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}