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

Distance & midpoint of complex numbers

Sal finds the distance between (2+3i) and (-5-i) and then he finds their midpoint on the complex plane. Created by Sal Khan.

Want to join the conversation?

  • ohnoes default style avatar for user Inspector Javert
    At , how is the distance equal to 4? Since that's on the imaginary axis, shouldn't it be equal to 4i? Then x^2 = 49 +16i^2, and since i^2 = -1 it would be x = √33 not √65.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user Keith
      Regarding your question as to whether 4 should've been 4i:

      We are looking at the magnitude of the number just to show the distance between the points. It is the same as when we get negstive numbers--we simply use the absolute value (which is the magnitude).
      (8 votes)
  • aqualine ultimate style avatar for user Justin McGriff
    at he says over 2 does that apply all the time or just for this instance?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user Aiyan Alam
    Can the distance formula be used in this situation? If not, why not?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user cossine
      If you know how to apply distance formula on the x-y number plane then you would know how to apply distance formula on the complex number plane.

      It is just Pythagoras. All you got to do is consider the real and imaginary components
      (5 votes)
  • blobby blue style avatar for user ✨ Sofia Utama 💯
    Hello! To find the midpoint of a complex number, can't we have just divided √65 by 2? (I'm using the example from the video.) Also, Sal said that 3-1=-2, which is wrong, at . Lastly, why does the formula work to find the midpoint? (Which Sal wrote at the bottom of his page in the video, at about .)
    (3 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user loumast17
      (√65)/2 would give the length from one point to the midpoint, but to find the midpoint you would need a bit more work. The way Sal did it is definitely pretty effective. Another way to think of it is to take the horizontal and vertical distances, so 7 and 4 respectively, cut them in half to get 7/2 and 2 respectively then add/subtract that to each part of one of the points. The leftmost point gets half the horizontal distance added to it while the rightmost point gets half the horizontal distance subtracted. so -5 + 7/2 = -3/2 and 2 - 7/2 = -3/2. Both get the same answer. Similarly the most vertical point gets half the horizontal distance subtracted, and lowest point gets it added. so 3-2 = 1 or -1 + 2 = 1. No matter how you do it you get the horizontal part of -3/2 and the vertical part equal to 1, so for a complex nuber that is -3/2 + i

      Sal may have said 3-1=-2 at , but then at he corrects himself.

      The formula that sal wrote works because he is taking one coordinate in one direction and adding the other, then dividing that by 2, which is how you take the average of two points. the average of two points is always the middle of the two. So then he gets the middle horizontally and then the middle vertically and just needs to plug them into the rectangular form of a complex number.

      It might make it easier to understand if you use the traditional coordinate pairs just to see things working. So if you had the points (2, 3) and (-5, -1) could you find the distance between the two and the midpoint? That's what this video is doing, but is putting those points into the form of two complex numbers on the complex plain.

      If my method didn't make sense either let me know, as well as where it didn't make sense.
      (6 votes)
  • leafers ultimate style avatar for user artgrohe
    What is the use of finding the midpoint of two complex numbers? just curious..
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Patrick Hearn
    There's a few questions on this, but I haven't seen an answer that nails it for me. I understand the method: so mod(3+4i) = √((3^2) + (4^2)) = 5

    What confuses me is if I have a complex number (0 + 1i) then it's distance is 1, which implies to me that i would be equal to 1 (which clearly it's not?)

    I wonder if you could say that the distance of 3+4i = 5 AND 5i...simultaneously? It seems almost arbitrary to me that we choose to represent the distance as a real rather than an imaginary number.
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      i has a magnitude of 1, that's correct. 1 also has a magnitude of 1, as does -1, 1/√2 +i/√2, and infinitely many other complex numbers. That does not mean that they are all the same number. They just have a property in common. (6 and 12 are both even numbers, but 6≠12.)

      We choose to represent magnitude as a real number because we're thinking of it as distance across the complex plane, and distances are always given by positive real numbers.
      (6 votes)
  • blobby blue style avatar for user ✨ Sofia Utama 💯
    Hello (again)! On a questions on the Practice: Midpoint of complex numbers, I got a question where it said to Express your answer in rectangular form- what does that mean?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Nathan
    Find the distance between z_1=(12+3i)and z_2=(10-5i)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Hecretary Bird
      Find the distance from one point to the other along the real axis, then find the distance along the imaginary axis. You have the legs of a right triangle. Compute the hypotenuse and solve for the distance.
      The above process can be shortened into taking the absolute value of the difference between the two numbers, since complex numbers are written as one expression when graphed.
      |10 - 5i - 12 - 3i|
      = |-2 -8i|
      = sqrt((-2)^2 + (-8)^2)
      = sqrt(4 + 64)
      = sqrt(68)
      = 2*sqrt(17)
      (2 votes)
  • starky sapling style avatar for user Nightmare252
    is the x-axis and the real axis exchangeable and the y axis and the imaginary axis interchangeable??
    (1 vote)
    Default Khan Academy avatar avatar for user
    • mr pink red style avatar for user andrewp18
      No. In the complex plane, you wouldn't refer to the horizontal axis as the 𝑥-axis, you would call it the real axis. Likewise, in the complex plane, you wouldn't call the vertical axis the 𝑦-axis, you would call it the imaginary axis. Labelling axes 𝑥 and 𝑦 are only standard for the real Cartesian plane.
      (2 votes)
  • starky sapling style avatar for user flashassasin14
    ok is anyone else confused on how he gets the x=sqrt of 49+16
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Voiceover:So we have two complex numbers here. The complex number z is equal to two plus three i and the complex number w is equal to negative five minus i. What I want to do in this video is to first plot these two complex numbers on the complex plane and then think about what the distance is between these two numbers on the plane and what complex number is exactly halfway between these two numbers or another way of thinking about it, what complex number is the midpoint between these two numbers. So I encourage you to pause this video and think about it on your own before I work through it. So let's first try to plot these on the complex planes. So let me draw, so right over here, let me draw our imaginary axis. So our imaginary axis, and over here let me draw our real axis. Real axis right over there, and let's first, let's see, we're gonna have it go as high as positive two in the real axis and as low as negative five along the real axis so let's go one, two, three, four, five. One, two, three, four, five. Along the imaginary axis we go as high as positive three and as low as negative one. So we could do one, two, three and we could do one, two, three and of course I could keep going up here just to have nice markers there although we won't use that part of the plane. Now let's plot these two points. So the real part of z is two and then we have three times i so the imaginary part is three. So we would go right over here. So this is two and this is three right over here. Two plus three i, so that right over there is z. Now let's plot w, w is negative five. One, two, three, four, five, negative five minus i, so this is negative one right over here. So minus i, that is w. So first we can think about the distance between these two complex numbers; the distance on the complex plane. So one way of thinking about it, that's really just the distance of this line right over here. And to figure that out we can really just think about the Pythagorean theorem. If you hear about the Distance Formula in two dimensions, well that's really just an application of the Pythagorean theorem, so let's think about that a little bit. So we can think about how much have we changed along the real axis which is this distance right over here. This is how much we've changed along the real axis. And if we're going from w to z, we're going from negative 5 along the real axis to two. What is two minus negative 5? Well it's seven, if we go five to get to zero along the real axis and then we go two more to get to two, so the length of this right over here is seven. And what is the length of this side right over here? Well along the imaginary axis we're going from negative one to three so the distance there is four. So now we can apply the Pythagorean theorem. This is a right triangle, so the distance is going to be equal to the distance. Let's just say that this is x right over here. x squared is going to be equal to seven squared, this is just the Pythagorean theorem, plus four squared. Plus four squared or we can say that x is equal to the square root of 49 plus 16. I'll just write it out so I don't skip any steps. 49 plus 16, now what is that going to be equal to? That is 65 so x, that's right, 59 plus another 6 is 65. x is equal to the square root of 65. Now let's see, 65 you can't factor this. There's no factors that are perfect squares here, this is just 13 times five so we can just leave it like that. x is equal to the square root of 65 so the distance in the complex plane between these two complex numbers, square root of 65 which is I guess a little bit over eight. Now what about the complex number that is exactly halfway between these two? Well to figure that out, we just have to figure out what number has a real part that is halfway between these two real parts and what number has an imaginary part that's halfway between these two imaginary parts. So if we had some, let's say that some complex number, let's just call it a, is the midpoint, it's real part is going to be the mean of these two numbers. So it's going to be two plus negative five. Two plus negative five over two, over two, and it's imaginary part is going to be the mean of these two numbers so plus, plus three minus one. Three minus one, minus one, over two times i and this is equal to, let's see, two plus negative five is negative three so this is negative 3/2 plus this is three minus 1 is negative, is negative two over two is let's see three, make sure I'm doing this right. Three, something in the mean, three minus one is two divided by two is one, so three plus three. Negative 3/2 plus i is the midpoint between those two and if we plot it we can verify that actually makes sense. So real part negative 3/2, so that's negative one, negative one and a half so it'll be right over there and then plus i so it's going to be right over there. And I'll just have to draw it perfectly to scale but this makes sense, that this right over here would be the midpoint.