In Exercises 65–68, find all the complex roots. Write roots in po...

Question

In Exercises 65–68, find all the complex roots. Write roots in polar form with θ in degrees.

The complex cube roots of 8(cos 210° + i sin 210°)

Accepted Answer

Welcome back. I am so glad you're here. We're asked to determine all of the complex roots for the given complex number express the complex ruts in the polar form using angles and degrees. And we're given the complex square ruts of 256 multiplied by the quantity of cosine 240 degrees plus I sine 240 degrees. Our answer choices are answer choice. A 256 multiplied by the quantity of cosine 120 degrees plus I sine 120 degrees and 256 multiplied by the quantity of cosine 300 degrees plus I sine 300 degrees. Answer choice B four multiplied by the quantity of cosine 120 degrees plus I sine 120 degrees and four multiplied by the quantity of cosine 300 degrees plus I sine 300 degrees. Answer choice C 16 multiplied by the quantity of cosine 240 degrees plus I sine 240 degrees and 16 multiplied by the quantity of cosine 300 degrees plus I sine 300 degrees. Answer choice D 16 multiplied by the quantity of cosine 120 degrees plus I sine 120 degrees and 16 multiplied by the quantity of cosine 300 degrees plus I sine 300 degrees. All right. So we're looking for the complex square ruts of our given polar form of a complex number. And so we recall from previous lessons, Devers theorem, which tells us first, we need to find our NN is the NTH root that we're taking, we're looking for square rots, square rots, that's an N of two. So we will now take the square rut, the nth rut. So here it's the second rut of our modulus. The modulus is the number in front of the parentheses. And we're also going to take our argument, that's the angle, the 240 degrees in this case. And we're going to divide it by N. So dividing 240 degrees by two, this is going to get us our Z subzero our first rut. So we can go ahead and do this simplification. The square rut of 256 is 16. So we have 16, multiplied by the quantity of and the cosine of degrees divided by two is the cosine of degrees. Then plus I multiplied by the sign of and again, 240 degrees divided by two is 120 degrees. And so there's our first one. Now for our Z sub one, we're going to build off of our Z sub zero. So we can bring down 16 multiplied by the quantity of the cosine of and it is still 120 degrees. But to that 120 degrees, we're going to add 360 degrees divided by N. So in this case, N is two, so 360 degrees divided by two, which is 180 degrees. So it'll be the cosine of 120 plus 180 degrees and then bring down plus I multiplied by the sign of N again, we're taking not just 120 degrees, but it's 120 degrees plus our 360 degrees divided by N which is two and that's 180 degrees. And so we can do that simplification, bring over our 16 multiplied by the quantity of 120 degrees plus 180 degrees 300 degrees. So it's the cosine of 300 degrees plus I sine of degrees. And those are, are our two ruts. So we look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

In Exercises 65–68, find all the complex roots. Write roots in polar form with θ in degrees. The complex cube roots of 8(cos 210° + i sin 210°)

Key Concepts

Complex Numbers

De Moivre's Theorem

Polar Coordinates

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