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Multiplying complex numbers in polar form

We can multiply two complex numbers in polar form by multiplying their moduli and adding their arguments. Created by Sal Khan.

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  • blobby green style avatar for user Ruan Carlos
    Thank You for the video, Sal!

    I'll show here the algebraic demonstration of the multiplication and division in polar form, using the trigonometric identities, because not everyone looks at the tips and thanks tab.

    But I also would like to know if it is really correct.

    w1 = a*(cos(x) + i*sin(x))
    w2 = b*(cos(y) + i*sin(y))

    Multiplication

    w1 * w2

    a*(cos(x) + i*sin(x)) * b*(cos(y) + i*sin(y))

    a*b*(cos(x) + i*sin(x))*(cos(y) + i*sin(y))

    ab*(cos(x) + i*sin(x))*(cos(y) + i*sin(y))

    ab*(cos(x)cos(y) + cos(x)*sin(y)*i + cos(y)*sin(x)*i + sin(x)*sin(y)*i^2)

    i^2 = -1

    ab
    (cos(x)cos(y) - sin(x)*sin(y) + i(cos(x)sin(y) + cos(y)*sin(x))

    ab
    (cos(x + y) + i*sin(x + y))


    Division

    w1/w2

    numerator: a*(cos(x) + i*sin(x))
    denominator: b*(cos(y) + i*sin(y))

    multiply both the numerator and denominator by (cos(y) - i*sin(y))

    numerator: a*(cos(x) + i*sin(x))*(cos(y) - i*sin(y))

    a*(cos(x)cos(y) - cos(x)*sin(y)*i + sin(x)*cos(y)*i - sin(x)*sin(y)*i^2)

    i^2 = -1

    a
    (cos(x)cos(y) + sin(x)*sin(y) + i(sin(x)cos(y) - cos(x)*sin(y)))

    a
    (cos(x - y) + i*sin(x - y))

    denominator: b*(cos(y) + i*sin(y))*(cos(y) - i*sin(y))

    Difference of squares

    b*((cos(y))^2 - i^2*(sin(y))^2)

    b*((cos(y))^2 + (sin(y))^2)

    The fundamental relation of trigonometry

    b*1


    numerator: a*(cos(x - y) + i*sin(x - y))
    denominator: b

    a*(cos(x - y) + i*sin(x - y))/b

    (a/b)*(cos(x - y) + i*sin(x - y))
    (23 votes)
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  • blobby green style avatar for user c howard
    At , he says in order to find the product of 120 degrees and 330 degrees, we simply add them. Why does 120 degrees x 330 degrees equal 450 degrees? Is there something special about multiplying angles that makes their sum the same as their product?
    (2 votes)
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Video transcript

- [Instructor] We're given two different complex numbers here, and we want to figure out what is the product? Pause this video and see if you can figure that out. All right, now let's work on this together. So we know from the form that it's written here that the modulus of w sub one is equal to three. And we know that the argument of w sub one is equal to 330 degrees. And by the same line of reasoning, we know that the modulus of w sub two is equal to two. And that the argument of w sub two is going to be equal to, we can see that right over here, 120 degrees. Now, when you multiply complex numbers you could view as one transforming the other. We've seen this in multiple examples. So let's imagine that we are transforming w two by multiplying it by w one. So what is going to happen? Well, let me write it here. So what's the resulting modulus of w one times w two? Well, we're just going to scale up w two's modulus by w one's modulus. Or essentially we're just going to multiply the two. So this is going to be equal to six, three times two. And then the argument of w sub one times w sub two, if we start at w sub two's argument, 120 degrees and then we rotate it by w sub one's argument, well then you're going to add these two angles, that gets you to 450 degrees. So this is equal to 450 degrees, which is more than a complete rotation. And so if we wanted to give it an angle between zero and 360 degrees, if we just subtract 360 from that, that is going to be equal to 90 degrees. And so we can rewrite this here, or we can rewrite the product as w sub one times w sub two is equal to its modulus six times cosine of its argument, 90 degrees. Plus i times sine of its argument. Now we know what the cosine and sine of 90 degrees is. Cosine of 90 degrees is equal to zero and sine of 90 degrees is equal to one. So all of the simplifies quite nicely. All you're left with is a six times I. So this is equal to six i, and we are done.