Trigonometry DeMoivre's Theorem

Hi we are working on some more examples of DeMoivre’s theorem.0000

Our first extra example here is to convert the complex number 2 (root 2) – 2 (root 2i) in polar form and then use DeMoivre’s theorem to calculate z⁵.0004

Let us first convert that into polar form, remember my equations x² + y², square of that used to be (r) and (theta)=arctan( y/x).0016

Sometimes you have to introduce the extra term plus pi, you have to do that when (x) is less than 0.0035

Here, my x=2(root 2), y=-2(root 2), r= x² + y².0045

2 (root 2) ² is 4 x(root 2)² is 2, 4 x 2=8, y=-2(root 2), r= square root of 8 + 8, square root of 16 is 4.0058

My (theta) is arctan(y/x) is -2(root)2/2(root)2 that is -1.0077

There is no fudge factor on this one because the (x) is positive, there is no plus pi.0089

Arctan(-1) that is a common value that I remember is –pi/4.0095

If you work that out on your calculator in degrees it will say -45, if you have your calculator in radian mode it will say –pi/4.0102

But you should not really need a calculator for that because that is a common value.0109

That is –pi/4, that is still not in the range that I like which is 0 to 2pi, I think what I’m going to do is I’m going to add 2pi to that and get 7pi/4.0115

That is now in the range between 0 and 2pi.0133

My z in polar form is 4e^7pi/4, that is the answer to the first part of the problem.0138

We have converted the complex number into polar form, DeMoivre’s theorem says we want to calculate z⁵ .0151

We can do that by z ⁿ=rⁿ x cos(n)( theta) + (i)sin(n)(theta).0159

That is how DeMoivre’s theorem applies to this one z⁵ = r², that is 4⁵ x cos(n)(theta), n=5, theta=7pi/4.0173

35 pi/4 + (i)sin(35)(pi/4).0194

Remember that 35/4 came from n(theta) which was 5 x 7 pi/4.0205

I forget to include there the (i) in the polar form, let me put that in, that is 7pi/4i.0213

There is supposed to be an (i) in there.0222

I do not like the fact that 35 pi/4 is not between 0 and 2 pi.0226

Let me see if I can simplify that down, 35 pi/4 if I subtract off 2 pi.0230

2pi is 8 pi/4, that would give me 27 pi/4, subtract off another 2 pi that would give me 19 pi/4.0242

Subtract off another 2 pi=27pi/4, subtract off another 2pi would give me 11pi/4.0259

It is still in the range but if I subtract off 2 pi again, that gives me 3pi/4.0265

This whole thing simplifies down a little bit 4⁵ is (4 x 4 x 4 x 4 x 4)= 1024.0274

Now the formula of 35pi/4 between 0 and 2pi is 3pi/4, so cos(3pi/4) + (i)sin (3pi/4).0287

To simplify that, it is helpful to draw a little unit circle and remember where 3pi/4 is.0307

That is right there on the unit circle, 3pi/4 between pi/2 and pi.0318

The cos and sin there of both (root 2)/2, you just got to figure out which one is positive and which one is negative.0326

But because it is in the second quadrant, the (x) is negative so that is (-root 2)/2, and (y) is positive so (root 2)/2.0337

We can simplify this down a little bit, this is -512(root 2) because we divided by 2 + 512 (root 2) and that is my answer.0352

Let us recap what we did for that one, we are given a complex number in rectangular form.0383

I took those at my (x) and (y) and I plugged them into my formulas for (r) and (theta) to find the polar form.0388

For the formula for (theta) the (x) was not zero so I had to include the- sorry, the (x) was positive, I did not have to include the fudge factor.0396

When I took the arctan I got –pi/4 so I added on 2pi to get it into the range of 0 to 2pi.0407

The polar form for that complex number is 4e^7pi/4i.0415

To raise it up to the nth power, I’m going to use DeMoivre’s theorem that says zⁿ is rⁿ, cos(n)(theta) +(i)(sin)(n)(theta).0423

(theta) was 7pi/4, the n=5, so I get 35pi/4, pretty nasty.0434

So I subtract off a bunch off 2 pi and get it down to 3pi/4, I fill in the cos and sin of 3pi/4, which I remember those are common values on the unit circle there.0441

Finally I simplify it down and I get my answer there.0458

For our last example, we are trying to find all complex 4th roots of the complex number -2-(2 root 3(i)).0000

From the beginning I should say we expect 4 answers because we are looking for 4th roots.0009

At the end of this, we should have 4 answers and I’m going to convert the number in polar form.0023

Let me remind you of the formulas for polar form, x² + y² and (theta)=arctan(y/x).0031

If (x) is less than 0 you have to introduce the fudge factor plus pi there and I will go ahead and find that right away.0042

R=square root of -2² is 4, 2 (root 3) ² is 2² x 3, that is 4 x 3 x 12.0051

That is the square root of 16 is 4, (theta)=arctan(y/x) so that is 2 (root 3)/2 (root 3).0069

Positive because we have two negatives, but (x) is negative so I do have to introduce the fudge factor of pi.0084

Arctan(root 3) that is a common value for arctan that is pi/3 + pi = 4pi/3.0092

I have got my (theta), (z)=re^i(theta), 4e ^{4pi/3 x (i)}.0108

That is the polar form for that complex number, we got the (r) and (theta), I have got polar form for the complex number.0125

Now I want to find all complex 4th roots of that, let me remind you what DeMoivre’s theorem says about this.0135

It says that z^1/n is equal to r^1/n x cos((theta + 2 k(pi/n)) + (i)sin((theta + 2 k(pi/n)).0144

You figure that out for each value of (k) from 0 to n -1, in this case (n)=4 so (k) goes from 0 to 3.0177

I want to make a chart of all the possibilities here, let us work out what (k) could be.0194

We already said that could be 0, 1, 2, and 3, now I want to figure out what this angle ((theta + 2k (pi/n)) is.0211

((theta + 2k (pi/n)) now (theta) =4pi/3 and n= 4, this is (4pi/3 + 2k pi/4) which simplifies down to 4pi/3/4 is just pi/3 + 2k pi/4 is just k pi/2.0227

Let us figure out what that value is for each value of (k), when (k)=0 that is just pi/3, when (k)=1 that is pi/3 + pi/2, the common denominator is 6, that is 2pi/6 + 3pi/6 = 5pi/6.0270

When (k)=2 this is pi/3 + 2pi/2 is pi which is 4pi/3, when (k)=3, we have pi/3 + 3pi/2, again common denominator is 6, we have 2pi/6 + 9pi/6= 11pi/6.0303

Those are the four angles we will be plugging into the sin and cos formula, if you think about that being alpha.0338

Next step is to figure out the sin and cos of alpha is, cos(alpha) + (i) sin (alpha).0346

When alpha is pi/3, let me draw a little unit circle to help me figure these values out.0364

When (alpha) = pi/3 that is up there, the cos and sin are ½ and root 3/2, they are both positive because we are in the first quadrant.0379

For 5pi/6, that is in the second quadrant over there, the sin and cos are root 3/2 and ½ and now the cos is negative, sin is still positive because the (x) is negative and the (y) is positive there.0406

4pi/3 is down here on the third quadrant, sin and cos are both negative now and its -1/2 – root 3/2.0429

And 11pi/6 is way over there, the sin and cos are root 3/2 and the sin is negative so –(i) x ½ there.0448

Those are the sin and cos of those four angles, by the way if you look at that unit circle that I drew.0477

Let me highlight where those points are, you will notice that they are exactly evenly spaced around the unit circle.0483

It is because we are adding on k pi/2 each time, we are adding on pi/2 each time.0492

We get these 4 angles that are exactly spaced out around the unit circle by enables of pi/2 and that is not an accident.0499

It is because we started out looking for 4th roots, we divide the unit circle into 4 parts and that is why you go around pi/2 each time.0509

That is no accident that those are evenly spaced out by the multiples of pi/2 and have we gone one more step we would have ended up at pi/3, where we started again.0514

Let me go back to finding these 4th roots, we find cos(alpha) + (i) sin(alpha).0533

The only thing we have to do left is to multiply on the r^1/n, let me figure that out.0540

R^1/n =4, so ¼, there is a clever thing we can do to the exponents here I know that 4 is 2² to the ¼ so that is 2^2/4, that is 2^1/2 which is root 2.0547

What we are doing here is we are finding r^1/n x cos(alpha) + (i) sin (alpha).0566

But the r^1/n is root 2, I’m just going to multiply root 2 by each of the complex numbers in the proceeding column of the chart.0578

Root 2 x the first complex number is root 2/2 + (i) x root 6/2, that is because root 2 x root 3 is root 6.0589

Next one is –root 6/2 + root 2/2 (i) – root 2/2 – root 6/2(i) and finally root 6/2 that is root 2 x root 3 – root 2/2(i).0604

I’m just going to box off this last column of the chart and call that my answers.0638

What that means is that these are four complex numbers if you take any one of these four complex numbers, those are my four answers.0649

If you raise any one to the 4th power, what you should get is the original complex number that we started with -2-(2 root 3(i)).0659

That is the end of our last problem, let us go back and recap and see what strategies we used to solve it.0672

We started with this number -2-(2 root 3(i)), I want to get that in polar form so I can use it to DeMoivre’s theorem.0677

DeMoivre’s theorem only works on complex numbers in polar form so I used my equation for polar form r=square root (x)² + (y)² and (theta)=arctan(y/x) plus the fudge factor of x is less than 0.0686

My (x) and (y) I got those from the original complex number and the (theta) I did have to use the fudge factor plus pi.0702

Arctan(root 3) is an angle that I know pi/3 and I get (theta) = 4 pi/3.0712

I got the polar form of the complex number, (r)(e) ^(i)(theta), r =4, e^(i)(theta)=e^4pi/3 x (i).0719

Now I used DeMoivre’s theorem, DeMoivre’s theorem says you look at z^1/n is r^1/n x cos(alpha) + (i) sin(alpha).0731

Where the (alpha) is this (theta) + 2k pi/n, and then you plug in different values of (k) going from 0 to n-1.0744

Where the n here is 4, so I run (k) from 0 to 3, that is n-1, and here I made a list of my different (k).0754

For each one of those I figured out that (theta) + 2kpi/n, that was the second column.0765

(theta) + 2k pi/n and that is what I was calling (alpha) so I plugged in (theta) = 4pi/3, n=4 and I plugged different values of (k) each time to get these 4 answers for (alpha).0772

To get there 4 answers for (alpha) DeMoivre’s theorem says you look at cos(alpha) + (i) sin(alpha).0788

I looked at cos (alpha) + (i) sin (alpha) in this next column of the chart.0795

To get those cos and sin I drew a little unit circle here and I read off the cos and sin of each one, those are common values so I do remember those.0801

Finally I have to multiply this by r^1/n, what I worked out over here is r ^1/n.0811

(r) was 4 and (n) was also 4, it is 4^1/4, y⁴ is 2 squared.0818

That gives you a quick way to figure out to simplify down 4^1/4.0825

Simplify that into square root of 2, so I multiply r^1/n square root of 2, that is here.0831

Multiply that by each of the values that we got in the proceeding column.0838

We multiply this by the square root of 2 and that finally gave me my 4 answers for the 4th roots of -2-(2 root 3(i)).0844

These are my 4 answers right here as my four 4th roots.0856

That is the end of our lecture on DeMoivre’s theorem to find nth powers and nth roots of complex numbers, in fact that is the end of all our lectures on trigonometry.0862

Thank you very much for watching, this is Will Murray for www.educator.com.0873

Hi, these are the trigonometry lectures on educator.com.0000

We're here today to talk about De Moivre's theorem.0004

The De Moivre's theorem is a little bit tricky.0010

The idea is that we're going to use the polar form of complex numbers to find nth powers and nth roots of complex numbers.0013

We start with a complex number z.0026

We write it in polar form re^iθ.0029

We learned in the previous lecture how to convert a complex number into polar form.0032

If that's a little bit unfamiliar to you, what you should really do is go back and review the previous lecture on how to convert a complex number into polar form, and how to convert it back into rectangular form.0038

There's several formulas that we're going to be using very heavily here.0052

One that we learned in the previous lecture is e^iθ=cos(θ)+isin(θ).0056

We're going to be using that really heavily.0064

Let's see how we can use that to find nth powers of complex numbers.0067

For trying to find zⁿ, we write that as re^iθ, to the nth power.0073

If you think about that, we can distribute this nth power onto the r and onto the e^iθn.0080

rⁿ, that just gives you rⁿ.0091

e^iθn, that's an exponent raised with an exponent.0099

You multiply the exponents.0106

That's where I get iθ×n here, so e^i×n×θ.0107

If you expand that, e^iθ=cos(θ)+isin(θ), e^inθ gives you cos(nθ)+isin(nθ).0116

That's where De Moivre's theorem comes in handy, and that's where it comes from.0133

You can expand this into rⁿ×cos(nθ)+isin(nθ).0137

Another way to start out with that is to expand e^iθ into cos(θ)+isin(θ).0145

The form we're going to be using most often is this form right here zⁿ=rⁿ×cos(nθ)+isin(nθ).0154

I know this looks like lots of stuff to remember here.0166

The key one that you want to memorize is this one right here, zⁿ=rⁿ×cos(nθ)+isin(nθ).0168

Memorize that one and we'll practice it during the examples.0180

The next step of using De Moivre's theorem is to, instead of finding nth powers, we're going to find nth roots.0185

For example, we'll find square roots and cube roots, and fourth roots of complex numbers.0194

This is quite a bit more tricky than it is with real numbers.0199

In fact, if you're looking for nth roots of a complex number, you always expect to find exactly n answers.0204

If a problem says find all the 8^th roots of a complex number, you better find 8 answers.0211

Unless of course the complex number happens to be 0, in which case the only roots are 0.0219

That's why I say every non-zero complex numbers has exactly n nth roots here.0224

Let me show you how to find them.0231

It's a little bit complicated.0232

First of all, we think about the nth root of z.0235

We write that as e^1/n.0236

That's re^iθn.0240

Remember, e^iθ is cos(θ)+isin(θ), to the 1/n.0248

Remember how when we were finding nth powers, we distributed the n into the r, we had r to the n.0254

Let me write this again.0265

zⁿ=rⁿ×e^niθ, which was rⁿ×cos(nθ), you multiply the angle by n,plus isin(θ).0266

With 1/n, replacing the n by 1/n, we get r^1/ncos(θ/n)+isin(θ/n).0296

We start out with just cos(θ/n)+isin(θ/n).0315

We look at (θ/n).0322

We have to find other nth roots as well.0324

Our first nth root is just (θ/n).0328

To find the other ones, what we add on is multiples of 2π/n.0329

That's why I say 2kπ/n.0340

We keep doing that for different values of k.0343

The reason we do that is we run all the values of k from 0 to n-1.0347

If we plugged in k=n into this formula, we get (θ/n)+(2nπ/n), which would be (θ/n)+(2π).0354

In terms of angles, that's the same as (θ/n) again.0375

That's why we stop at n-1.0379

We don't go to k=n, because when we get to k=n, we're repeating ourselves again.0383

Essentially, what we're doing here is we're breaking up the unit circle into multiples of (θ/n).0388

θ/n, (θ+2π)/n, (θ+4π)/n, we're just taking all these angles around the unit circle until we get back to θ/n.0401

This is a little bit tricky.0423

What we do is we find one answer for each value of k.0426

One nth root for each value of k.0432

Since we run k from 0 to n-1, that's a total of n nth roots.0445

That's worth remembering.0470

We'll practice these with the examples.0471

Anytime you have to find nth roots, the r part is easy, you do r^1/n, but then you have to find this cosine and i-sine formula for each angle, for each value of k from 0 to n-1.0476

You run this formula separately, n times over, and at the end, you have n complex numbers as your answers.0494

We'll check that out with some examples and you'll get the hang of it.0501

The first example here, we have to convert the complex number z equals negative root 3 plus i into polar form, then use De Moivre's theorem to calculate z⁷.0505

Remember to convert it into polar form.0518

You do r equals the square root of x²+y², θ=arctan(y/x).0524

Sometimes you have to modify that θ formula, sometimes you have to add on a π.0532

You know you have to do that when x is negative.0538

You do that if x is negative.0540

Let's find our r in our θ here.0542

Let me graph that thing.0544

Graph it just so that we'll be able to check whether our answer's possible.0557

Negative square root of 3 on the x-axis, i on the y-axis, that's about right there.0560

Let me calculate r and θ to see if it's possible.0568

r is the square root of x²+y², x²=3, y²=1, square root of 4 is 2.0574

θ is arctangent of 1 over negative square root of 3 which is negative root 3 of 3, that's a common value.0585

The arctangent of that is -π/6.0598

There's this fudge factor that I have to include here.0606

The x < 0 here, I have to add on a π.0610

I add on a π and I get 5π/6.0615

That does check with my little graph here because that really is 5π/6, the angle over there, and the radius does indeed look like about 2.0620

That's reassuring.0630

z=re^iθ, that's 2e^(5π/6)i.0632

We have converted a complex number into a polar form, that was the first part of the exercise.0642

The main part here is to use De Moivre's theorem to calculate z to the seventh.0650

Let's work that out.0654

z⁷, the whole point is we're going to use the polar form to find z⁷, so this is 2e^(5π/6)i7.0655

That's 2⁷.0671

I have each one exponent raised to an exponent.0675

I just want to multiply those two exponents e^7×5, is (35π/6)i.0678

35π/6 is a little cumbersome, that's not in between 0 and 2π.0696

I'll work on that a little bit.0703

In the meantime, 2⁷ is 128.0704

35π/6, how can I simplify that?0711

35π/6, let me subtract 2π, 12π/6, that's 23π/6, that's still not in my range between 0 and 2π.0715

Let me subtract another 2π, that gives me another 12π/6 off, is 11π/6.0732

That is in the range between 0 and 2π.0743

This is the same as e^(11π/6)i.0745

I want to convert that into rectangular form.0751

It's very good to remember this formula e^iθ is cos(θ)+isin(θ).0755

You can also use x=rcos(θ), y=rsin(θ).0766

I prefer the e^iθ form.0771

This is equal to 128cos(11π/6)+isin(11π/6).0774

11π/6, where is that on the unit circle?0797

That's just π/6 short of 2π.0801

That's down there.0808

That's a common value, I know what the sine and cosine are.0811

It's root 3 over 2 and 1/2.0814

Root 3 over 2 is positive, the sine is negative because it's below the x-axis so it's -1/2.0820

I have 128/2, that's 64 root 3 minus 64i.0831

That's what that simplifies down to.0842

Let's review how we did that one.0848

We start out with a complex number, and we have to convert into polar form.0850

I look at my formulas for r and θ including the fudge factor for θ if x < 0.0857

Run that through, my x and y are negative root 3 and 1.0861

I get an r.0866

I get a θ including the fudge factor.0867

That gives me re^iθ.0870

I've got my polar form.0872

To raise it up to the 7^th power, De Moivre's theorem says if you use polar form, then you just put 2⁷, then nθ.0874

This is the nθ.0889

That reduces down by subtracting 2π at a time, e^(11π/6)i.0890

This is really nθ here, cos(nθ), sin(nθ), although we reduced down by subtracting off 2π at a time.0899

We get 128 times the cosine and sine of 11π/6.0909

That's a common value.0914

I look at my unit circle to remember my sine and cosine of 11π/6.0915

I plug them in and I get my answer.0923

Second example here, we find to find all complex 8^th roots of the number 16.0928

In order to find all complex 8^th root, we have to think about 16 being a complex number.0934

Of course 16 is just the same as 16+0i, that is a complex number.0944

We're asked for 8^th roots.0952

Let me remind you that because we're asked for 8^th roots, we expect 8 different answers.0956

We have to find 8 answers here.0968

I'm going to try and write 16 in polar form.0976

r is equal to the square root of x²+y².0981

θ=arctan(y/x), plus π if the x happens to be negative.0987

My r is the square root of 16²+0², that's just 16.1001

θ, my y=0, arctan(0)=0.1007

My z=16e⁰ⁱ.1016

I could have worked that out, certainly 16e⁰=16 just by itself because e⁰=1.1027

That's nice to check our work.1034

z^1/8.1036

According to De Moivre's theorem, let me remind you what De Moivre's theorem said about complex nth roots.1044

It said that, you do r^1/n×cos((θ+2kπ)/n)+isin((θ+2kπ)/n).1050

You run this for different values of k.1086

k is equal to 0, 1, 2, up to n-1.1089

Here, n=8, so z^1/8, r=16^1/8.1096

Cosine is, θ=0, (0+2kπ)/8, plus isin(0+2kπ)/8.1113

There's going to be lots of values for k here.1144

Maybe I should make a little chart for what k is, and then the different angles that we have for each value of k, the different angles that we're going to be plugging into De Moivre's formula there.1147

(0+2kπ)/8 which actually simplifies down to just kπ/4.1165

For k goes from 0, 1, 2, 3, 4, 5, 6, you run it to n-1, and n=8, so 0 through 7 there.1182

We'll have 0, π/4, 2π/4 is π/2, 3π/4, 4π/4 is π, 5π/4, 6π/4 is 3π/2, and 7π/4.1194

For each one of these angles we're going to plug it in and we're going to get an answer.1224

I also have to simplify 16^1/8.1229

Let me see if I can do something with that.1233

16^1/8, I know that 16=2⁴, that's 2^4/8, 2^1/2 is square root of 2.1235

I'm going to multiply the square root of 2, that's 16^1/8 times cos(θ)+isin(θ) for each one of these values of θ.1250

Let me not reuse the same Greek letter θ, I'll use α.1264

For each one of these values of α here ...1280

Let me write down what cos(α)+isin(α) is for each one of these values of α.1288

Cos(0)+isin(0), cos(0)=1, plus isin(0)=0.1291

Cos(π/4)+isin(π/4), cosine and sine of (π/4) are both square root of 2 over 2.1311

That's k=1.1324

For π/2, the cosine is 0, the sine is 1.1329

For 3π/4, we have the cosine is negative root 2 over 2, the sine is positive root 2 over 2.1337

I think this will be easy to work out if I draw a unit circle, so that I can easily and quickly find the sines and cosines.1353

They're all common values but it helps to draw a unit circle to remember where things are positive and negative.1362

What I started with is π/4, there's π/2, there's 3π/4, π, we're going to move on to 5π/4, 3π/2, move on to 7π/4, and we started out as 0.1369

Let's see.1398

We've already hit 3π/4 moving on to π now.1400

Cosine and sine is -1+0i.1404

5π/4, cosine and sine are both negative root 2 over 2.1410

They're both negative.1418

3π/2, the cosine is 0 again, the sine is -1.1422

Finally, 7π/4, cosine and sine are root 2 over 2, but the cosine is positive and the sine is negative.1430

For our answers, what we have to do is multiply root 2 by each one of these.1441

Let me multiply the root 2 by each one of these.1447

The first one you just get square root of 2 times 1+0.1450

Multiplying root 2 times root 2 over 2, that gives me 2/2, which gives me 1.1462

Plus i to the same thing, so just i.1470

Multiply root 2 times 0+i, gives me root 2i.1471

Multiply root 2 by this, we get -1+i.1478

Multiply root 2 here, we get negative root 2.1485

Multiply root 2 here, -1-i, remember root 2 times root 2 over 2, is 2/2, which simplifies to 1.1489

Multiply root 2 here, we get negative root 2i.1502

Multiply root 2 here, we get 1-i.1507

We are finally done here.1514

We get 8 different answers, 8 different complex numbers here.1518

Each one of these complex numbers has the property that if you raised it up to the eighth power, it will come out to be exactly 16.1529

Let me write down what we found here.1541

Each one satisfies w⁸=16.1545

If you multiply them up by themselves eight times, you'll get back to 16.1555

That will be pretty messy I'm not going to check that here, but you can check it on your own if you like.1561

Let me recap how we found that.1569

We started out with the complex number 16, think about it as 16+0i.1570

We wanted to write that in polar form, so I founded r and a θ.1580

r=16, and θ=0, z=16e⁰ⁱ.1585

Then I use De Moivre's theorem which says that you get complex nth roots by doing r^1/n, that's where the 16^1/8 came from.1590

Then cosine plus isine of these angles (θ+2kπ)/n.1600

That's why I started to make this chart (θ+2kπ)/8.1609

You run the k from 0 to n-1, that's why I ran the k from 0 to 7.1613

For each one of those I got an angle that I called α, then I worked out cos(α)+isin(α) for each one of those.1621

That's where I got these sectional values here, and for that it was really helpful to plot my α's on the unit circle here.1630

Remember what the sines and cosines of each one of them was.1641

Those are common values so I didn't need to calculate to look those up.1645

Multiply each one by 16^1/8.1648

A little cleverness with the laws of exponents tells me that that's root 2.1654

Finally, I multiply each one of those by root 2, and I get 8 different answers, each one of them is an eighth root of 16 and complex numbers.1659

We're now going to find all complex cube roots of -1.1670

Cube roots, that's a third root, we expect 3 answers here because we're looking for cube roots.1675

I want to put that complex number into polar form first.1690

I'm going to use my r equals square root of x²+y², and θ=arctan(y/x), plus π if x < 0.1695

Negative 1, think of that as -1+0i, the x=-1, y=0.1713

My r is square root of x²+y², that is square root of 1, r=1.1720

θ=arctan(y/x), that's arctan(0), x < 0, so I have to add π.1729

Arctan(0)=0, θ=π.1741

z=re^iθ, 1e^iπ.1746

I've got my complex number into polar form, now I'm going to use De Moivre's theorem.1757

Let me remind you how that goes, it says r^1/n×cos(θ+2kπ)/n+isin(θ+2kπ)/n.1763

The key thing here is k=0, 1, 2, up to n-1.1790

Here we're finding cube roots, our n=3.1797

Let me make a little chart again of the angles.1803

n=3, θ=π, I'll make a chart of k and (θ+2kπ)/3 for each value of k here.1809

k goes from 0 to n-1, that's 0, 1, and 2.1837

(θ+2kπ)/3, when k=0, that's just, θ=π, π/3.1846

When k=1, that's (π+2π)/3, which is 3π/3, which is π.1856

When k=2, this is (π+4π)/3, which is 5π/3.1869

The three angles we're going to be looking at are π/3, π, and 5π/3.1879

Let me also work out r^1/n.1887

r^1/n, r=1, raised to 1/3 power is just 1.1890

That part is very easy.1897

We have cos(α)+isin(α) for each one of these α's.1899

Cos(π/3)+isin(π/3) ...1913

Let me draw out where that would be. π/3 is about right there.1922

π is right there.1927

5π/3 is down there.1931

Those are the three angles I'm going to be looking at.1932

Let me go ahead and include these on my chart.1936

Cos(α)+isin(α), cos(π/3)=1/2, isin(π/3) is root 3 over 2, that's because (π/3) is right there.1940

π, the cosine is -1, the x-coordinate, isine is zero.1967

5π/3,that's down here, cosine is 1/2, the sine is negative root 3 over 2.1980

That was cos(α)+isin(α).1992

r^1/n×cos(α)+isin(α) is kind of anticlimatic because we already figured out that r^1/n=1.1997

We're just multiplying each of these by 1.2012

We get 1/2 plus i root 3 over 2, -1 and 1/2 minus i root 3 over 2 as our three answers.2015

Remember we're looking for cube roots, so we did expect to find 3 answers, it's reassuring here that we found our 3 answers.2029

Let me remind you how we did that.2038

First of all, we were given a complex number.2040

We had to convert it inot plar form so I found my r and my θ using the standard formulas.2042

I did have to include the fudge factor plus π here because the x < 0.2049

Arctan(0) gave me 0, gave me θ=π.2050

My θ was π, my r was 1, so I get 1e^iπ.2057

I go to De Moivre's theroem which says, r^1/n×cos(θ+2kπ)/n, isin(θ+2kπ)/n.2063

I made a little chart of the different values of k.2075

You go from 0 to n-1.2079

For each one, I figured out (θ+2kπ)/n.2081

That gave me the π/3, π, and 5π/3.2087

I found the cosine plus isine of each one.2090

I multiplied those by r^1/n.2095

That gave me my 3 answers.2098

Those are the three complex numbers that are cube roots of 1.2100

Each one satisfies wr³=-1.2106

If you worked out wr³ for any of these complex numbers, you'd get -1.2111

We've got some more examples for you later.2116

Try them out on your own and then we'll work through them together.2118