Practice English Speaking&Listening with: De Moivre's Theorem Roots of Polar Complex Numbers

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BAM!!! Mr. Tarrou. Ok, we are going to do three examples of using De Moivre's Theorem

for finding complex roots that are in polar form. So, we have z sub k is equals the nth

root of r times the cosine of theta plus 2pi k over n plus i times sine theta plus 2pi

k over n, where k is a counting value starting at zero and going up to n minus one. Now I

have this formula written in radians where it is 2pi, instead of 360k. You can alternate

back and forth, it is fine. I think I am going to do examples in both radians and degrees.

So, let's get started with our first example. So our first example that I am going to take

a look at, is I want all the complex roots of 27. This is going to be z equals 27 times

the cosine of 210 plus i times the sine of 210. So it is a pretty straight forward formula.

The only problem is they are a little bit long because you have more than just one complex

root. If we are going to find a, oops, I didn't say which root I wanted. We are finding cube

roots. So, if we want all of the cube roots, this says that we are going to from zero,

one, two, etc all the way up to n minus one. So k starts at zero and if we are going to

find all of the cube roots that means n is equal to three. So we are going to use k values

of zero, one, and two. Let's get this started. So we have got the cube root of 27 times the

cosine of, now this is in degrees and not radians so let me put some degree marks in

there, so it is going to be the cosine of theta which is 270 plus 360...I know I wrote

2pi but I am going to examples in both degrees and radians. Right now I am doing degrees

which is 360 degrees for a full rotation...times k. Now k starts at zero. So we have 360 times

zero divided by n, the root that you are taking. I am taking the cube root, so n equals three.

Now plus i sine of the same thing, so theta plus 360 times the initial value of k which

is zero over n and we are taking the third root. We are probably going to have to erase

as I go because I am not going to fit all three up here at once. Or maybe I will just

change this as I go. The cube root of 27 is 3. We have the cosine, now we have 360 times

zero of course which is zero, so 210 divided by three is seventy. This plus i times sine

of 210 plus zero divided by three and 210 divided by 3 is 70 again. That is the first

cube root of 27 times the cosine of 210 plus i sine of 210. But the trouble is see, at

least as far as my space is concerned, is that there is two more answers. Let me try

to squeeze one more down here. The next one is going to be the cube root of 27 times the

cosine of 210 plus 360, but now we are going to let k step up one to one so,360 times one

divided by n which is three. This plus i times sine of 210 plus 360 times one all over three.

We are going to get again 3 times the cosine of 210 and 360, so we have 360...560...570,

and 570 divided by three is 190. There we go, ok. We have the first root with and angle

of 70. We have another root which is three times the cosine of 190 plus i times sine

of 190. Now we are going to find the third root. Well for the third root, I am just going

to do this again. I am going to write it for just the purpose of teaching, but you will

notice a pattern. Especially if you have three, or five, or six roots, you will notice a pattern

of how these angles are growing. You might want to short cut it, just make sure that

you include all of the roots. If you are taking the fifth root, you will have five roots for

example. So we are going to have as the last root, the cube root of 27 times the cosine

of again 210 plus 360 only now it is going to be multiplied by two, we are letting k

go up to two, over three the third root again. This plus i times the sine of 210 plus 360

now multiplied by two over three. That is going to be three times the cosine, again

just for practice this is 720...920...930 and 930 divided by 3 is 310. Yep, I am just

making sure. This plus i times sine of 310 again. Good. Now, I said you would notice

a pattern to how these angles grow. This is seventy and now you go to 190. 70 and 190,

that is a difference of 120. If you add 120 again to the 190, 190 and 120 makes up the

310. Now I am showing all of the work because I am assuming you are just learning how to

find the roots of complex numbers in polar form using DeMoivre's Theorem so I am just

showing you all of the steps. But, if you start to notice the pattern, the difference

between the angles, you can short cut this a little bit. Ok, let's do another example.

In this example, I would like to find all the cube roots of just 27. Now we are using the DeMoivre's

Theorem to find the roots, all of the roots, of complex numbers in polar form. Here I am

just giving you twenty-seven. The cube root of twenty-seven is three. But that is just

one of them. What about the others, the imaginary roots? You know, that is not even a complex

number. Really, I just don't have it written as a complex number. Let's write twenty-seven

as a complex number where there is a real and an imaginary part. So this is equal to

27 plus 0i. Now it is a complex number and let's get that complex number into polar form.

Well, you should always include a picture, so we have the point that is 27 units to right

on the real axis. That means that the answer is z equals twenty-seven because we are 27

units away from the origin times the cosine of zero...I am writing theta:/...plus i times

sine zero. Now we are going to find all of the cube roots of that number number, that

complex number in polar form. So that is going to be the cube root of 27 again time the cosine

of, I am going to do radians this time because my notes are in radians, zero plus 2pi times

zero over three because we are finding the cube roots. This plus i times sine of zero

plus 2pi times zero over three. The cube root of 27 again is three. We have zero divided

three, so we get the cosine of zero plus i times sine of zero. So, obviously that's kind

of like the real cube root of twenty-seven because we still have angle measures of zero

so we are looking at the 27 and the cube root of 27 is three and we are still along the

real axis. Let's find the other two. Just to be lazy, I am going to let k become now

one. So, we are going to find now z sub one. Those of you watching the video just realized

what I did right? We just found the real cube root and instead of letting k equal zero,

I am letting k be one and I am too lazy to write that all over again. So now the second

root of this complex number in polar form, and it will be a complex root, is we have

2pi times times one so it is going to be the cosine of 2pi over three because it is zero

plus 2pi times one over three, 2pi/3. This plus i times sine of 2pi over three. And 2pi/3,

pi over three is sixty degrees right, so the 2pi over three is over here. We are kind of

like, I am not going to do this exactly, but here is where my estimate of where three is

and then now there is going to be another complex root over here in quadrant two on

the imaginary coordinate system. Now we have found z sub zero, z sub one, and now we are

going find z sub 2. We are finding again three complex roots because we are finding the cube

root. If we were doing the fourth root, we would be finding four answers. Now let's let

k equal two, and humor me for my laziness here not wanting to write all of this over

again. Well, so the cube root of twenty-seven is still three. Zero plus 2pi times two, two

times two is four, so 4pi over three and again 4pi/3. So, there is another complex root down

here somewhere where 4pi/3 is. One movement is 60 degrees, another rotation of 60 degrees,

another rotation of 60 degrees, and finally...so one, two, three, 4pi/3. There is the three

complex roots of twenty-seven. Didn't know there was so many did you:) Ok, let's do one

last example and call it a day with this video, and just call it a day for me period! Ok,

we want to find the complex third roots of

one minus i. Well, again I have given you an example that is not already in polar form.

So, we need to get this complex number in polar form and then find the complex roots

of it. Draw it out. Real and imaginary, right and down. Now let's be honest. I am doing

examples that I can quickly do in my head. I can figure out the angle by knowing or using

my knowledge of the unit circle. But yet it may be a bit more complex for the type of

questions that are in your book, I hope so anyway. r is going to be the square root of

one squared plus negative one squared, so we know of course that is square root of two.

We are going to set up the tangent of theta just for practice. We know that it is 315

degrees or 7pi/4. But that is going to be negative one over one. Again, if you are using

or given a point that you can't just look at and tell what the angle and you are using

your calculator, theta equals the inverse tangent of negative one is going to give you

negative 45 degrees. Well, negative forty-five degrees is in quadrant four like the point

is, but I want a rotation...I want an angle measure that is between zero and 2pi. I want

manipulate the number out of the calculator a little bit and make sure that I have or

I am using an angle that is appropriate for the question that my textbook is asking or

wants the format of the question in. Let's see, my notes are in degrees as well. Let's

say that theta is equal to 315 degrees. (360-45) So we have a complex number that is square

root of two, in polar form, it is the square root of two times the cosine of 315 degrees

plus i times sine of 315 degrees. We are doing again the third root because I don't want

to do five or six different roots for the video. We will find the third root again.

That is going to be z sub zero is equal to the third root of two...square root of two...

Actually, let's write that appropriately. I almost botched this up a little bit. The

third root. I can't just put a three in the index, it is already a square root. So, the

square root of two raised to the one-third power. The square root of two now taking another

third root of it is going to be equal to.... Let's just figure that out so that I can use

it through the rest of the question. The square root of two is two to the one-half power.

Then that is raise the the one-third power because we are finding the third roots. One-half

times one-third makes one-sixth or the sixth root of two. So, when I was making my scratch

work I missed that. Hopefully or thankfully I caught that mistake for the video. So, the

third root of the square root of two is the sixth root of two and that is why. Let's go

ahead and just write that answer so that I get that in here. The sixth root of the square

root of two times the cosine of 315 plus 360 times zero divided by three because we are

doing cube roots. This plus i times sine of 315 plus 360 times zero over three because

we are doing the third root. That comes out to be the sixth root of two times the cosine

of... 360 times zero is zero so 315 divided by three is 105, now plus i times sine of

105. That is our first third root. Now for space, let's just come in here and change

that zero to one. Now we are going to find z sub one instead of z sub zero. I forgot

to fill that in. Of course, all that is going to do is change the size of the angle. So,

360 is now 660 and 670...675. Now 675 divided by three is 325. No it is not...haha 675 divided

by three is 225. Just a little brain fade there. So 675 divided by three is 225. So

that is the first and the second root, let's find the third root. Since I am getting a

little air headed here, that will be good that the video is almost over! Let's find

z sub two. We have 720...1020...1035, 1035 divided by three, I am just going to cheat

this time to make sure that I write the right answer the first time, is 345 degrees. There

are three examples of how to find roots of complex numbers in polar form. I am Mr. Tarrou.

BAM!!! Go Do Your Homework:) Thank you very much for watching. I hope you are finding

these to be helpful.

The Description of De Moivre's Theorem Roots of Polar Complex Numbers