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## Precalculus

### Course: Precalculus > Unit 3

Lesson 2: Distance and midpoint of complex numbers# Distance & midpoint of complex numbers

CCSS.Math:

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?

- At3:15, 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.(85 votes)
- 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.(48 votes)

- Can the distance formula be used in this situation? If not, why not?(5 votes)
- 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(4 votes)

- at4:52he says over 2 does that apply all the time or just for this instance?(0 votes)
- The midpoint formula is ((x1+x2)/2,(y1+y2)/2). This applies all the time.(14 votes)

- 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, at5:25. Lastly, why does the formula work to find the midpoint? (Which Sal wrote at the bottom of his page in the video, at about5:50.)(2 votes)
- (√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 at5:25, but then at5:35he 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.(5 votes)

- What is the use of finding the midpoint of two complex numbers? just curious..(2 votes)
- The midpoint of two complex numbers is their arithmetic mean.(3 votes)

- 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.(0 votes)- 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.(5 votes)

- 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?(1 vote)- It means in the standard a+bi format, as opposed to, say, polar form.(3 votes)

- 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?(1 vote)- 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.(3 votes)

- is the x-axis and the real axis exchangeable and the y axis and the imaginary axis interchangeable??(1 vote)
- 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)

- Why didn't he say in distance formula that

X^2= (7)^2+(4i)^2

X^2= 49-16=33

X=√33(1 vote)- Change in y axis is 4 not 4i.

The y axis is the imaginary axis. You plot any coefficient of i in that axis. As in the equation the coefficient of the imaginary part is 4.

And also √33 ≈ 5.744 which is clearly smaller that 7. That is not possible cause the hypotenuse is the largest side in a right triangle.(2 votes)

## 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.