If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Intro to the imaginary numbers

Learn about the imaginary unit, "i", a unique number defined as the square root of -1. It's a key part of complex numbers, which are in the form a + bi. The powers of "i" cycle through a set of values. Created by Sal Khan.

Want to join the conversation?

  • leaf blue style avatar for user fuller.jeremiah
    What happens when you put i to the power of i?
    (518 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Eric.Bannatyne
      If you've learned about Euler's formula, you'd know that e^(ix) = cos(x) + i*sin(x) (Khan has videos about this formula). Say we want to find a value for x which makes this equal to i, so want cos(x) = 0 and sin(x) = 1. x = pi/2 is a value which satisfies this. So e^(i*pi/2) = i.

      So now that we have this seemingly more complicated way of writing i, we can substitute it for the base of our exponent and raise it to the power of i. So we get (e^(i*pi/2))^i.

      Recall that (a^b)^c = a^(b*c), so we can apply the same idea here. (e^(i*pi/2))^i = e^(i*i*pi/2) = e^(i^2 * pi/2).

      Now by definition i^2 = -1 so that cancels out and we get i^i = e^(-pi/2), which is approximately equal to about 0.207...

      Interesting how an imaginary number raised to the power of an imaginary number results in a real number.
      (505 votes)
  • orange juice squid orange style avatar for user Madeline
    How can i^2 be equal to a negative number if a square cannot equal a negative? Why is i useful? How come it was invented?
    (159 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user cmiddleman8
      You're question is the answer. There is no "real" number that you could find on the number line that when squared would equal a negative number, so mathematicians created (in their diabolical laboratory of weird and mystical things) a number that would fulfill those properties. In terms of usefulness, you'll see i pop up in quadratics that don't touch the x-axis and later if you do work in the complex plane. It's also in some advanced calculus and other topics, so don't think you can run away from it just because it's not real ;)!
      (305 votes)
  • leaf grey style avatar for user cmaryk12296
    How could I apply Imaginary Numbers to the real world, what fields require knowledge of these numbers?
    (70 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Jackson Wilcox
    Whats an example of a real life problem related to Imaginary numbers? Something usefull as apposed to a crazy theory someone came up with because they didn't understand negative roots...
    (67 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user blahdee327
      complex numbers(numbers with i in them)are great for dealing with things that are periodic due to the properties mentioned in this vid.
      For example, something oscillating back and forth on a spring can be thought of in terms of complex numbers. On the other hand, you can also describe it using sines and cosines... but sometimes it just works out simpler to do it the complex way (as the above poster mentioned, there are cases where the only way to get the answer is going through complex analysis, and arriving at a real answer).
      (61 votes)
  • eggleston green style avatar for user Σthan
    Is there any way to find a pattern in the power of i? For example, if a friend were to ask me to find the answer for i to the power of 542, is there a pattern in the power I can follow, or is the only way to count up by multiples of 4 (since the pattern is 1, i, -1, -i, repeat).
    (10 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      Counting up by multiples of 4 can be achieved by dividing by 4.
      The pattern is i, i^2=-1, i^3=-i, i^4=1
      When you divide by 4, the remainder will always be either 0.25, 0.5, 0.75, or 0.
      0.25 means i
      0.5 means i^2=-1
      0.75 means i^3=-i
      and no remainder (0) means i^4 = 1
      So, given i^542, divide 542 by 4, to get 135.5, so the answer is i^2=-1
      Lets try i^333. 333/4= 83.25, so the answer is i.
      and i^7, we have 7/4=1.75, so the answer is i^3=-i.
      Hope that helps.
      (34 votes)
  • male robot hal style avatar for user Kunal Chugh
    At Sal says E. What is that number and can i have a link to it?
    (7 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      e is a constant that comes up in math and science all the time. It is an irrational and transcendental number the first few digits of which are 2.71828...

      e is officially defined as:
      lim h→0 (1+h)^(1/h)
      This same definition can also be expressed as:
      lim h→ ∞ (1+1/h)^(h)
      (19 votes)
  • primosaur seedling style avatar for user Harley-Payj
    @-58 Sal discusses:
    i^4=(i)(i^3)=(i)(-i)=(-1)(i)(i)...
    -How does the equation go from (i)(-i) to (-1)(i)(i)?
    I tried to grasp the concept by studying how he defines (i^3) but it isn't quite clicking? If someone could further break this down for me that would be wonderful.
    (10 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Daniel Choi
      To explain this, first you must know that i^2 is equal to -1 because i=√(-1) and √(-1)* √(-1) cancels out the √ and therefore gives you a -1. So back to the question, i^4 can be broken down into (i)(i^3) because if you multiply (i) with (i^3) since they have the same base it would be just be i^(1+3) which would give i^4. The (i)(i^3) can be broken down into (i)(-i) because remember i=√(-1), if you multiply √(-1) with √(-1) ( which we got -1 before) and multiply it once more with i, then you get a negative i (because -1*i= -i). Then, afterwards you can break down the -i to become (-1)(i) and you can add the part above which was i to give you an answer (-1)(i)(i). There are two i’s and as explained above i^2=-1. To find the final answer, by multiplying the -1 with the -1, you will get an answer of 1. Hope this helps!
      (9 votes)
  • blobby green style avatar for user Stephanie  Warner
    when do i use this in real life?
    (9 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Kunal Chugh
    At Sal says E. Now I know what it is @ Just Keith. But when and why do you use it? Is it like pi (3.1415926535897932)
    (2 votes)
    Default Khan Academy avatar avatar for user
  • hopper happy style avatar for user Tofu
    so is the pattern like...

    1
    i
    -1
    -i
    1
    i
    -1
    i
    Because its like a cycle.
    (8 votes)
    Default Khan Academy avatar avatar for user

Video transcript

In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. "i" is defined as the number whose square is equal to negative 1. This is the definition of "i", and it leads to all sorts of interesting things. Now some places you will see "i" defined this way; "i" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, but I just want to make you a little bit careful, when you do this some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of "i" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of "i" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with "i" to the zeroth power. And so you might say, anything to the zeroth power is one, so "i" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including "i" is one. Then you say, ok, what is "i" to the first power, well anything to the first power is just that number times itself once. So that's justgoing to be "i". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have "i" to the second power. "i" to the second power, well by definition, "i" to the second power is equal to negative one. Lets try "i" to the third power ill do this in a color i haven't used. "i" to the third power, well that's going to be "i" to the second power times "i" And we know that "i" to the second power is negative one, so its negative one times "i" let me make that clear. This is the same thing as this, which is the same thing as that, "i" squared is negative one. So you multiply it out, negative one times "i" equals negative "i". Now what happens when you take "i" to the fourth power, I'll do it up here. "i" to the fourth power. Well once again this is going to be "i" times "i" to the third power. So that's "i" times "i" to the third power. "i" times "i" to the third power Well what was "i" to the third power? "i" to the third power was negative "i" This over here is negative "i". And so "i" times "i" would get negative one, but you have a negative out here, so its "i" times "i" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is "i" times negative "i", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times "i" times "i". "i" times "i", by definition, is negative one. Negative one times negative one is equal to positive one. So "i" to the fourth is the same thing as "i" to the zeroth power. Now lets try "i" to the fifth. "i" to the fifth power. Well that's just going to be "i" to to the fourth times "i". And we know what "i" to the fourth is. It is one. So its one times "i", or it is one times "i", or it is just "i" again. So once again it is exactly the same thing as "i" to the first power. Lets try again just to see the pattern keep going. Lets try "i" to the seventh power. Sorry, "i" to the sixth power. Well that's "i" times "i" to the fifth power, that's "i" times "i" to the fifth, "i" to the fifth we already established as just "i", so its "i" times "i", it is equal to, by definition,"i" times "i" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of "i" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of "i", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. "i" to the seventh power is equal to "i" times "i" to the sixth power. "i" to the sixth power is negative one. "i" times negative one is negative "i". And if you take "i" to the eighth, once again it'll be one, "i" to the ninth will be "i" again, and so on and so forth.