In Exercises 69–76, find all the complex roots. Write roots in re...

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In Exercises 69–76, find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth.The complex sixth roots of 64

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Welcome back. I am so glad you're here. We are asked to determine all of the complex sixth roots of 729 express the roots in rectangular form and round to one decimal if needed. All right. So when we're finding the complex roots of a number, we're going to use Demore's theorem. And we recall from previous lessons that demos theorem when we are in Radians, which is what we're going to do here. That's given to us with the equation Z sub K equals the nth rut of R. It almost looks like an N here. The nth rut of R multiplied by the quantity of the cosine of in the numerator. We have theta plus two pi K all divided by N plus I. A sign of the same thing. We have theta plus two pi K all divided by N. And so we just need to figure out a few things. We need to figure out N, we need to figure out our K values. We need to figure out R and we need to figure out theta. So starting with NN is pretty quick to figure out that's the roots that we're taking we're taking the six roots. So N equals six K is pretty closely related to NK, begins with zero. And then it's the series of numbers all the way up to N minus one, N is six. So six minus one is five. So our K values will be 01234 and five. We'll run the equation for each one of those K values. So we'll run it six times now. R and theta, we don't know those yet. We take a look at what we're finding the roots of. It's 729 which is the same as 729 plus zero eyes. So here this is something in rectangular form to find R and theta, we'll need to convert that to polar form. So this is in format of A plus B I, we see that our A value here is 729 and our B value equals zero. And if we're to envision this on the complex plane real quick, if we've got a vertical imaginary axis and a horizontal real axis for our real numbers are a, that's 729 units and then we're going nowhere on our imaginary axis. So we're somewhere over here and this is exactly on the positive part of the real axis. And that'll be important in a minute. So now let's convert, we're going to get this into the format for polar form of R multiplied by the quantity of cosine theta plus I sine theta. We recall from previous lessons that R equals the square root of A squared plus B squared. So that's the square root of A is 729 squared plus B squared. B is zero. So that's zero squared. And I'm just going to keep that zero squared is nothing. So I'm keeping this as the square rut of 729 squared, which is 729. There's our R. Now for theta, we know that the tangent of theta equals B divided by A B here is zero, divided by A is 729. So the tangent of theta here equals zero. This is true when theta equals either zero or pi. But as we see, we're over here, this is at zero radiance, not pi radiance because it's the positive part of the real axis. So here we can disregard the pi theta equals zero radiance. Great. We know what everything is. Now. It's just a matter of plugging it in six times. So this will take a little while, but that's going to take the rest of the video. OK. So our first K value that we're going to use is zero. So we have Z subzero equals the NTH root N is six. The sixth root of RR is 729 multiplied by the quantity of the cosine of theta is zero plus two pi multiplied by KK right now. Is zero, all divided by N and A six plus I sign of the same thing. Zero plus two pi multiplied by zero, all divided by six. And now it's just simplifying every single time we do this, the sixth threat of 729 is going to be 33 multiplied by the quantity of the cosine of. I'm going to simplify what is in the cosign and the sign very quickly. If you need to pause the video and enter it into your calculator or figure it out by hand, that's great. So zero plus two pi multiplied by zero, all divided by six is just zero. So we've got the cosine of zero plus I sine of the same thing zero. Now, what we'll be doing is finding the exact values for our cosines and signs. So there's still three multiplied by the quantity of and for the cosine of zero. The exact value of that looking at the charts in the textbook and recalling from previous lessons is one plus the sign of zero that is going to be zero. So there's zero is now distributing the 33 multiplied by one is three and three multiplied by zero, I is zero. So Z subzero, our first one here is three because we're doing so many, I'm going to copy this down here for when we scroll down and probably lose this one or when we put this to a different part of the screen. So Z subzero is equal to three. All right, that's 15 more to go Z sub one. That's our next K value one still the six thre of 729 multiplied by the quantity of the cosine of still zero plus. But now it's two pi multiplied by the new K value of one all still divided by six plus I sine of the same thing zero plus two pi multiplied by the new K value of one all divided by six. Simplifying six thread of 720 minus three, multiplied by the quantity of the cosine of zero plus two. Pi multiplied by one all divided by six. Simplifies to be pi divided by three. So the cosine of pi divided by three plus I sine of the same thing pi divided by three. And so this is three multiplied by the quantity of the exact value for the cosine of pi divided by three is one half plus the exact value for the sine of pi divided by three is square at three divided by two. Don't forget the I now distributing the +33 multiplied by one half is three halves plus three multiplied by square at three, divided by two is three square at three divided by two. Don't forget the I, there's a couple different ways to express that you can have the three I square root three divided by two. That's just how I'm going to write it for the sake of the space that I'm trying to take up. All right. So we've got two of these down. Z sub one is three halves plus three square root three divided by two. I now Z sub two. All right. So we're still taking the sixth threat of 729 multiplied by the quantity of the cosine of zero plus two pi but now are two pi is multiplied by the new K value of two all divided by six plus I sign of zero plus two. Pi multiplied by two all divided by six, six out of 729. Still three and multiplied by the quantity of the cosine of zero pi or excuse me, zero plus two pi multiplied by two all divided by six is going to simplify to be two pi divided by three plus I sign of the same thing. Two pi divided by three. Now there's something a little bit different here. It's still three multiplied by the quantity of now for our exact values for two pi divided by three, take a look at the unit circle that is in quadrant two. So in quadrant two cosine is going to be negative and sine is going to be positive. Our reference angle for two pipi divided by three in quadrant one is going to be pi divided by three. And that's what we've already been using is pi divided by three. So we know that the cosine of pi divided by three is one half. And the sine of pi divided by three is square root three divided by two. But now for those exact values, since it's the cosine of two pi divided by three in quadrant two, that's equal to negative one half. And then plus the sine of two pi divided by three is square root three divided by two. And then our I now we just distribute the 33 multiplied by negative one half is negative three halves plus three, multiplied by square root. Three, divided by two is the three square root three divided by two. And then our, I afterward there's our Z sub two. We're halfway there. All right. Now, I'm going to move this to a different part of the screen. So we got a little more room. OK? Our next K value is four Z, excuse me, it's our fourth K value. Our next K value is three three ZZ sub three equals the sixth threat of 729 multiplied by the quantity of the cosine of zero plus two pi multiplied by our K value of three all divided by six plus I sign of same thing zero plus two pi multiplied by three, all divided by six. And simplifying sixth root of 729 remains three, multiplied by the quantity of the cosine of zero plus two. Pi multiplied by three divided by six. Simplifies to be pi plus I sign of pi. Now the exact values for the cosine of pi and the sine of pi. So we've got three multiplied by the quantity of the exact value of the cosine of pi is negative one. And the exact value for the sin of pi is zero. So plus zero, I distributing the 33 multiplied by negative one is negative 33 multiplied by zero I is zero. So we've just got for our Z sub three, this is equal to negative three. All right, two more KZ sub four, our 5th, 1/5 K value. So we're still taking the sixth threat of 729 multiplying that by the quantity of the cosine of zero plus two pi now multiplied by the K value of four all divided by six plus I sine of the same thing zero plus two pi multiplied by the K value of four all divided by six, six thread of 729. 3 multiplied by the quantity of the cosine of zero plus two pi multiplied by four, all divided by six is going to be four pi divided by three plus I sign of four pi divided by three. All right. So it's three multiplied by the quantity of now for our exact values of four pi divided by three, that one, you look at your unit circle, that one's in quadrant three. So that means our cosine and our s are both negative. The reference angle in U in quadrant 144 pi divided by three is again pi divided by three. So the cosine of pi divided by three is one half and the sine of pi divided by three is square root three divided by two. So when we're substituting in the exact value for the cosine of four pi divided by three, and quadrant three is negative one half plus I. And then the exact value for the sine of four pi divided by three is negative square root three divided by two. Now distributing our 33 multiplied by negative one half is negative three halves and three multiplied by negative square root. Three divided by two is going to be minus three square root three divided by two. And then our I, so we're Z sub four has been found. We've just got one more. OK. Well, we have room on the screen. Let's see. All right, Z sub our final K value is five equals the sixth rut of 729 multiplied by the quantity of the cosine of zero plus two pi multiplied by our final K value of five all divided by six plus I sine of the same thing. We've got zero plus two pi multiplied by five all divided by six. So the sixth thread of 729 surprises, nobody is three multiplied by the quantity of the cosine of zero plus two pi multiplied by five all divided by six. Simplifies to be five pi divided by three plus I sign of the same thing. Five pi divided by three. And so that's three multiplied by the quantity of. Now, let's take a look at our five pi divided by three, look at the unit circle that's in quadrant four. So in quadrant four, cosine is going to be positive sine is going to be negative. Our reference angle in quadrant one for five pi divided by three is still pi divided by three. So the cosine of pi divided by three is still one half. The sine of pi divided by three is still square root three divided by two. So this three is multiplied by cosine of five pi divided by three is positive one half plus our I and the sine of five pi divided by three is negative square at three divided by two. Now distributing three multiplied by one half is three halves. N three multiplied by negative square root. Three divided by two is minus three square root three divided by two. Don't forget our I and there's our Z sub five and there's all of our roots, all six of them. Well done. We'll catch you on the next one.

In Exercises 69–76, find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth.The complex sixth roots of 64

Key Concepts

Complex Numbers

Roots of Complex Numbers

Rectangular Form

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