what would -i^-i be, would it just be 2^2

it would be i^3(i^3) = -1^(-1) = 1/-1 = -1

What is the real world application for this??

design, simulation, analysis of normal and semiconductor circuits, acoustics and speakers, physics., mechanical system vibration, automotive exhaust note tuning, guitar pickups and boutique high power tube/solid state amplifiers, chemical engineering linear/non linear flows, financial modeling, statistics and big data,

Can you have different answers to simplifying depending on what numbers you take from the original, or would those be wrong? For example: Problem 3, instead of using 4 and 6 I used 8 and 3 and it came out to be 2i x square of 2 x square of 2 x square of 3, but it was counted as wrong. Was it wrong because it wasn't what Kahn had, or because it was just wrong?

They were asking for the square root. The square root of 4 is 2 so you would have 2i sqrt(6) ... The cubed root of 8 is 2 not the square root.

Does it matter if the i is in front or behind of the solution.

As long as it is clear what the i is affecting, you can do both. EG (2 + 3i) + (4 + 5i) = (2 + 4) + i(3 + 5) or (2 + 4) + (3 + 5)i However, there are conventions. When we simplify the above we would normally write 6 + 8i, not 6 + i8, but both are fine, but the second one just looks weird. For example, you are used to the notation "1 + 2", but the following notations "+ 1 2" or "1 2 +" are also acceptable in many situations, through they probably looks weird to you now. (The 1st is Polish Notation, the 2nd Reverse Polish Notation) Another convention is to place the i before the radical, eg i√8. If you want to place it after, make sure to use parenthesis: (√8)i or √8(i), so as to avoid confusion. If you write √8i, do you mean (√8)i or √(8i)? As you keep studying, you will get more and more exposure to the notation conventions we use.

If imaginary numbers aren't real, how is it possible to use them in real life? You can't count things that don't exist so how do you use them?

None of the numbers you use in life are real. Can you show me a 3? Not a drawing or a representation of a 3, but the actual number 3? Of course not. It's just an abstraction. You mention counting, but most numbers aren't used for counting either. You can't have exactly √2 apples, or any irrational number of apples. That would require splitting atoms and quarks in impossible ways. Yet a vast majority of the real numbers are irrational. They're not about counting either. Numbers are just concepts that follow certain rules. The misleadingly-named real numbers are defined as a complete ordered field. The word "field" just means that they follow 9 certain rules, like "for every real number x, x+0=x" Likewise, "ordered" just adds about 3 more rules, and "complete" adds one more. Any relation to real life is just the result of people applying these abstractions to real-world problems. To get the complex numbers, we do a similar thing. Take the real numbers and add in 1. Every real number is complex. 2. There is a complex number i such that i²= -1. 3. The sum of two complex numbers is complex. 4. The product of two complex numbers is complex. 5. For any two complex numbers a and b, a^b is complex. Now we have this concept of "the complex numbers" that we can further explore. Application to reality is not necessary.

what is the cube root of -1?

The cube root of -1 is -1. -1^3=-1. Since the power is an odd number, the sign does not change. For example, *-1* to the power of *1/9*, *1/11*, or any odd number would be -1.

Why write sqrt(-5) as i*sqrt(5) but not sqrt(5)*i ?

Why not? You can write it both ways, thanks to the commutative property:```x * y = y * x```

What genius actually had the time to create i and discover its properties?

It was primarily Vieta, Cardan, Tartaglia, and their colleagues/rivals in the mid-1500s. At the time, searching for reliable ways to find the roots of third-degree polynomials was a popular problem, and the complex numbers fell out of their work. Even negative numbers were widely hated and rejected at the time, and the idea of complex numbers even more so, but they couldn't get around the fact that they _worked_. Eventually, they began to study them in their own right, and not just as tools to solve polynomials.

Main content

Algebra 2

Course: Algebra 2 > Unit 2

Lesson 1: The imaginary unit i

Intro to the imaginary numbers

CCSS.Math:

HSN.CN.A.1

Google Classroom

Learn about the imaginary unit i, about the imaginary numbers, and about square roots of negative numbers.

In your study of mathematics, you may have noticed that some quadratic equations do not have any real number solutions.

For example, try as you may, you will never be able to find a real number solution to the equation x, squared, equals, minus, 1. This is because it is impossible to square a real number and get a negative number!

However, a solution to the equation x, squared, equals, minus, 1 does exist in a new number system called the complex number system.

The imaginary unit

The backbone of this new number system is the imaginary unit, or the number i.

The following is true of the number i:

i, equals, square root of, minus, 1, end square root
i, squared, equals, minus, 1

The second property shows us that the number i is indeed a solution to the equation x, squared, equals, minus, 1. The previously unsolvable equation is now solvable with the addition of the imaginary unit!

Pure imaginary numbers

The number i is by no means alone! By taking multiples of this imaginary unit, we can create infinitely many more pure imaginary numbers.

For example, 3, i, i, square root of, 5, end square root, and minus, 12, i are all examples of pure imaginary numbers, or numbers of the form b, i, where b is a nonzero real number.

Taking the squares of these numbers sheds some light on how they relate to the real numbers. Let's investigate this by squaring the number 3, i. The properties of integer exponents remain the same, so we can square 3, i just as we'd imagine.

$\begin{aligned}(3i)^2&=3^2i^2\\ \\ &=9{i^2}\\\\ \end{aligned}$

Using the fact that i, squared, equals, minus, 1, we can simplify this further as shown.

$\begin{aligned}\phantom{(3i)^2} &=9\goldD{i^2}\\\\ &=9(\goldD{-1})\\\\ &=-9 \end{aligned}$

The fact that left parenthesis, 3, i, right parenthesis, squared, equals, minus, 9 means that 3, i is a square root of minus, 9.

Check your understanding

What is left parenthesis, 4, i, right parenthesis, squared?

[I need help!]

Which of the following is a square root of minus, 16?

[I need help!]

In this way, we can see that pure imaginary numbers are the square roots of negative numbers!

Simplifying pure imaginary numbers

The table below shows examples of pure imaginary numbers in both unsimplified and in simplified form.

Unsimplified form	Simplified form
square root of, minus, 9, end square root	3, i
square root of, minus, 5, end square root	i, square root of, 5, end square root
minus, square root of, minus, 144, end square root	minus, 12, i

But just how do we simplify these pure imaginary numbers?

Let's take a closer look at the first example and see if we can think through the simplification.

Original equivalence	Thought process
$\begin{aligned}\sqrt{-9} = 3i \end{aligned}$	The square root of minus, 9 is an imaginary number. The square root of 9 is 3, so the square root of negative 9 is start text, 3, end text imaginary units, or 3, i.

The following property explains the above "thought process" in mathematical terms.

For a, is greater than, 0, square root of, minus, a, end square root, equals, i, square root of, a, end square root

If we put this together with what we already know about simplifying radicals, we can simplify all pure imaginary numbers. Let's look at an example.

Example

Simplify square root of, minus, 18, end square root.

Solution

First, let's notice that square root of, minus, 18, end square root is an imaginary number, since it is the square root of a negative number. So, we can start by rewriting square root of, minus, 18, end square root as i, square root of, 18, end square root.

Next we can simplify square root of, 18, end square root using what we already know about simplifying radicals.

The work is shown below.

\begin{aligned}\sqrt{-18}&=i\sqrt{18}&&\small{\gray{\text{For $a>0$, $\sqrt{-a}=i\sqrt{a}$}}}\\\\ &=i\cdot\sqrt{9\cdot 2}&&\small{\gray{\text{$9$ is a perfect square factor of $18$}}}\\\\ &=i\sqrt{9}\cdot\sqrt{2}&&\small{\gray{\sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \text{ when } a, b\geq0}} \\\\ &=i\cdot 3\cdot \sqrt2&&\small{\gray{\sqrt{9}=3}}\\\\ &=3i\sqrt{2}&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}

So it follows that square root of, minus, 18, end square root, equals, 3, i, square root of, 2, end square root.

[I'm still confused. I'd like to see another example.]

Let's practice some problems

Problem 1

Simplify square root of, minus, 25, end square root.

[I need help!]

Problem 2

Simplify square root of, minus, 10, end square root.

[I need help!]

Problem 3

Simplify square root of, minus, 24, end square root.

[I need help!]

Why do we have imaginary numbers anyway?

The answer is simple. The imaginary unit i allows us to find solutions to many equations that do not have real number solutions.

This may seem weird, but it is actually very common for equations to be unsolvable in one number system but solvable in another, more general number system.

Here are some examples with which you might be more familiar.

With only the counting numbers, we can't solve x, plus, 8, equals, 1; we need the integers for this!
With only the integers, we can't solve 3, x, minus, 1, equals, 0; we need the rational numbers for this!
With only the rational numbers, we can't solve x, squared, equals, 2. Enter the irrational numbers and the real number system!

And so, with only the real numbers, we can't solve x, squared, equals, minus, 1. We need the imaginary numbers for this!

As you continue to study mathematics, you will begin to see the importance of these numbers.

Want to join the conversation?

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Dlnickelson17
Posted 7 years ago. Direct link to Dlnickelson17's post “what would -i^-i be, woul...”
what would -i^-i be, would it just be 2^2
Button navigates to signup pageComment on Dlnickelson17's post “what would -i^-i be, woul...”
(79 votes)
Answer
- ArjanK
  Posted 5 months ago. Direct link to ArjanK's post “it would be i^3(i^3) = -1...”
  it would be i^3(i^3) = -1^(-1) = 1/-1 = -1
  Comment on ArjanK's post “it would be i^3(i^3) = -1...”
  (6 votes)
Michael Mendoza
Posted 7 years ago. Direct link to Michael Mendoza's post “What is the real world ap...”
What is the real world application for this??
Button navigates to signup pageComment on Michael Mendoza's post “What is the real world ap...”
(52 votes)
Answer
- kurt westphal
  Posted 7 years ago. Direct link to kurt westphal's post “design, simulation, analy...”
  design, simulation, analysis of normal and semiconductor circuits, acoustics and speakers, physics., mechanical system vibration, automotive exhaust note tuning, guitar pickups and boutique high power tube/solid state amplifiers, chemical engineering linear/non linear flows, financial modeling, statistics and big data,
  Comment on kurt westphal's post “design, simulation, analy...”
  (149 votes)
Daisy Dukealoopakiss
Posted 7 years ago. Direct link to Daisy Dukealoopakiss's post “Can you have different an...”
Can you have different answers to simplifying depending on what numbers you take from the original, or would those be wrong? For example: Problem 3, instead of using 4 and 6 I used 8 and 3 and it came out to be 2i x square of 2 x square of 2 x square of 3, but it was counted as wrong. Was it wrong because it wasn't what Kahn had, or because it was just wrong?
Button navigates to signup pageComment on Daisy Dukealoopakiss's post “Can you have different an...”
(20 votes)
Answer
- jesse.l.kent
  Posted 7 years ago. Direct link to jesse.l.kent's post “They were asking for the ...”
  They were asking for the square root. The square root of 4 is 2 so you would have 2i sqrt(6) ... The cubed root of 8 is 2 not the square root.
  Comment on jesse.l.kent's post “They were asking for the ...”
  (34 votes)
Sarah Myers
Posted 7 years ago. Direct link to Sarah Myers's post “Does it matter if the i i...”
Does it matter if the i is in front or behind of the solution.
Button navigates to signup pageComment on Sarah Myers's post “Does it matter if the i i...”
(11 votes)
Answer
- Stefen
  Posted 7 years ago. Direct link to Stefen's post “As long as it is clear wh...”
  As long as it is clear what the i is affecting, you can do both.
  EG (2 + 3i) + (4 + 5i) = (2 + 4) + i(3 + 5) or (2 + 4) + (3 + 5)i
  However, there are conventions.
  When we simplify the above we would normally write 6 + 8i, not 6 + i8, but both are fine, but the second one just looks weird. For example, you are used to the notation "1 + 2", but the following notations "+ 1 2" or "1 2 +" are also acceptable in many situations, through they probably looks weird to you now. (The 1st is Polish Notation, the 2nd Reverse Polish Notation)
  
  Another convention is to place the i before the radical, eg i√8. If you want to place it after, make sure to use parenthesis: (√8)i or √8(i), so as to avoid confusion. If you write √8i, do you mean (√8)i or √(8i)?
  
  As you keep studying, you will get more and more exposure to the notation conventions we use.
  Button navigates to signup page
  (30 votes)
Syeda5
Posted 6 years ago. Direct link to Syeda5's post “If imaginary numbers aren...”
If imaginary numbers aren't real, how is it possible to use them in real life? You can't count things that don't exist so how do you use them?
Button navigates to signup pageComment on Syeda5's post “If imaginary numbers aren...”
(5 votes)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “None of the numbers you u...”
  None of the numbers you use in life are real. Can you show me a 3? Not a drawing or a representation of a 3, but the actual number 3? Of course not. It's just an abstraction.
  
  You mention counting, but most numbers aren't used for counting either. You can't have exactly √2 apples, or any irrational number of apples. That would require splitting atoms and quarks in impossible ways. Yet a vast majority of the real numbers are irrational. They're not about counting either.
  
  Numbers are just concepts that follow certain rules. The misleadingly-named real numbers are defined as a complete ordered field. The word "field" just means that they follow 9 certain rules, like "for every real number x, x+0=x" Likewise, "ordered" just adds about 3 more rules, and "complete" adds one more. Any relation to real life is just the result of people applying these abstractions to real-world problems.
  
  To get the complex numbers, we do a similar thing. Take the real numbers and add in
  1. Every real number is complex.
  2. There is a complex number i such that i²= -1.
  3. The sum of two complex numbers is complex.
  4. The product of two complex numbers is complex.
  5. For any two complex numbers a and b, a^b is complex.
  
  Now we have this concept of "the complex numbers" that we can further explore. Application to reality is not necessary.
  Comment on kubleeka's post “None of the numbers you u...”
  (41 votes)
Jimmy Ye
Posted 9 days ago. Direct link to Jimmy Ye's post “I got this tysm so much s...”
I got this tysm so much sal u changed my life more than dhar man
Button navigates to signup pageComment on Jimmy Ye's post “I got this tysm so much s...”
(18 votes)
Answer
- McBoyle, Joseph
  Posted 2 days ago. Direct link to McBoyle, Joseph's post “That's so crazy.”
  That's so crazy.
  Button navigates to signup page
  (1 vote)
Turtlewolf
Posted 13 days ago. Direct link to Turtlewolf's post “this is easy yay omg I go...”
this is easy yay omg I got scared when I saw i
Button navigates to signup pageButton navigates to signup page
(11 votes)
Answer
Selena Bean UwU
Posted 5 years ago. Direct link to Selena Bean UwU's post “What genius actually had ...”
What genius actually had the time to create i and discover its properties?
Button navigates to signup pageComment on Selena Bean UwU's post “What genius actually had ...”
(2 votes)
Answer
- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “It was primarily Vieta, C...”
  It was primarily Vieta, Cardan, Tartaglia, and their colleagues/rivals in the mid-1500s. At the time, searching for reliable ways to find the roots of third-degree polynomials was a popular problem, and the complex numbers fell out of their work. Even negative numbers were widely hated and rejected at the time, and the idea of complex numbers even more so, but they couldn't get around the fact that they worked. Eventually, they began to study them in their own right, and not just as tools to solve polynomials.
  Button navigates to signup page
  (15 votes)
jarron eastman
Posted 4 years ago. Direct link to jarron eastman's post “what is the cube root of ...”
what is the cube root of -1?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- KaiLiu21
  Posted 4 years ago. Direct link to KaiLiu21's post “The cube root of -1 is -1...”
  The cube root of -1 is -1. -1^3=-1. Since the power is an odd number, the sign does not change. For example, *-1* to the power of 1/9, 1/11, or any odd number would be -1.
  Button navigates to signup page
  (11 votes)
Apichat Klang
Posted a month ago. Direct link to Apichat Klang's post “Why write sqrt(-5) as i*s...”
Why write sqrt(-5) as i*sqrt(5) but not sqrt(5)*i ?
Button navigates to signup pageComment on Apichat Klang's post “Why write sqrt(-5) as i*s...”
(3 votes)
Answer
- HSstudent16
  Posted a month ago. Direct link to HSstudent16's post “Why not? You can write i...”
  Why not? You can write it both ways, thanks to the commutative property:
  x * y = y * x
  Comment on HSstudent16's post “Why not? You can write i...”
  (9 votes)