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### Course: Precalculus>Unit 3

Lesson 2: Distance and midpoint of complex numbers

# 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?

• 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.
• Thats a good question. I think that since we are working with the complex plane the letter i simply indicates the vertical direction rather than representing the square root of -1. So for example (2 + 4i) and (3 + 6i) represent the points (2,4) and (3,6) on the complex plane, and the distance between (2 + 4i) and (3 + 6i) on the complex plane would be the same as the distance between (2,4) and (3,6) on the real plane. Hope this helps.
• What is the real world application of this? How do you even utilize imaginary numbers in the real world?
• Quantum mechanics, sines and cosines and stuff. Waves
• Can the distance formula be used in this situation? If not, why not?
• 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
• at he says over 2 does that apply all the time or just for this instance?
• The midpoint formula is ((x1+x2)/2,(y1+y2)/2). This applies all the time.
• 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 .)
• (√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.
• 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?
• It means in the standard a+bi format, as opposed to, say, polar form.
• 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)
• 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.
• Hello,
Sal is pretty clear about it all.
What I find hard to digest is that we move onle 4 on the Imaginary line, and not 4i. Why not 4i so that sq(x) = sq(7) + sq(4i), with sq(4i) = -16.
I understand the video, it's just I don't like that the Imaginary line is in i units, then we drop that later in compuations. Or I am missing something?
• Someone many many years ago determined that the "y" axis is redefined as the "i" axis for the complex plane and it is in units of "i". This was likely done because any number line uses real numbers (not imaginary). You can't find "i" itself on a number line. By making the number line as units of "i", it resolves this issue.