For more information, please see full course syllabus of Trigonometry
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Polar Form of Complex Numbers
Main definition and formulas:
 Complex numbers can be written in rectangular form z = x + yi, representing the rectangular coordinates of the point.
 They can be also be written in polar form z = re^{iθ}, representing the polar coordinates of the point.
 r represents the distance from the origin and θ represents the angle it makes with the positive xaxis.
 Conversions:
r =
√x^{2} + y^{2}θ = arctan y x, if x > 0 (Quadrants I and IV) π + arctan y x, if x < 0 (Quadrants II and III) x = r cosθ y = r sinθ  Usually, r ≥ 0, but not necessarily.
 Usually, 0 ≤θ< 2π , but not necessarily.
 Euler's formula:
e^{iθ} = cosθ + i sinθ  If z = r_{1}e^{iθ1} and w = r_{2}e^{iθ2} are two complex numbers given in polar form, then we can easily multiply them:
zw = r_{1}e^{iθ1}r_{2}e^{iθ2} = r_{1}r_{2} e^{i(θ1+θ2)}
Example 1:
Convert the following complex numbers from rectangular form to polar form: z = −√ 3 + i, w = 6√ 2 − 6√ 2iExample 2:
Convert the following complex numbers from polar form to rectangular form: z = 4e^{[−2π /3]i}, w = 2e^{[3π/4]i}Example 3:
Perform the following multiplication by first converting each of the complex numbers to polar form. Check your answer by multiplying them directly in rectangular form.

Example 4:
Convert z = −√ 2 −√ 2i from rectangular to polar form, and w = 6 e^{[5π /6]i} from polar to rectangular form.Example 5:
Simplify the expression (1+i)^{7} by converting to polar form, performing the exponentiation, and converting back to rectangular form.Polar Form of Complex Numbers
 x = 3√2, y = − 3√2, r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
 Polar Form: re^{iθ}
 r = √{(3√2 )^{2} + ( − 3√2 )^{2}} = √{36} = 6
 θ = arctan[( − 3√2 )/(3√2 )] = − [(π)/4] + 2π = [(7π)/4]
 x = − 2√2, y = 1, r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
 Polar Form: re^{iθ}
 r = √{( − 2√2 )^{2} + (1)^{2}} = √9 = 3
 θ = arctan[1/( − 2√2 )] + π = 2.8
 x = − √3, y = − √3, r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
 Polar Form: re^{iθ}
 r = √{( − √3 )^{2} + ( − √3 )^{2}} = √6
 θ = arctan[( − √3 )/(√3 )] + π = [(π)/4] + π = [(5π)/4]
 x = rcosθ, y = rsinθ
 x = 8cos([(5π)/6])
 x = 8( − [(√3 )/2])
 x = − 4√3
 y = 8sin([(5π)/6])
 y = 8([1/2])
 y = 4
 x = rcosθ, y = rsinθ
 x = 6cos( − [(3π)/4])
 x = 6( − [(√2 )/2])
 x = − 3√2
 y = 6sin([(5π)/6])
 y = 6( − [(√2 )/2])
 y = − 3√2
 x = rcosθ
 x = 8cos([(3π)/4]) ⇒ x = 8( − [(√2 )/2]) ⇒ x = − 4√2
 y = rsinθ
 y = 8sin([(3π)/4]) ⇒ y = 8([(√2 )/2]) ⇒ y = 4√2
 r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π
 Polar Form: re^{iθ}
 e^{iθ} = cosθ+ isinθ
 r = √{1^{2} + 1^{2}} = √2
 θ = arctan1 = [(π)/4]
 Polar Form: √2 e^{[(π)/4]i}
 Exponentiation Form: (1 + i)^{6} = (√2 e^{[(π)/4]i})^{6} (1 + i)^{6} = (√2 )^{6}e^{[(6π)/4]i} (1 + i)^{6} = 8e^{[(3π)/2]i}
 r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π
 Polar Form: re^{iθ}
 e^{iθ} = cosθ+ isinθ
 r = √{1^{2} + ( − 1)^{2}} = √2
 θ = arctan( − 1) = − [(π)/4] + 2π = [(7π)/4]
 Polar Form: √2 e^{[(7π)/4]i}
 Exponentiation Form: (1 − i)^{5} = (√2 e^{[(7π)/4]i})^{5} ⇒ (1 − i)^{5} = (√2 )^{5}e^{[(35π)/4]i} ⇒ (1 − i)^{5} = 4√2 e^{[(3π)/4]i}
 r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π
 Polar Form: re^{iθ}
 e^{iθ} = \cosq + isinθ
 r = √{( − √3 )^{2} + ( − 1)^{2}} = 2
 r = √{( − 2)^{2} + (2√3 )^{2}} = 4
 θ = arctan([( − 1)/( − √3 )]) + π = [(π)/6] + π = [(7π)/6]
 θ = arctan([(2√3 )/( − 2)]) + π = − [(π)/3] + π = [(2π)/3]
 Polar Form: (2e^{[(7π)/6]i})(4e^{[(2π)/3]i}) = 8e^{[(7π)/6]i + [(2π)/3]i} = 8e^{[(11π)/6]i}
 8e^{[11p/6]i} = 8(cos[(11π)/6] + isin[(11π)/6]) = 8([(√3 )/2] + i( − [1/2]))
 r = √{x^{2} + y^{2}}
 θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π
 Polar Form: re^{iθ}
 e^{iθ} = cosθ+ isinθ
 r = √{(2)^{2} + (2)^{2}} = 2√2, r = √{(3)^{2} + ( − 3)^{2}} = 3√2
 θ = arctan([2/2]) = [p/4], θ = arctan([( − 3)/3]) + π = − [(π)/4] + 2p = [(7π)/4]
 Polar Form: (2√2 e^{[(π)/4]i})(3√2 e^{[(7π)/4]i}) = 12e^{[(π)/4]i + [(7π)/4]i} = 12e^{2πi}
 12e^{2πi} = 12(cos2π+ isin2π) = 12(1 + i(0))
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Polar Form of Complex Numbers
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Polar Coordinates
 Polar Form Conversion
 Multiplying Two Complex Numbers in Polar Form
 Example 1: Convert Rectangular to Polar Form
 Example 2: Convert Polar to Rectangular Form
 Example 3: Multiply Two Complex Numbers
 Extra Example 1: Convert Between Rectangular and Polar Forms
 Extra Example 2: Simplify Expression to Polar Form
 Intro 0:00
 Polar Coordinates 0:49
 Rectangular Form
 Polar Form
 R and Theta
 Polar Form Conversion 2:27
 R and Theta
 Optimal Values
 Euler's Formula
 Multiplying Two Complex Numbers in Polar Form 6:10
 Multiply r's Together and Add Exponents
 Example 1: Convert Rectangular to Polar Form 7:17
 Example 2: Convert Polar to Rectangular Form 13:49
 Example 3: Multiply Two Complex Numbers 17:28
 Extra Example 1: Convert Between Rectangular and Polar Forms
 Extra Example 2: Simplify Expression to Polar Form
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
Transcription: 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 x^{2} of x^{2} + y^{2}.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 (x^{2} + y^{2}).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(x^{2} + y^{2}) 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 2^{7}, 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 (2^{7}) 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 xcoordinate and a ycoordinate.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
re^{i}θ, 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 xaxis.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 x^{2}+y^{2}.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, r_{1}×e_{1}^{iθ}, it's got an r and a θ, and r_{2}×e_{2}^{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 r_{1}×r_{2}.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, iθ_{1}+iθ_{1}, just gives you i(θ_{1})+θ_{2}).0409
You add the exponents there.0426
You end up just multiplying the r's and adding the angles θ_{1}+θ_{2} 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 x^{2}+y^{2}.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 x^{2}, negative root 3 squared is just 3, +y^{2} 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 xcoordinate 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 r_{2}e^{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 x^{2}, 6 root 2 squared is 36, times 2 is 72, +y^{2} 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 45degree angle on the lefthand 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 lefthand 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, ycoordinate 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 x^{2}+y^{2}.1071
Let me write these formulas generically, x^{2}+y^{2}, 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 1^{2} 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 xcoordinate 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 30degree angle south of the xaxis.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 12i, 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, i^{2}, but i^{2}=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, +2i6i, 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 x^{2}+y^{2}.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
1 answer
Last reply by: Dr. William Murray
Sun May 3, 2015 7:29 PM
Post by David Wu on April 28, 2015
Hiï¼ŒProfessor
At 10:48 you said pi/4 is a positive term so we don't have to add pi to it, but it seems to me that pi/4 is a negative term, wold you explain why?
Thank you
1 answer
Last reply by: Dr. William Murray
Sun Apr 14, 2013 6:40 PM
Post by enya zh on April 12, 2013
I don't understand exactly what 'e' stands for in the formula re^theta*i. Is it the natural number? Please help me understand.
Thanks!
1 answer
Last reply by: Dr. William Murray
Sun Apr 14, 2013 6:40 PM
Post by Dave Seale on April 6, 2013
To further help me visualize the concepts of complex numbers in rectangular and polar form can you provide a few word problems and a couple real world applications where we can utilize these conversions please. I found it extremely helpful to grasp the concepts we are practicing from earlier lectures once I went through the real world applications for trigonometry lecture. Also, these conversions look like something that will show up in a tricky word problem in class, if you could walk us through 1 or 2 that would be fantastic. The word problems give me a sense of when I will need to go straight to conversions in the real life instances as well, I think they kind of go hand in hand and really solidify a strong understanding of these neat formulas and conversions. Thanks for all the help Dr. Murray, you do a great job of explaining advanced math concepts and I feel very prepared to take on AP trig next semester!
1 answer
Last reply by: Dr. William Murray
Sun Dec 9, 2012 7:36 AM
Post by valtteri viinikainen on December 9, 2012
4. (a) Given v = âˆ’3i and w = 1+ i express the product
v^3 times w^2 in polar form and
exponential form and show it is real.
(b) Find the solutions of z^3 = 1âˆ’ i
and sketch the roots in the complex plane.
1 answer
Last reply by: Dr. William Murray
Sun Dec 9, 2012 7:41 AM
Post by varsha sharma on June 9, 2011
For extra example 2 , we could have used pascals triangle for binomial expansion. that was another way to do.
But converting to polar form was fun and i learnt it.
Thanks.
Varsha