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.                 _x³ − (1 + i√3 = 0

Accepted Answer

Welcome back. I am so glad you're here. We're asked to represent the solutions in polar and rectangular form. After solving the given equation in the complex number system, our equation is X cubed minus the quantity of one plus I equals zero. All right. So we're solving for X here and we're gonna start off as one would solving for X. We want to get it by itself on one side of the equation. So we'll add the quantity of one plus I to both sides of the equation. Noting that we're adding a complex number. Then on the left, all we have is X cubed and that's equal to on the right, a positive one plus I now to get X by itself, we want to take the cubed rut of both sides. But we notice on the right, we're taking the cubed rut of a complex number. And so we want to pause and we want to recall previous lessons on taking the rut of a complex number. So step one, taking the rut of a complex number is we want to make sure it's in polar form. This one here is not. So we recall from those same previous lessons that to get a complex number here, we've got one plus I and we want to get into polar form that's going to be R multiplied by the quantity of the cosine of theta plus I sign theta where R equals the square root of A squared plus B squared. And for finding theta, we know that the tangent of theta equals B divided by A but where are A and B? Well, we look at our complex number one plus I and that's in the format of A plus BI A is going to be the number without the I attached to it. So here A equals a positive one. B is going to be the number being multiplied by the I and also B equals the positive one. So now we can solve for R and for theta for R, we've got the square rot of an A squared, that's going to be one squared plus B squared. That's also one squared, one squared is one and one squared is also one. So we've got the square root of one plus one. That's the square rut of two. So our, our value is the square rut of two. Now let's find theta for the tangent of theta, we know that it's B divided by A B is one divided by A is also one, this is one. So when is the tangent of theta equal to one? And the tangent of theta is equal to one. When theta equals pi divided by four, we'll just keep this in radiant for this problem. So for polar form, we've got the square root of two multiplied by the quantity of the cosine of not theta, but pi divided by four plus I sine of pi divided by four. So we can go ahead and plug that back in over here. What we're doing is we're taking the cubed rut of the square root of two multiplied by the quantity of the cosine of pi divided by four plus I sine pi divided by four. And that's going to help us solve for X now because it's the cubed root. This tells us that our N is three. And now we call recall from previous lessons, Devers theorem which tells us when we're taking the end threat. So here, the Q threat we're going to and in summary of the simpler form of this, we're going to be taking the nth rut of our R. So it might be easier to think of this as we've got the square root of two. That's the same as two raise to the one half power. And we're raising that further to the one third power by taking the cubed root of it. So we're going to multiply our exponents one half multiplied by one third when we have a power raise to a power. So this is two raised to the 1/6 power, which is the same as the sixth rut of two. So we're taking the nth root of our R and then we've got to find three different solutions here. So that's going to be our Z subzero. We're going to have the sixth row of two. That's the cub of our R multiplied by. And then we're going to take our arguments. That's our, theta values here, pi divided by four. And we're going to divide those by our N. So we're taking pi divided by four, divided by N which is three, which is the same as multiplying pi divided by four by one third, which is pi divided by 12. So we have for our Z subzero that's equal to the sixth root of two, multiplied by the quantity of the cosine of pi divided by 12 plus I sine pi divided by 12. That gets us our first root here. And now for our Z sub one and our Z sub two, what we're going to be doing is we're going to take for our theta values for our arguments. We're going to add a full rotation. So that's two pi ingredients divided by N and here is three. So adding a two pi divided by three. So I'm gonna bring down, we keep our six thru of two, the same, that's multiplied by the cosine of now we have pi divided by 12 plus two pi divided by three. Well, two pi divided by three. If we want that with the denominator of 12, we're going to multiply that by four fourths. So we'll be adding in the numerator one pie plus eight pie, which is nine pie that's divided by 12. And that one reduces to three pi divided by four. So the cosine of three pi divided by four plus I sine of three pi divided by four. And now for the next one, we're going to add another full rotation divided by N. So adding another two pi divided by three. So now we're taking three pi divided by four plus two pi divided by three, getting those. So they have a common denominator. We're adding nine pi divided by 12 plus eight pi divided by 12. And in the numerator, nine pi plus eight pi is 17 pi that's divided by 12. So we bring down the six th threat of two multiplied by the quantity of the cosine of now 17 pi divided by 12 plus I sign of 17 pi divided by 12. And those are our answers in polar form, but we also need them in rectangular form. So highlighting these for a moment, so let's convert these to rectangular form. When we are converting to rectangular form, essentially, all we're doing is distributing the R value. So here we're distributing the sixth rut of two to the exact values of what's on the inside. And for this, you're gonna want to use a calculator because these aren't on the unit circle. So for rectangular form for our Z subzero, you can plug that into a calculator, the sixth row of two or two rays to the 1/6 power multiplied by the cosine of pi divided by 12. Make sure you're in radiance and you're going to get 1.1 and then distributing the six thread of two multiplied by the sign of pi divided by 12. You'll get plus 0.3. And then don't forget the I. Now for Z sub one, distributing that six thread of two to the cosine of three pi divided by four, we get a negative 0.8. And then for distributing the six thread of two to the sign of three pi divided by four, we get a positive 0.8. And don't forget the I for our Z sub two, distributing the six thread of two to the cosine of 17 pi divided by 12. We get a negative 0.3 and plus distributing the sixth thread of two to the sign of 17 pi divided by 12. We get, excuse me, that's minus, we'll have a negative 1.1. And then don't forget the I and those are our answers in rectangular form. 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. _x³ − (1 + i√3 = 0

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