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Intro to complex number conjugates

Current time:0:00Total duration:8:04
Learn about the conjugate of complex numbers, how to add and divide using conjugates, and the key property that multiplying a complex number by its conjugate produces a real number. Conjugates are very useful for simplifying division of complex numbers. Created by Sal Khan.
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  • leaf grey style avatar for user sandeep walia
    why does sal uses a Z wih a dashed line? is it a rule for complex no.
    (15 votes)
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    • piceratops ultimate style avatar for user Hack Saw
      Do you mean the Z with the bar over the top, or the dash across the middle of the Z?

      The bar over the top (in this case) means that we are talking about the conjugate of Z, not Z itself.

      If you talking about the dash across the middle of the Z, that's a way to make it clear that it's a Z, and not a 2, which with messy handwriting in math can be a problem.
      (70 votes)
  • piceratops tree style avatar for user Chandana Deeksha
    at the end of the video. why did sal say that any complex number multiplied by its conjugate is equal to the magnitude of the complex number squared??
    i just lost my understanding there....please help....
    (24 votes)
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    • male robot hal style avatar for user Figgy Madison
      Because it is true. The magnitude of z=a+bi is written |z|=√(a²+b²) from the distance formula. A complex number multiplied by its own conjugate (a+bi)(a-bi)=a²+(bi)²=a²+b². But if you take the square root of that, you get the magnitude. So any number times its complex conjugate is equal to the square of its magnitude (which is a real number).
      (4 votes)
  • leaf green style avatar for user Brillante Wang
    At , why is a^2 + b^2 = |z|^2 ?
    (19 votes)
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  • boggle blue style avatar for user Miss Daily
    Sal references a previous video in which "imaginary parts" and functions to obtain them are discussed. If one is following the College Algebra coursework, however, that video does not appear. Where would I be able to find it?
    (7 votes)
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  • starky sapling style avatar for user Claudia
    What does conjugate mean?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      If you have a number like 2+3√3, the conjugate is what you get by flipping the sign on the irrational or imaginary term.

      So the conjugate of 2+3√3 is 2-3√3.
      The conjugate of π-4i is π+4i.
      The conjugate of 5 is 5 again.

      If you add or multiply a number with its conjugate, you get a rational or real number back.
      (8 votes)
  • blobby green style avatar for user Marcus
    At time stamp why does the speaker state that I squared is negative one for five I sqared. He did not stat that for the numberator.
    (4 votes)
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  • piceratops tree style avatar for user Lyric
    Umm... at , why does Sal write a^2 + b^2 = |z|^2? I don't understand that line. Can someone explain it? Thank you.
    (3 votes)
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    • blobby green style avatar for user Gavin Anderson
      Starting from (a+bi)(a-bi), a factored difference of two squares, here are some steps in the simplification process that got left out in the video:

      (a+bi)(a-bi)

      From here, this can either be distributed the long way, or it can be recognized that this is in the form "(x+y)(x-y)," which can be rewritten in the form "x²-y²," shown here:

      a²-(bi)²

      Now, here is what got left out. An expression in the form "(xy)²" can be rewritten as "x²y²," shown bellow:

      a²-(b²i²)
      (The parentheses around "b²i²" are just for clarity).

      Since i=√(-1), the following holds true:

      a²-{b²[√(-1)]²}

      The subtraction of any number "b" from any number "a," which is a-b, can be rewritten using addition as a+(-1)*b. So, the expression can be rewritten as the following:

      a²+(-1){b²[√(-1)]²}

      This can be further rewritten as this:

      a²+{(-1)*(b²)*[√(-1)]²}

      Since the square of a square root is just equal to whatever is under the square root symbol, this can be simplified to this:

      a²+[(-1)*(b²)*(-1)]

      Switching the order of two of the terms yields the following:

      a²+[(-1)*(-1)*(b²)]

      Of course, (-1)*(-1)=1; so, the expression simplifies to this:

      a²+(1*b²)

      Finally, because any number "a" multiplied by "1" is simply equal to "a," this becomes the expression bellow:

      a²+b²

      Now, I'm not sure why there are absolute value bars around "Z," since squaring it will make it positive without them, and I'm not sure why a²+b²=|Z|². However, I hope this helps.
      (4 votes)
  • starky ultimate style avatar for user Adrienne
    at sal multiples the equation by the conjugate of the dominator, but he only changes the sign on the imaginary number, not the real number. why did he not change he sign of the real number?
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      Rationalizing the denominator with complex numbers is the same as rationalizing the denominator when there is a square root in the denominator. When there are 2 terms in the denominator, you need to create a difference of 2 squares: a^2-b^2. The factors of a difference of 2 squares need to fit the pattern:(a-b)(a+b). Each factor has the same terms but opposite signs in the middle. Anything other than this pattern will not eliminate any square roots or "i" in the denominator.

      If you take your approach, you are basically multiplying the original denominator by -1. So, you are essentially doing: -1(a+b)^2 = -a^2-2ab-b^2. The middle term will still contain "i", so it will not have been eliminated.

      Here's the math to show you the difference.
      Sal has: (4+5i)(4-5i) = 16-20i+20i-25i^2 = 16+25 = 41
      Your version: (4+5i)(-4-5i) = -16-20i-20i-25i^2 = -16-40i+25 = 9-40i

      Because (-4-5i) does not fit the pattern to create a difference of 2 squares, the middle term is created and "i" is not eliminated.

      Hope this helps.
      (3 votes)
  • blobby green style avatar for user Jermin Hu
    What’s the relationship between complex numbers and vectors? And which video was it first talked about in?
    (3 votes)
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  • male robot johnny style avatar for user kponi
    i have a question about the last formula that there was. how does that "i" that was in the (a^2) - (bi^2) just disappear in the next part of the formula? the a^2+b^2 part? Thanks..!
    (1 vote)
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Video transcript

I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z, and it's equal to a plus bi, I used a word. And I have to be careful about that word, because I used in the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I did make it clear that if you were to see the function the real part of z, this would spit out the a. And the function the imaginary part of z, this would spit out-- and we talked about this in the first video-- it would spit out the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word "imaginary part" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd be denoted with z with a bar over it. Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here. And so we can actually look at this to visually add the complex number and its conjugate. So we said these are just like position vectors. So if we were to add z and its conjugate, we could essentially just take this vector, shift it up here, do heads to tails. So this right here, we are adding z to its conjugate. And so this point right here, or the vector that specifies that point, is z plus z's conjugate. And you can see right here, just visually, this is going to be 2a. And to do that algebraically. If we were to add z-- that's a plus bi-- and add that to its conjugate, so plus a minus bi, what are we going to get? These two guys cancel out. We're just going to have 2a. Or another way to think about it-- and really, we're just playing around with math-- if I take any complex number, and to it I add its conjugate, I'm going to get 2 times the real part of the complex number. Oh, and this is also going to be 2 times the real part of the conjugate because they have the exact same real part. Now with that said, let's think about where the conjugate could be useful. So let's say I had something like 1 plus 2i divided by 4 minus 5i. So it's no real obvious way to simplify this expression. Maybe I don't like having this i in the denominator. Maybe I just want to write this as one complex number. If I divide one complex number by another, I should get another complex number. But how do I do that? Well, one thing to do is to multiply the numerator and the denominator by the conjugate of the denominator, so 4 plus 5i over 4 plus 5i. And clearly, I'm just multiplying by 1, because this is the same number over the same number. But the reason why this is valuable is if I multiply a number times its conjugate, I'm going to get a real number. So let me just show you that here. So let's just multiply this out. So we're going to get 1 times 4 plus 5i is 4 plus 5i. And then 2i times 4 is plus 8i. And then 2i times 5i-- that would be 10i squared, or negative 10. And then that will be over-- now, this has the form a minus b times a plus b. Well, a plus b times a minus b is a squared minus b squared. So it's going to be equal to 4 squared, which is 16, minus 4 squared-- oh, why did I have 4 plus 4i here? This should be 4 plus 5i. What am I doing? 4 plus 5i, the same number over the same number. This was a 10 right over there. This is the conjugate. I don't know. My brain must have been thinking in 4's. So obviously, I don't want to change the number-- 4 plus 5i over 4 plus 5i. So let's multiply it. This is a minus b times a plus b, so 4 times 4. So this is going to be 4 squared minus 5i squared. And so this is going to be equal to 4 minus 10. Let's add the real parts. 4 minus 10 is negative 6. 5i plus 8i is 13i. Add the imaginary parts. And then you have 16 minus 5i squared. Well, 5i squared-- i squared is negative 1. 5 squared's just going to be negative 25. The negative and the negative cancel out, so you have 16 plus 25. So that is 41. So we can write this as a complex number. This is negative 6/41 plus 13/41 i. We were able to divide these two complex numbers. So the useful thing here is the property that if I take any complex number, and I multiply it by its conjugate-- and obviously, the conjugate of the conjugate is the original number. But I would take any complex number and I multiply it by its conjugate, so this would be a plus bi times a minus bi. I'm going to get a real number. It's going to be a squared minus bi squared, difference of squares, which is equal to a squared-- now, this is going to be negative b squared. But we have a negative sign out here, so they cancel out-- a squared plus b squared. And just out of curiosity, this is the same thing as the magnitude of our complex number squared. So this is a neat property. This is what makes conjugates really useful, especially when you want to simplify division of complex numbers. Anyway, hopefully, you found that useful.