Complex Numbers
Complex numbers are a set of numbers that consist (individually) of a real part and an imaginary part
To better understand this let's say we have a quadratic equation
..x² + x +1 =0 by method of formula
The quadratic formula is given as.......
X =( -b ± √b2 - 4ac)
We want to see if we can use the quadratic formula to solve that quadratic equation
x² + x + 1= 0
Using the quadratic formula
The root of this equation is
X =( -b ± √b2 - 4ac)
_________________
2a
a is the coefficient of the term with the highest power (1x²)
b is the coefficient of the x term (1x) ..and c is the constant (which happens to be 1 also)..
Applying this formula we can notice that we would have to take the square root of b2 - 4ac which is a negative number or more precisely we can say that for this equation
b2 - 4ac is < 0...
To solve this problem..
The concept of using imaginary numbers that represented the square root of negative 1 was introduced..
Imaginary numbers were written mostly as i or j
now that we have Introduced imaginary numbers we can conveniently write the following..
The roots of the equation can now be expressed as
-0.5 ± 1.5j
This way of writing roots is what we regard to as a complex number..
Note that this root has a real part of (-0.5) and an imaginary part of......?
1.5j correct!!
lets now go on into the complex number system..
Addition subtraction multiplication and division of complex numbers
Now say we have two complex numbers we wish to add together..
Z1 = 1 - 2j
Z2 = 3 + 3j
Z1 + z2 is simply gotten by collecting like terms and adding them up
Collect the real numbers and add them up.. And add the imaginary parts and add them up
This becomes
1 + 3 -2j + 3j
Which is.....??
4 + j
Subtraction is done in the same way....
Now try this..
Z1 = 3 + j
Z2 = 2 - j
Before we continue, there's something i should mention, and that is multiples or powers of imaginary numbers
We've already established that an imaginary number j = √-1
j² = (√-1)² ,the square root and square cancel out leaving
j² = -1
j³ = -j
j⁴ = j² * j² = -1 * -1 = 1
Keep j⁴ = 1 to heart you'll use it to simply imaginary numbers with higher powers
say j²⁰ , this could be simplified as (j⁴)⁵
since we know that j⁴ = 1
∴ we have that j²⁰ = 1⁵ = 1
Now that we know this,
Let's move on to multiplication
To multiply two complex numbers
Z1 ND z2..
It's the same way you would two algebraic expressions
If z1 = 3-7j
Z2 = 2 + j4
And we need to find z1z2 our answer would be 34-2j
Alright moving on to division
Complex numbers can be divided by a real number.. And by another complex number as well these two cases - we will treat
If a complex number is put in division with a real number..
We simply divide the real party of the complex number by the real number divisor and the coefficient of the imaginary part by the divisor as well
For example
If z1 = 3 - j7
And we divide it by 2
We would have
(3 - j7)/2 = 3/2 - j(7/2)
This would be.....??
For example..
If z1 = 3-j7 and z2 = 2 - j
To divide z1 by z2
We need to multiply the numerator ND denominator by the conjugate of the denominator
That'd be:
2 + j
So we have:
Z1/ z2 =
(3 + j)/(2 - j) * (2+j)/(2+j)
as shown below:
Alright then We move on to:
Z = a + jb
Another way of representing complex number z is by plotting it on an Argand diagram.. Which is simply the x-y plane.. Where the real part of the the complex number is plotted on the x axis and the imaginary Part on the y axis
Below is an example of an Argand Diagram
Note that the real part 'a' of the complex number is on the x axis and the imaginary part on the y axis..for example to represent the complex no z = 1 - 2j on an Argand diagram it'd be..??
In the next part ...we'll cover even more ways of representing complex other than using Argand diagrams
To better understand this let's say we have a quadratic equation
..x² + x +1 =0 by method of formula
The quadratic formula is given as.......
X =( -b ± √b2 - 4ac)
_________________
2a
Alright pick up your Pens and jotters.. We want to see if we can use the quadratic formula to solve that quadratic equation
x² + x + 1= 0
Using the quadratic formula
The root of this equation is
X =( -b ± √b2 - 4ac)
_________________
2a
a is the coefficient of the term with the highest power (1x²)
b is the coefficient of the x term (1x) ..and c is the constant (which happens to be 1 also)..
Applying this formula we can notice that we would have to take the square root of b2 - 4ac which is a negative number or more precisely we can say that for this equation
b2 - 4ac is < 0...
To solve this problem..
The concept of using imaginary numbers that represented the square root of negative 1 was introduced..
Imaginary numbers were written mostly as i or j
now that we have Introduced imaginary numbers we can conveniently write the following..
The roots of the equation can now be expressed as
-0.5 ± 1.5j
This way of writing roots is what we regard to as a complex number..
Note that this root has a real part of (-0.5) and an imaginary part of......?
1.5j correct!!
lets now go on into the complex number system..
Addition subtraction multiplication and division of complex numbers
Now say we have two complex numbers we wish to add together..
Z1 = 1 - 2j
Z2 = 3 + 3j
Z1 + z2 is simply gotten by collecting like terms and adding them up
Collect the real numbers and add them up.. And add the imaginary parts and add them up
This becomes
1 + 3 -2j + 3j
Which is.....??
4 + j
Subtraction is done in the same way....
Now try this..
Z1 = 3 + j
Z2 = 2 - j
Before we continue, there's something i should mention, and that is multiples or powers of imaginary numbers
We've already established that an imaginary number j = √-1
j² = (√-1)² ,the square root and square cancel out leaving
j² = -1
j³ = -j
j⁴ = j² * j² = -1 * -1 = 1
Keep j⁴ = 1 to heart you'll use it to simply imaginary numbers with higher powers
say j²⁰ , this could be simplified as (j⁴)⁵
since we know that j⁴ = 1
∴ we have that j²⁰ = 1⁵ = 1
Now that we know this,
Let's move on to multiplication
To multiply two complex numbers
Z1 ND z2..
It's the same way you would two algebraic expressions
If z1 = 3-7j
Z2 = 2 + j4
And we need to find z1z2 our answer would be 34-2j
Alright moving on to division
Complex numbers can be divided by a real number.. And by another complex number as well these two cases - we will treat
If a complex number is put in division with a real number..
We simply divide the real party of the complex number by the real number divisor and the coefficient of the imaginary part by the divisor as well
For example
If z1 = 3 - j7
And we divide it by 2
We would have
(3 - j7)/2 = 3/2 - j(7/2)
This would be.....??
For example..
If z1 = 3-j7 and z2 = 2 - j
To divide z1 by z2
We need to multiply the numerator ND denominator by the conjugate of the denominator
That'd be:
2 + j
So we have:
Z1/ z2 =
(3 + j)/(2 - j) * (2+j)/(2+j)
as shown below:
Alright then We move on to:
Argand Diagram
Complex numbers as I stated earlier on can be represented in a number of ways.... One of which we have become quite acquainted with is the standard formZ = a + jb
Another way of representing complex number z is by plotting it on an Argand diagram.. Which is simply the x-y plane.. Where the real part of the the complex number is plotted on the x axis and the imaginary Part on the y axis
Below is an example of an Argand Diagram
Note that the real part 'a' of the complex number is on the x axis and the imaginary part on the y axis..for example to represent the complex no z = 1 - 2j on an Argand diagram it'd be..??
In the next part ...we'll cover even more ways of representing complex other than using Argand diagrams
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