WEBVTT mathematics/trigonometry/murray
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Hi, these are the trigonometry lectures on educator.com.
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We're here today to talk about De Moivre's theorem.
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The De Moivre's theorem is a little bit tricky.
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The idea is that we're going to use the polar form of complex numbers to find nth powers and nth roots of complex numbers.
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We start with a complex number z.
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We write it in polar form re^iθ.
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We learned in the previous lecture how to convert a complex number into polar form.
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If that's a little bit unfamiliar to you, what you should really do is go back and review the previous lecture on how to convert a complex number into polar form, and how to convert it back into rectangular form.
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There's several formulas that we're going to be using very heavily here.
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One that we learned in the previous lecture is e^iθ=cos(θ)+isin(θ).
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We're going to be using that really heavily.
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Let's see how we can use that to find nth powers of complex numbers.
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For trying to find z^n, we write that as re^iθ, to the nth power.
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If you think about that, we can distribute this nth power onto the r and onto the e^iθ⁺n.
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r^n, that just gives you r^n.
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e^iθ⁺n, that's an exponent raised with an exponent.
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You multiply the exponents.
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That's where I get iθ×n here, so e^i×n×θ.
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If you expand that, e^iθ=cos(θ)+isin(θ), e^inθ gives you cos(nθ)+isin(nθ).
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That's where De Moivre's theorem comes in handy, and that's where it comes from.
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You can expand this into r^n×cos(nθ)+isin(nθ).
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Another way to start out with that is to expand e^iθ into cos(θ)+isin(θ).
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The form we're going to be using most often is this form right here z^n=r^n×cos(nθ)+isin(nθ).
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I know this looks like lots of stuff to remember here.
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The key one that you want to memorize is this one right here, z^n=r^n×cos(nθ)+isin(nθ).
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Memorize that one and we'll practice it during the examples.
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The next step of using De Moivre's theorem is to, instead of finding nth powers, we're going to find nth roots.
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For example, we'll find square roots and cube roots, and fourth roots of complex numbers.
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This is quite a bit more tricky than it is with real numbers.
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In fact, if you're looking for nth roots of a complex number, you always expect to find exactly n answers.
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If a problem says find all the 8th roots of a complex number, you better find 8 answers.
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Unless of course the complex number happens to be 0, in which case the only roots are 0.
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That's why I say every non-zero complex numbers has exactly n nth roots here.
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Let me show you how to find them.
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It's a little bit complicated.
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First of all, we think about the nth root of z.
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We write that as e^1/n.
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That's re^iθ⁺n.
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Remember, e^iθ is cos(θ)+isin(θ), to the 1/n.
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Remember how when we were finding nth powers, we distributed the n into the r, we had r to the n.
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Let me write this again.
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z^n=r^n×e^niθ, which was r^n×cos(nθ), you multiply the angle by n,plus isin(θ).
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With 1/n, replacing the n by 1/n, we get r^1/ncos(θ/n)+isin(θ/n).
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We start out with just cos(θ/n)+isin(θ/n).
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We look at (θ/n).
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We have to find other nth roots as well.
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Our first nth root is just (θ/n).
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To find the other ones, what we add on is multiples of 2π/n.
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That's why I say 2kπ/n.
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We keep doing that for different values of k.
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The reason we do that is we run all the values of k from 0 to n-1.
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If we plugged in k=n into this formula, we get (θ/n)+(2nπ/n), which would be (θ/n)+(2π).
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In terms of angles, that's the same as (θ/n) again.
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That's why we stop at n-1.
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We don't go to k=n, because when we get to k=n, we're repeating ourselves again.
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Essentially, what we're doing here is we're breaking up the unit circle into multiples of (θ/n).
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θ/n, (θ+2π)/n, (θ+4π)/n, we're just taking all these angles around the unit circle until we get back to θ/n.
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This is a little bit tricky.
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What we do is we find one answer for each value of k.
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One nth root for each value of k.
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Since we run k from 0 to n-1, that's a total of n nth roots.
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That's worth remembering.
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We'll practice these with the examples.
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Anytime you have to find nth roots, the r part is easy, you do r^1/n, but then you have to find this cosine and i-sine formula for each angle, for each value of k from 0 to n-1.
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You run this formula separately, n times over, and at the end, you have n complex numbers as your answers.
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We'll check that out with some examples and you'll get the hang of it.
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The first example here, we have to convert the complex number z equals negative root 3 plus i into polar form, then use De Moivre's theorem to calculate z⁷.
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Remember to convert it into polar form.
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You do r equals the square root of x²+y², θ=arctan(y/x).
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Sometimes you have to modify that θ formula, sometimes you have to add on a π.
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You know you have to do that when x is negative.
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You do that if x is negative.
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Let's find our r in our θ here.
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Let me graph that thing.
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Graph it just so that we'll be able to check whether our answer's possible.
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Negative square root of 3 on the x-axis, i on the y-axis, that's about right there.
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Let me calculate r and θ to see if it's possible.
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r is the square root of x²+y², x²=3, y²=1, square root of 4 is 2.
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θ is arctangent of 1 over negative square root of 3 which is negative root 3 of 3, that's a common value.
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The arctangent of that is -π/6.
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There's this fudge factor that I have to include here.
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The x < 0 here, I have to add on a π.
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I add on a π and I get 5π/6.
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That does check with my little graph here because that really is 5π/6, the angle over there, and the radius does indeed look like about 2.
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That's reassuring.
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z=re^iθ, that's 2e^(5π/6)i.
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We have converted a complex number into a polar form, that was the first part of the exercise.
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The main part here is to use De Moivre's theorem to calculate z to the seventh.
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Let's work that out.
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z⁷, the whole point is we're going to use the polar form to find z⁷, so this is 2e^(5π/6)i⁷.
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That's 2⁷.
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I have each one exponent raised to an exponent.
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I just want to multiply those two exponents e^7×5, is (35π/6)i.
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35π/6 is a little cumbersome, that's not in between 0 and 2π.
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I'll work on that a little bit.
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In the meantime, 2⁷ is 128.
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35π/6, how can I simplify that?
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35π/6, let me subtract 2π, 12π/6, that's 23π/6, that's still not in my range between 0 and 2π.
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Let me subtract another 2π, that gives me another 12π/6 off, is 11π/6.
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That is in the range between 0 and 2π.
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This is the same as e^(11π/6)i.
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I want to convert that into rectangular form.
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It's very good to remember this formula e^iθ is cos(θ)+isin(θ).
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You can also use x=rcos(θ), y=rsin(θ).
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I prefer the e^iθ form.
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This is equal to 128cos(11π/6)+isin(11π/6).
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11π/6, where is that on the unit circle?
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That's just π/6 short of 2π.
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That's down there.
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That's a common value, I know what the sine and cosine are.
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It's root 3 over 2 and 1/2.
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Root 3 over 2 is positive, the sine is negative because it's below the x-axis so it's -1/2.
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I have 128/2, that's 64 root 3 minus 64i.
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That's what that simplifies down to.
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Let's review how we did that one.
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We start out with a complex number, and we have to convert into polar form.
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I look at my formulas for r and θ including the fudge factor for θ if x < 0.
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Run that through, my x and y are negative root 3 and 1.
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I get an r.
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I get a θ including the fudge factor.
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That gives me re^iθ.
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I've got my polar form.
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To raise it up to the 7th power, De Moivre's theorem says if you use polar form, then you just put 2⁷, then nθ.
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This is the nθ.
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That reduces down by subtracting 2π at a time, e^(11π/6)i.
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This is really nθ here, cos(nθ), sin(nθ), although we reduced down by subtracting off 2π at a time.
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We get 128 times the cosine and sine of 11π/6.
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That's a common value.
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I look at my unit circle to remember my sine and cosine of 11π/6.
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I plug them in and I get my answer.
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Second example here, we find to find all complex 8th roots of the number 16.
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In order to find all complex 8th root, we have to think about 16 being a complex number.
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Of course 16 is just the same as 16+0i, that is a complex number.
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We're asked for 8th roots.
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Let me remind you that because we're asked for 8th roots, we expect 8 different answers.
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We have to find 8 answers here.
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I'm going to try and write 16 in polar form.
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r is equal to the square root of x²+y².
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θ=arctan(y/x), plus π if the x happens to be negative.
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My r is the square root of 16²+0², that's just 16.
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θ, my y=0, arctan(0)=0.
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My z=16e^0i.
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I could have worked that out, certainly 16e⁰=16 just by itself because e⁰=1.
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That's nice to check our work.
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z^1/8.
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According to De Moivre's theorem, let me remind you what De Moivre's theorem said about complex nth roots.
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It said that, you do r^1/n×cos((θ+2kπ)/n)+isin((θ+2kπ)/n).
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You run this for different values of k.
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k is equal to 0, 1, 2, up to n-1.
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Here, n=8, so z^1/8, r=16^1/8.
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Cosine is, θ=0, (0+2kπ)/8, plus isin(0+2kπ)/8.
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There's going to be lots of values for k here.
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Maybe I should make a little chart for what k is, and then the different angles that we have for each value of k, the different angles that we're going to be plugging into De Moivre's formula there.
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(0+2kπ)/8 which actually simplifies down to just kπ/4.
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For k goes from 0, 1, 2, 3, 4, 5, 6, you run it to n-1, and n=8, so 0 through 7 there.
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We'll have 0, π/4, 2π/4 is π/2, 3π/4, 4π/4 is π, 5π/4, 6π/4 is 3π/2, and 7π/4.
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For each one of these angles we're going to plug it in and we're going to get an answer.
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I also have to simplify 16^1/8.
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Let me see if I can do something with that.
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16^1/8, I know that 16=2⁴, that's 2^4/8, 2^1/2 is square root of 2.
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I'm going to multiply the square root of 2, that's 16^1/8 times cos(θ)+isin(θ) for each one of these values of θ.
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Let me not reuse the same Greek letter θ, I'll use α.
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For each one of these values of α here ...
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Let me write down what cos(α)+isin(α) is for each one of these values of α.
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Cos(0)+isin(0), cos(0)=1, plus isin(0)=0.
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Cos(π/4)+isin(π/4), cosine and sine of (π/4) are both square root of 2 over 2.
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That's k=1.
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For π/2, the cosine is 0, the sine is 1.
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For 3π/4, we have the cosine is negative root 2 over 2, the sine is positive root 2 over 2.
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I think this will be easy to work out if I draw a unit circle, so that I can easily and quickly find the sines and cosines.
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They're all common values but it helps to draw a unit circle to remember where things are positive and negative.
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What I started with is π/4, there's π/2, there's 3π/4, π, we're going to move on to 5π/4, 3π/2, move on to 7π/4, and we started out as 0.
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Let's see.
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We've already hit 3π/4 moving on to π now.
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Cosine and sine is -1+0i.
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5π/4, cosine and sine are both negative root 2 over 2.
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They're both negative.
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3π/2, the cosine is 0 again, the sine is -1.
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Finally, 7π/4, cosine and sine are root 2 over 2, but the cosine is positive and the sine is negative.
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For our answers, what we have to do is multiply root 2 by each one of these.
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Let me multiply the root 2 by each one of these.
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The first one you just get square root of 2 times 1+0.
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Multiplying root 2 times root 2 over 2, that gives me 2/2, which gives me 1.
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Plus i to the same thing, so just i.
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Multiply root 2 times 0+i, gives me root 2i.
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Multiply root 2 by this, we get -1+i.
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Multiply root 2 here, we get negative root 2.
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Multiply root 2 here, -1-i, remember root 2 times root 2 over 2, is 2/2, which simplifies to 1.
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Multiply root 2 here, we get negative root 2i.
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Multiply root 2 here, we get 1-i.
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We are finally done here.
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We get 8 different answers, 8 different complex numbers here.
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Each one of these complex numbers has the property that if you raised it up to the eighth power, it will come out to be exactly 16.
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Let me write down what we found here.
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Each one satisfies w⁸=16.
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If you multiply them up by themselves eight times, you'll get back to 16.
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That will be pretty messy I'm not going to check that here, but you can check it on your own if you like.
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Let me recap how we found that.
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We started out with the complex number 16, think about it as 16+0i.
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We wanted to write that in polar form, so I founded r and a θ.
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r=16, and θ=0, z=16e^0i.
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Then I use De Moivre's theorem which says that you get complex nth roots by doing r^1/n, that's where the 16^1/8 came from.
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Then cosine plus isine of these angles (θ+2kπ)/n.
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That's why I started to make this chart (θ+2kπ)/8.
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You run the k from 0 to n-1, that's why I ran the k from 0 to 7.
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For each one of those I got an angle that I called α, then I worked out cos(α)+isin(α) for each one of those.
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That's where I got these sectional values here, and for that it was really helpful to plot my α's on the unit circle here.
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Remember what the sines and cosines of each one of them was.
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Those are common values so I didn't need to calculate to look those up.
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Multiply each one by 16^1/8.
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A little cleverness with the laws of exponents tells me that that's root 2.
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Finally, I multiply each one of those by root 2, and I get 8 different answers, each one of them is an eighth root of 16 and complex numbers.
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We're now going to find all complex cube roots of -1.
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Cube roots, that's a third root, we expect 3 answers here because we're looking for cube roots.
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I want to put that complex number into polar form first.
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I'm going to use my r equals square root of x²+y², and θ=arctan(y/x), plus π if x < 0.
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Negative 1, think of that as -1+0i, the x=-1, y=0.
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My r is square root of x²+y², that is square root of 1, r=1.
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θ=arctan(y/x), that's arctan(0), x < 0, so I have to add π.
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Arctan(0)=0, θ=π.
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z=re^iθ, 1e^iπ.
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I've got my complex number into polar form, now I'm going to use De Moivre's theorem.
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Let me remind you how that goes, it says r^1/n×cos(θ+2kπ)/n+isin(θ+2kπ)/n.
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The key thing here is k=0, 1, 2, up to n-1.
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Here we're finding cube roots, our n=3.
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Let me make a little chart again of the angles.
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n=3, θ=π, I'll make a chart of k and (θ+2kπ)/3 for each value of k here.
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k goes from 0 to n-1, that's 0, 1, and 2.
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(θ+2kπ)/3, when k=0, that's just, θ=π, π/3.
00:30:56.000 --> 00:31:09.000
When k=1, that's (π+2π)/3, which is 3π/3, which is π.
00:31:09.000 --> 00:31:19.000
When k=2, this is (π+4π)/3, which is 5π/3.
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The three angles we're going to be looking at are π/3, π, and 5π/3.
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Let me also work out r^1/n.
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r^1/n, r=1, raised to 1/3 power is just 1.
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That part is very easy.
00:31:39.000 --> 00:31:53.000
We have cos(α)+isin(α) for each one of these α's.
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Cos(π/3)+isin(π/3) ...
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Let me draw out where that would be. π/3 is about right there.
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π is right there.
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5π/3 is down there.
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Those are the three angles I'm going to be looking at.
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Let me go ahead and include these on my chart.
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Cos(α)+isin(α), cos(π/3)=1/2, isin(π/3) is root 3 over 2, that's because (π/3) is right there.
00:32:47.000 --> 00:33:00.000
π, the cosine is -1, the x-coordinate, isine is zero.
00:33:00.000 --> 00:33:12.000
5π/3,that's down here, cosine is 1/2, the sine is negative root 3 over 2.
00:33:12.000 --> 00:33:17.000
That was cos(α)+isin(α).
00:33:17.000 --> 00:33:32.000
r^1/n×cos(α)+isin(α) is kind of anticlimatic because we already figured out that r^1/n=1.
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We're just multiplying each of these by 1.
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We get 1/2 plus i root 3 over 2, -1 and 1/2 minus i root 3 over 2 as our three answers.
00:33:49.000 --> 00:33:58.000
Remember we're looking for cube roots, so we did expect to find 3 answers, it's reassuring here that we found our 3 answers.
00:33:58.000 --> 00:34:00.000
Let me remind you how we did that.
00:34:00.000 --> 00:34:02.000
First of all, we were given a complex number.
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We had to convert it inot plar form so I found my r and my θ using the standard formulas.
00:34:09.000 --> 00:34:10.000
I did have to include the fudge factor plus π here because the x < 0.
00:34:10.000 --> 00:34:17.000
Arctan(0) gave me 0, gave me θ=π.
00:34:17.000 --> 00:34:23.000
My θ was π, my r was 1, so I get 1e^iπ.
00:34:23.000 --> 00:34:35.000
I go to De Moivre's theroem which says, r^1/n×cos(θ+2kπ)/n, isin(θ+2kπ)/n.
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I made a little chart of the different values of k.
00:34:39.000 --> 00:34:41.000
You go from 0 to n-1.
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For each one, I figured out (θ+2kπ)/n.
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That gave me the π/3, π, and 5π/3.
00:34:50.000 --> 00:34:55.000
I found the cosine plus isine of each one.
00:34:55.000 --> 00:34:58.000
I multiplied those by r^1/n.
00:34:58.000 --> 00:35:00.000
That gave me my 3 answers.
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Those are the three complex numbers that are cube roots of 1.
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Each one satisfies wr³=-1.
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If you worked out wr³ for any of these complex numbers, you'd get -1.
00:35:16.000 --> 00:35:18.000
We've got some more examples for you later.
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Try them out on your own and then we'll work through them together.