Trigonometry Polar Form of Complex Numbers

We are working on some more examples of polar form of complex numbers.0000

Remember the equations for polar form of complex numbers are exactly the same as the equations for polar coordinates that we have learned before in the previous lecture.0004

If you are having any trouble with these examples, you might want to review the previous lecture on polar forms of polar coordinates.0015

Once you understand polar coordinates really well, then the conversions for complex numbers into polar form uses exactly the same equations, even the same special cases.0025

They should make more sense for you.0035

In this example, we are converting one complex number from rectangular to polar form and another one from polar to rectangular form.0038

Let me remind you what the conversion formulas are, r=(square root) of x² of x² + y².0048

(theta)=arctan(y/x) and then just like with polar coordinates, sometimes you have to add on an extra pi.0057

The time when you have to do that is if x is less that 0, if x is bigger than 0 then you just stick with the actan(y/x).0069

I will go ahead and show you the conversions on the other direction.0078

For me the easiest one to remember is e^(i)(theta) = cos (theta) + (i) sin (theta).0082

If you do not want like that one you can also work everything out from x=arcos(theta) and y=arcsin(theta), either one of those will work for you.0092

Let us look at (z) now, (z) is (–root 2) – ((root 2(i)), my(r) is the square root of (root 2)² is just 2 + (root 2)² is 2 again.0104

This just simplifies down to the square root of 4 which is 2, (theta) =arctan(y/x), arctan(root 2)/(root 2) because the negative is cancelled.0122

But we still have (x) less than 0 so I have to add on a pi, arctan(1)+pi.0139

arctan(1) that is a common value, I know that is pi/4 + pi which is 5pi/4.0145

My complex number is (re)^(i)(theta), 2e^5pi/4(i).0166

It helps to check that graphically, if you graph ( –root 2) – ((root 2 (i)), the circle is a little bit loft sided, but that is not important.0179

(-root 2)-((root 2 (i)) that is down on the third quadrant, that is somewhere down here and if you look that really is at an angle of 5pi/4.0198

That checks that we probably got the right angle there.0212

On the other one, 6e^(5pi/6(i) that is the polar form, I’m supposed to convert it to rectangular form.0218

W=6, I’m going to use this form e^sin(theta)=cos(theta) + (i) sin (theta).0225

So 6 x cos(5pi/6) + (i) sin (5pi/6), if you do not like that you can also use x=arcos(theta), y=arcsin(theta), you will get to exactly same formula.0233

In fact there would not even me more steps, they would be just about the same number steps.0251

Which ever one you is more comfortable to you, feel free to use that one.0254

I’m drawing my unit circle and find 5pi/6 on it, 5pi/6 down there, it is just a little bit short of pi and that is a common value.0261

I know that is the one with the square root of 3/2 and 1/2 , we just got to figure out which one is positive and which one is negative.0273

The (x) one is negative, so this is 6 x (-root3/2) + (i) sin (5pi/6) is +1/2 because the y is positive.0281

This simplifies down to 6/2 is 3, -3(root 3) + 3 (i).0296

Each one of those was a pretty straight forward application of the formulas, one we had r=square root of (x² + y²).0312

And then the arctan formula for (theta), remembering that you put a correction if the (x) is less than 0, and we found our (r), we found our (theta), we did put on the correction.0321

By the way,I have been doing all these in terms of radians, if you found arctan(1) in terms of degrees, if your calculator was in degree form it would have given you 45 and you would have to correct that in radians.0334

In this case, I did not even use a calculator because arctan(1) is a common value, I remember that was pi/4.0349

Add on my correction term of pi and I get 5pi/4 and so (z)=re^(i)(theta) so 2 x 5pi/4(i).0358

On the other one we have to convert from polar to rectangular form, the polar form was 6e^5pi/6(i).0370

You could use the conversion formula x=arcos(theta), y=arcsin(theta) or you can use the formula e^(i)(theta) = cos(theta) + (i)sin(theta).0378

I really like that one so I plugged that one in, drew a unit circle to remind me where 5pi/6 is and what is sin and cos are, fill those in and I got the rectangular coordinates for that complex number.0389

Ok now we are asked to simplify the expression 1 + (i)⁷.0000

Now that would be a really nasty one if we had to multiply all that out to the 7th power.0005

Instead, what we are going to do is convert to polar form and hopefully the exponentiation will be easier in polar form and after we expand it out in polar form we will convert it back to rectangular form.0009

Let us see how it goes, remember r=square root(x² + y²) and (theta) = arctan(y/x).0022

Sometimes you have to add on an extra pi there, you do that when (x) is less than 0.0036

1 + (i) that means that x is 1 and my y is one, r = square root (1+1) which is square root(2).0043

(theta) is arctan(1), that is a common value pi/4 and I do not have to introduce the fudge factor this time because the x is positive.0059

You can check that on the unit circle 1 + (i) is right there and that does check that the radius is the square root of 2 and the angle is pi/4, that checks my work here.0074

What we have here is square root of 2 x e^pi/4(i) that is the polar form of the complex number 1 + (i).0096

We want to raise that to the 7th power, so we raise both sides to the 7th power.0117

Now that looks pretty horrible but this just turns into square root of 2⁷, now pi ⁴⁽ⁱ⁾.0123

e to 1 power raised to another power, you just multiply the exponents, that just turns into e^7th(i)/4(i) .0134

That is the beauty of the polar form is the exponents just multiply or add instead of making it really difficult in multiplying lots of things together.0148

The (e) part is already raised to the 7th power, (root 2)⁷ might take a little bit of work.0158

Let me look at this, I will write that as 2^1/2 to the 7th power and then you could write that as 2^{7 ½}.0165

2^{7 ½} is the same as 2^3½, 7 ½ is the same as 3½.0178

That is the same as 2 cubed x 2^1½, law of exponents there and 2 cubed is 8, 2^{1 ½} is root 2 there.0186

Root (2⁷) is 8 (root 2), so this whole thing turns into 8 (root 2) x e2^{7 pi/4(i)}.0199

I like to expand out e^{7 pi/4(i)}, convert that back into rectangular form.0214

For me, the easiest way to do that is with the formula e^(i)(theta) = cos(theta) + (i)sin(theta).0223

That one works really well for me, but you can also use x=arcos(theta) and y=arcsin(theta) if you like.0234

But I’m going to try the cos(theta) + (i)sin(theta), let me find where 7pi/4 is.0243

That is just short of 2pi, it is down there 7pi/4, pi/4 short of 2 pi.0251

That is a common value, I know the sin and cos of that.0261

There is no (i) there, plus (i) sin (7pi/4).0272

That is 8 (root 2) I know the sin and cos of 7pi/4 they are both (square root of 2)/2.0281

I just have to figure out which one is positive and which one is negative and since the y coordinate is negative there, we are below the axis.0288

The sin(1) is negative, the cos(7pi/4) is +root2/2.0298

The sin is –root 2/2, and now I just have to simplify this, 8 x (root 2) x (root 2)/2 that is 4 x (root 2) x root, that is 4 x 2.0306

4 x 2 – (i) x 4 x 2, where I am getting 4 x 2 that is 8/2 and then (root 2) x (root 2) so this is 8 – 8(i).0325

That was a little bit long but if you think about it, figuring out 1 + (i) ⁷ directly would also be long because we have to multiply complex numbers together and they get bigger and bigger.0344

It would have been pretty complicated if we did it in rectangular form.0355

Let me recap the steps that we did to solve this problem, we converted it into polar form, first of all.0359

We have to find (r) and (theta), we found (r) using square root of (x ² + y ²).0366

(theta) was just arctan(y/x) and we did not have to introduce the fudge factor because the x was positive.0373

Arctan(1) that was a common value, I know that it is pi/4.0380

This number converted in to the polar form (root 2) x e^pi/4(i) and we have to raise that up to the 7th power.0385

We got (root 2) ⁷ and e^pi/4(i) raise to the 7th power, you just multiply the exponents.0395

That was the real time saving step there was just multiplying the exponents and there was a little work of figuring out what (root 2) ⁷ was.0404

Right that is 2^{1/2 to the 7th}, 2 ^7/2, 2 ^{3 ½} and then separate that into 2 cubed is 8 and 2 ½ is root 2.0414

To figure out e^7pi/4(i), you could use arcos(theta) and arcsin(theta), but I like to use e^(i)(theta) is cos(theta) + (i)sin(theta).0425

That is what I’m doing here, I found the sin and cos of 7pi/4, that is a common value.0437

I figured out which was positive and which was negative and then I just multiplied it through to get my answer.0444

Simplifying, the (root 2) is cancelled and I got 8 – 8(i) as my answer there.0450

So we really did get some mileage out of converting into polar form.0455

That is the end of our lecture on complex numbers in polar form, these are the trigonometry lectures on www.educator.com.0461

Hi, these are the trigonometry lectures on educator.com and today we're going to talk about polar form of complex numbers.0000

A lot of what we're learning in this lecture is very directly related to polar coordinates.0007

If you're a little rusty on polar coordinates, what you might want to do is go back and review what you learned about polar coordinates before we learn about polar forms of complex numbers.0015

In particular, the main formulas for converting a complex number into polar form, they're exactly the same formulas that you learned for polar coordinates.0023

They should be familiar to you when we go through them now.0034

If they're very rusty, you might want to go back and practice those formulas for converting a point into polar coordinates and back, because they'll be really helpful in this section of polar forms of complex numbers.0037

Let's start out there.0050

Complex numbers can be written in rectangular form, z=x+yi.0052

That represents, if you graph it, then you have an x-coordinate and a y-coordinate.0058

We write the rectangular formula complex number as x+yi.0070

Just like with points, you would give the coordinates as (x,y), with complex numbers, we give the form as (x+yi).0075

They can also be written in polar form, z=re^iθ.0085

That represents the polar coordinates of the same point.0093

reⁱθ, sometimes people write it as re^θi.0098

That represents the polar coordinates of the point, r is the radius from the origin going diagonally instead of going in a rectangular fashion.0109

θ represents the angle that makes with the positive x-axis.0121

Just like we've had polar coordinates rθ who have the polar form of a complex number re^iθ.0127

The conversion's back and forth between those two forms are exactly the same as what we've had for polar coordinates.0137

Let's check those out.0145

The conversion for r is square root of x²+y².0147

That comes straight from the Pythagorean theorem.0152

The conversion for θ is a little more complicated and it's got the same kind of subtleties and nuances that it had with polar coordinates.0155

θ is either arctan(y/x) or π+arctan(y/x).0164

The way you know which one of these formulas to use is you check the sine of x.0172

This is when x is greater than 0.0178

This is when x is less than 0.0183

Another way to remember that is to ask whether the point is in quadrant 1, 2, 3, or 4.0187

Remember arctangent will always give you a value in quadrants 1 or 4.0200

If you start out in quadrant 1 or 4, then you just want to use the arctangent function directly.0206

If you're looking for point in quadrants 2 or 3, then the arctangent will not give you the right value, that's why you add π to it.0213

That's the tricky one.0235

x and y, same formulas as we had for polar coordinates before, rcos(θ) and rsin(θ).0237

We'll try to use values of r that are positive, but that's not absolutely essential.0245

We'll try to use values of θ that are between 0 and 2π, but that's not absolutely essential.0249

Let me give you one more formula that's very very useful in working out conversions between rectangular and polar coordinates.0255

We write re^iθ as x+yi.0265

I'll write that as iy.0277

Polar form is re^iθ, rectangular form is x+iy.0280

If you convert that, the x is rcosθ, iy is irsin(θ).0291

If you factor out an r there, we get r×cos(θ)+isin(θ).0304

If you just take r=1, if you factor out the r from both sides, what you get here is the e^iθ=cos(θ)+isin(θ).0315

That is an extremely useful formula in converting complex numbers to polar coordinates.0328

That one is probably worth memorizing as well.0337

c=cos(θ)+isin(θ).0338

Let me decorate that a little bit, illustrate how important it is.0344

e^iθ=cos(θ)+isin(θ), that's definitely worth remembering.0351

We'll be using it on some of the examples.0359

Let's go ahead and practice doing some conversions here.0362

One more thing that I need to show you before we practice that.0367

Multiplying two complex numbers in polar form.0371

If we have two complex numbers in polar form, r₁×e₁^iθ, it's got an r and a θ, and r₂×e₂^iθ.0375

There's a very easy way to multiply them.0386

If we multiply these together, what we do is we just multiply the r's together r₁×r₂.0390

Remember the laws of exponents x^a×x^b=x^a+b.0399

Here, we have x or e^iθ₁×e^iθ₂, you add the exponents, iθ₁+iθ₁, just gives you i(θ₁)+θ₂).0409

You add the exponents there.0426

You end up just multiplying the r's and adding the angles θ₁+θ₂ because they're in the exponents.0429

Now let's try some examples.0436

We're going to convert the following complex numbers from rectangular form to polar form.0438

Let's start out with -3+i.0445

The -3 is x and y is 1 there.0449

We want to find the r and θ, r is the square root of x²+y².0454

Let me write this at the top page so I don't have to keep rewriting it.0460

θ=arctan(y/x), that's if x > 0, or we might have to add π to that if x < 0.0464

In this case, our r is the square root of x², negative root 3 squared is just 3, +y² is 1, that simplifies down to 2.0484

θ=arctan(y/x), y=1, 1 over negative root 3, which is arctan of negative root 3 over 3.0498

That's one of my common values.0523

I know what the arctan negative root 3 over 3 is, it's -π/6.0525

My x-coordinate was negative there so I haven't actually been using the right formula, I have to add π to each of these, +π.0536

You almost always use radians and not degrees here.0546

If you do happen to plug this into your calculator, make sure your calculator is in radian mode.0549

I didn't have to use my calculator on this one because negative root 3 over 3 is a common value.0556

(-π/6)+π=5π/6.0562

My polar form for that complex number is r₂e^iθ, so e^(5π/6)i.0572

Let's keep going with the next one.0591

6 root 2, that's my x, -6 root 2, that's my y, r is the square root of x², 6 root 2 squared is 36, times 2 is 72, +y² is 6 root 2 again, 72, square root 144 is 12.0595

θ=arctan(y/x), that's negative 6 root 2 over 6 root 2, which is arctan(-1) which is -π/4.0626

My x in this case was positive so I don't have to introduce that correction term.0642

I get w=re^iθ=12e^(-π/4).0650

I don't really like that negative value of π/4, so what I'm going to do is to make it positive, to get it into the range, 0 to 2π, I'll add 2π to it.0661

I'll write that as 12e, I need an i there, ex^(7π/4)i.0672

You can also understand these things graphically.0685

Let me draw a unit circle here.0693

Negative root 3 plus i, that means my x is negative root 3 and my y is 1.0707

I recognize that as a multiple of root 3 over 2 and -1.0717

I recognize that as being over here.0724

That's z with radius of 2, because it's 2 times root 3 over 2 and 1/2, I know that that's 5π/6.0726

That's the way to kind of check graphically that my z is 2×e^(5π/6)i.0742

For w, 6 root 2 minus 6 root 2, I know that's 12 times root 2 over 2 root 2 over 2, except the y is negative.0750

That value is 7π/4, that's kind of a little graphical check that we have the right polar form for the complex numbers.0774

Let's go back and recap what we did for that problem.0788

We're converting complex numbers from rectangular form to polar form, really just boils down to these two conversion formulas for r and θ.0791

r gives you the magnitude, θ gives you the angle.0801

The problem though is that this θ formula is little bit tricky.0802

It has this two cases depending on whether x is positive or negative.0808

If x is negative then you have to add an extra π to it, that's what we did here, we were adding an extra π to the value of θ.0811

Once you find r and θ, you just plug them into this form re^iθ.0820

That's how we got the answers for each of those.0825

For the next one, we're converting from polar form to rectangular form.0830

We're given z=4e^(-2π/3)i, and w=2e^(3π/4)i.0834

Let me write down the conversion formulas.0845

x=rcos(θ), y=rsin(θ).0847

For the first one, x=4cos(-2π/3).0859

Let me graph that quickly on the unit circle.0871

-2π/3 is down here, it's the same as 4π/3.0878

The cosine is -1/2, that's a common value.0883

This is 4×-1/2, which is -2.0890

The y there is 4sin(-2π/3), the sine of that is negative root 3 over 2.0894

This is 4 times negative root 3 over 2, which is -2 root 3.0913

We're going for the form x+yi, our z is equal to x=-2, +yi, -2 root 3, i.0925

For the second one, we have 2e^(3π/4)i.0943

I'll graph that on the unit circle to help me find the sine and cosine.0951

3π/4 is over there, it's 45-degree angle on the left-hand side.0954

I know the sine and cosine very quickly.0958

x is equal to r which is 2, cosine of 3π/4, which is 2, cosine of that is negative because it's on the left-hand side.0962

2 times negative root 2 over 2, which is just negative root 2.0977

y=2sin(3π/4), which is 2 times positive root 2 over 2 because we're in that second quadrant, y-coordinate is positive.0987

x+yi is negative root 2 plus root 2i.1000

That one wasn't too bad, it was simply a matter of remembering x=rcos(θ), y=rsin(θ), then putting those into x+yi.1020

For finding the sines and cosines, it helps if you graph the angle in each case.1034

Once you remember those formulas, you just work in through arccos(θ) and arcsin(θ) in each case.1041

For the third example, we're going to use polar form in an application.1050

We're going to perform a multiplication by converting each one of the complex numbers to polar form, then we're going to check the answer by multiplying them directly in rectangular form.1054

-1 plus root 3i, I'm going to figure out my r there.1065

My r is equal to square root of x²+y².1071

Let me write these formulas generically, x²+y², θ=arctan(y/x), that's if x is bigger than 0, we'll have to add on a π, the fudge factor π if x < 0.1078

In the first one, r is equal to 1² plus root 3 squared, that's 3, which is 2 square root of 1 plus 3.1104

θ is equal to arctan negative root 3 over 1.1113

Let me write that as root 3 over -1.1123

I have to add on a π because the x is negative.1129

Arctan of negative root 3 is negative π/3+π, that was a common value that I remembered there.1134

Plus π gives me 2π/3.1144

That tells me my r and my θ for the first one.1151

Let me go ahead and figure them out for the second one before I plug them in.1152

For the second one, we have r is equal to the square root of 2 root 3 squared.1155

2 root 3 squared is, 4 times 3, is 12, plus 2 squared is 4, 12+4=16, that gives me root 16 is 4.1163

θ is arctan 2 over 2 root 3, but the x-coordinate was negative, I have to add a π, so this is arctan 1 over root 3, is root 3 over 3 plus π.1177

Again, that's a common value, so arctan of root 3 over 3, I remember that's a common value, that's (π/6)+π=7π/6.1205

If I convert one to each one of these numbers into polar form, this one is 2e^(2π/3)i.1220

This one is 4e^(7π/6)i.1235

I want to multiply those, but multiplying numbers in polar form is very easy.1247

First, you multiply one radius by the other one, that's 2×4=8, then you add the angles e^{((2π/3)+(7π/6))i}.1255

You just add the angles, you multiply their radius by the other one and then you add the angles.1272

That's 8e to the, let's see (2π/3)=4π/6, you get (11π/6)i.1277

I want to convert that back into rectangular form.1287

I forgot to put my e in there.1293

I'm going to use this formula e^iθ=cos(θ)+isin(θ).1296

That one's really useful, definitely worth remembering.1302

This is 8cos(11π/6)+isin(11π/6).1306

You could also use x=rcos(θ), y=rsin(θ), you'll end up with the same formula at the end, either way works.1322

Let me draw on the unit circle to remind where 11π/6 is.1329

11π/6 is just short of 2π, it's right there.1336

It's a 30-degree angle south of the x-axis.1341

The cosine there is root 3 over 2, it's positive because we're on the right hand side.1345

The sine is -1/2.1358

What we get there that simplifies down to 4 root 3 minus 4i.1366

Now we've done it.1383

We've converted each number into polar form.1385

We multiplied them in polar form which is very easy, then we converted the polar form back into rectangular form to give us our answer.1388

It says we have to check our answer by multiplying them directly in rectangular form.1398

Let's do the check here, we'll FOIL the multiplication out.1404

I'll do the check over here.1410

I'll do the check in blue.1411

Foiling it out, my first terms give me -1 times 2 root 3, that's positive root 3.1414

My outer terms give me -1-2i, so +2i.1421

My inner terms give me -2 times root 3 times root 3, that's 6i.1427

Those are my inner terms, I'm doing FOIL here.1437

First outer, inner, and my last terms are root 3i minus 2i, that's -2 root 3, i², but i²=-1, this counts as +2 root 3.1440

If we simplify that down, we get 2 root 3 plus 2 root 3, 4 root 3, +2i-6i, is -4i.1459

That does indeed check with the answer we got by converting into polar form.1471

That was kind of a long one.1477

Let's recap what we did there.1478

We had these two complex numbers.1480

We wanted to convert each one into polar form.1482

For each one, I found my r, and I used square root of x²+y².1484

I found my θ by using arctan(y/x), [intelligible 00:24:55] each one the x's were less than 0, so I had to add on this fudge factor plus π to get me into the right quadrant.1491

I found my r, my θ, another r, my other θ.1504

I converted there each one into re^iθ form.1509

To multiply them together, you multiply the r's but then you add the θ's because they're up in the exponents.1514

That's the law of exponents there, so we added the θ's.1516

We got a simplified polar form and then we converted back into rectangular form using either the iθ=cos(θ)+isin(θ).1524

You could also use x=rcos(θ), y=rsin(θ), you'll get to exactly the same place.1535

I know my cosine and sine of 11π/6, that's a common value.1542

I get the answer there.1547

To check it, I skipped all the polar forms.1548

I just multiplied everything out using FOIL, simplified it down and it did indeed check with the answer that I've got using the polar form.1550

We'll try some more examples later.1563

You should try them on your own first and then we'll work on them together.1565