In Exercises 81–86, solve each equation in the complex number sys...

Question

In Exercises 81–86, solve each equation in the complex number system. Express solutions in polar and rectangular form.x⁶ − 1 = 0

Accepted Answer

Welcome back. I am so glad you're here. We are asked to represent the solutions in polar and rectangular form. After solving the given equation equation in the complex number system, our equation is X rays to the sixth power minus equals zero. All right, if we were in the regular number system to solve this, we would start by adding 64 to both sides of the equation. And then on the left, we'd have x-rays to the sixth power and that would be equal to on the right 64. Then it would be as simple as taking the sixth thru of both sides of the equation. And on the right. Well, in the regular number system, we can just go ahead and take the sixth thread of 64 and we're done, but we're in the complex number system. So we need to take the root of a complex number, but 64 doesn't look complex. It looks regular. Well, we can use 64 as a complex number. Typically complex numbers are written in the format of A plus B I. Here 64 is the A term. So A equals 64 and then plus there are zero, I's so 64 plus zero I or B is equal to zero. So there's our complex number. We're taking the sixth rut of 64 plus zero. I's, and we can do that. Now, we know that when we're taking the nth rut of a complex number here, our N is equal to six, we can utilize Demos theorem in order to find all of the ruts. So the first step in order to do that is to get our complex number into polar form. We recall from previous lessons that polar form for complex numbers are multiplied by the quantity of the cosine of theta plus I sine theta where R equals the square root of A squared plus B squared. And we can find theta because we know that the tangent of theta equals B divided by A. We know A and B. So let's first solve for R, we're taking the square root of instead of A squared, that's 64 squared plus instead of B squared, that's zero squared, we know that zero squared is zero. So if we take 64 and square it and then take the square root, we're just going to be left with an R value of 64. Now let's figure out the data, the tangent of theta is equal to B divided by A B is zero, divided by A is 64 0, divided by 64 is zero. So where is it true that the tangent of theta equals zero. Well, we recall from previous lessons and from the tables in the book that that's true in two different places. That's true when theta equals zero and when theta equals pi, but which one are we talking about here? Well, we know that A is the horizontal component and A is the only one with value here. B is A zero. And so we know that when A is positive, we're talking about to the right of the origin, well, an angle of zero is to the right of the origin, an angle of pi is to the left of the origin. So here we're talking in about a theta value of zero. So for our polar form, we have, we have 64 multiplied by the quantity of the cosine of zero plus I sine zero. So now we can take the nth rut the sixth rut of our polar form of our complex number. Now following Demaris theorem to find the first value following the book, we'll call this RZ subzero. We are going to take the N th threat. So the six th threat of our modulus, that's the R value in front multiply that by the cosine of, we take our argument that's theta and divide that by N which is six and plus I sign of and again, take our argument zero and divide that by N which is six. So simplifying this to get Rz subzero, the six th threat of 64 is two multiplied by the quantity of the cosine of zero divided by six is the cosine of zero plus I sine of again zero. Now to get the following five ruts, there are going to be six of them total because N is equal to six, we are going to add a full rotation which is two pi divided by N which is six. We can simplify that two pi divided by six is pi divided by three. So we're going to add to each argument. Our pi divided by three for Z sub one that's still equal to two multiplied by the cosine of or the quantity of the cosine of and zero plus pi divided by three which is pi divided by three plus I sign of zero plus pi divided by three which is pi divided by three or Z sub two. We still have two multiplied by the quantity of the cosine of pi divided by three plus pi divided by three is two pi divided by three plus I sine two pi divided by three or Z sub 32, multiplied by the quantity of the cosine of two pi divided by three plus pi divided by three is just pi plus I sine pi. That sounds like I don't know. I'm just hungry for pie with all this pie. All right, Z sub four equals two multiplied by the quantity of the cosine of pi plus pi divided by four is four pi divided by three plus I sign four pi divided by three. And the last one, our Z sub five, the sixth right here. Two multiplied by the quantity of the cosine of four pi divided by three plus pi divided by three is five pi divided by three plus I sine five pi divided by three. OK. Those are our ruts in polar form. So this is part of the answer. Now, we just need to find them all in rectangular form which isn't too bad. There's a couple ways to do it. So for these, these cosine of zero and the cosine of pi divided by three, et cetera, those angles are all on the unit circle. So if you are solid with the unit circle, you can just say all right, the cosine of zero is this and then you can take that. And what you're gonna do is you're distributing that two in front of the parentheses to both the cosine value and the sine value. And then that'll be your answer. You can also do this using a calculator. So use, whichever method feels more comfortable for you. If you take two multiplied by the cosine of zero, you're going to get a positive two and then you take two multiplied by the sign of zero. For that one, you're going to get a zero. So that would be two plus zero. And don't forget the I and two plus zero is, is just two. If we take two multiplied by the cosine of pi divided by three, we are going to get a positive one and then distribute two multiplied by the sign of pi divided by three, we get a positive square at three. And then don't forget the I very often the I is written first so that everyone knows it's not underneath the radical. So you can write one plus I square at three. And that's the second value. The 3rd 12 multiplied by the cosine of two pi divided by three, you'll get a negative one and two multiplied by the sign of two pi divided by three, you are going to get a square at three again. So this one is plus I square at three. Now for the 4th 12 multiplied by the cosine of pi is a negative two and then two multiplied by the sign of pi is zero. So negative two plus zero is equals negative two. For the 5th 12 multiplied by the cosine of four pi divided by three is negative one and two multiplied by the sign of four pi divided by three is a negative square at three. So we can have minus I square at three. And for the last 12 multiplied by the cosine of five pi divided by three. That's a positive one and two multiplied by the sign of five pi divided by three is a negative square at three. So one minus I square at three. And those are our solutions. Who knew that the sixth row of 64 had so many solutions. All right. Well done. We'll catch you on the next one.

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In Exercises 81–86, solve each equation in the complex number system. Express solutions in polar and rectangular form.x⁶ − 1 = 0

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