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Precalculus
Course: Precalculus > Unit 3
Lesson 8: Multiplying and dividing complex numbers in polar form- Multiplying complex numbers in polar form
- Dividing complex numbers in polar form
- Multiply & divide complex numbers in polar form
- Taking and visualizing powers of a complex number
- Complex number equations: x³=1
- Visualizing complex number powers
- Powers of complex numbers
- Complex number polar form review
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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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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))(12 votes)
Hi! Nice job on the explanation and proofs. It looks correct to me. I have no trouble with the explanation or anything really, but I would say, just putting both those equations labeled at the end of your answer again just so it is easier for others to read.(5 votes)
- At1:28, 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)
1:28Video 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.