Complex Numbers

Imaginary Numbers

OK kiddies let’s learn section 2.4

Previously we have learned that there are only ‘real numbers’. Which makes sense how can there be numbers that are non-real or imaginary? Well in fact there are both real and imaginary numbers.

How can this be … Well how does one explain x= sqrt{-1} ?

No number we have learned can be inputted for x to solve this problem. So how can we square a number and get a negative result?

But if we "imagine" that we can … and it turns out that such a number, which may seem impossible, is actually useful and can solve real problems.

So… this is where imaginary numbers come in.

Imaginary numbers are: A number that when squared gives a negative result.

The "unit" imaginary numbers (the same as "1" for Real Numbers) is \[\sqrt{-1}\]  (the square root of negative one), and its symbol is i.

To solve follow this formula:

\[\sqrt{-a}\]= \[\sqrt{-1*a}\]= \[\sqrt{-1}\] * \[\sqrt{a}\]= a+bi where a is real and bi is imaginary

Example:

\[\Sqrt{-9}\] = \[\sqrt{9}\] * \[\sqrt {-1}\] = 3 and I written in a = bi form is 3i.

or

\[\Sqrt{-100}\] = \[\sqrt{100}\] *\[\sqrt{-1}\] = 10i.

So to review this is a breakdown of all numbers we have learned up to now…

Real numbers;

Rational or Irrational

Terminating decimals or Repeating decimals

Integers or Non integers

Whole numbers or Negative numbers

Natural numbers or zero

But this leaves us with one more dilemma what is a number that is compose of both a real and imaginary number such as

\[\Sqrt{-75}\] = 5i \[\sqrt{3}\]

Which is it real or imaginary?

Well it is both we call these numbers Complex numbers, they are a larger overarching group that includes all numbers.

So now the classifications look like:

Complex Numbers

Real numbers or imaginary

Rational or Irrational

Terminating decimals or Repeating decimals

Integers or Non integers

Whole numbers or Negative numbers

Natural numbers or zero

How to simplify complex numbers:

1)      i¹ = i

2)      i² = sqrt{-1} = -1

3)      i³ = i² * i¹ = -i

4)      i⁴= i² * i² = 1

This pattern of i, -1, -i, 1 repeats in this order for all degrees, so

what is i²²³ = ???

To solve divide 223/4 = 55 r.3
which means you can go through that repeating pattern 55 times and will stop on the third on or -i

So i²²³ = -i

 Post By: Tommy Scheidt