\cdot{i^2}+12\cdot{i^2}=(-39)+49\cdot{i}+4\cdot{i^2}=(-39)-4+49\cdot{i}=(-43)+49\cdot{i}

So we have to:

z=z_1\cdot(z_2+z_3)=(-43)+49\cdot{i}