For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

### DeMoivre's Theorem

**Main formulas**:

- If the complex number
*z*is written in polar form*z*=*re*^{iθ}, then we can find*n*-th powers as follows:*z*^{n}= (*re*^{iθ})^{n}= [*r*(cosθ +*i*sinθ )]^{n}=*r*^{n}(cos*n*θ +*i*sin*n*θ ) =*r*^{n}*e*^{i nθ} - Every nonzero complex number has exactly
*n**n*-th roots. - We can find
*n*-th roots as follows:

where*n*√*z*= *z*^{[1/(n)]}= (*re*^{iθ})^{[1/(n)]}= [ *r*(cosθ +*i*sinθ )]^{[1/(n)]}= *r*^{[1/(n)]}( cos θ *+**2**k*π*n**+**i*sinθ *+**2**k*π*n*) = *r*^{[1/(n)]}*e*^{i [(θ + 2kπ)/n]}*k*= 0,1,2,...,*n*− 1.

**Example 1**:

*z*= −√ 3 +

*i*into polar form and then use DeMoivre's Theorem to calculate

*z*

^{7}.

**Example 2**:

**Example 3**:

**Example 4**:

*z*= 2√ 2 − 2√ 2

*i*into polar form and then use DeMoivre's Theorem to calculate

*z*

^{5}.

**Example 5**:

*z*= − 2 − 2√ 3

*i*.

Give me another blank page here.

### DeMoivre's Theorem

^{5}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{( − 1)
^{2}+ (√3 )^{2}} = 2 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([(√3 )/( − 1)]) + π = − [(π)/3] + π = [(2π)/3]
- w
^{5}= (2e^{[(2π)/3]i})^{5}= 2^{5}e^{[(10π)/3]i}= 32e^{[(4π)/3]i} - 32(cos[(4π)/3] + isin[(4π)/3]) = 32[ ( − [1/2]) + i([( − √3 )/2]) ]

^{6}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{( − 4√3 )
^{2}+ (4)^{2}} = 8 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([4/( − 4√3 )]) + π = − [(π)/6] + π = [(5π)/6]
- w
^{6}= (8e^{[(5π)/6]i})^{6}= 8^{6}e^{[(30π)/6]i}= 262144e^{5πi}= 262144e^{πi} - 262144(cosπ+ isinπ) = 262144[ ( − 1) + i(0) ]

^{7}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{( − 2)
^{2}+ (2√3 )^{2}} = 4 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([(2√3 )/( − 2)]) + π = − [(π)/3] + π = [(2π)/3]
- w
^{7}= (4e^{[(2π)/3]i})^{7}= 4^{7}e^{[(14π)/3]i}= 16384e^{[(2π)/3]i} - 16384(cos[(2π)/3] + isin[(2π)/3]) = 16384[ ( − [1/2]) + i([(√3 )/2]) ]

^{4}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{( − 3√2 )
^{2}+ ( − 3√2 )^{2}} = 6 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([( − 3√2 )/( − 3√2 )]) + π = [(π)/4] + π = [(5π)/4]
- w
^{4}= (6e^{[(5π)/4]i})^{4}= 6^{4}e^{5πi}= 1296e^{5πi}= 1296e^{πi} - 1296(cosπ+ isinπ) = 1296[ ( − 1) + i(0) ]

^{3}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{(4)
^{2}+ (4)^{2}} = 4√2 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([4/4]) = [(π)/4]
- w
^{3}= (4√2 e^{[(π)/4]i})^{4}= (4√2 )^{3}e^{[(3π)/4]i}= 128√2 e^{[(3π)/4]i} - 128√2 (cos[3p/4] + isin[3p/4]) = 16384[ ( − [(√2 )/2]) + i([(√2 )/2]) ]

^{4}

- DeMoivre's Theorem: z
^{n}= (re^{iθ})^{n}= [ r(cosθ+ isinθ) ]^{n}= r^{n}(cosnθ+ isinnθ) = r^{n}e^{inθ} - r = √{x
^{2}+ y^{2}} - r = √{(2)
^{2}+ ( − 2√3 )^{2}} = 4 - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- θ = arctan([( − 2√3 )/2]) = − [(π)/3] + 2π = [(5π)/3]
- w
^{4}= (4e^{[(5π)/3]i})^{4}= 4^{4}e^{[(20π)/3]i}= 256e^{[(2π)/3]i} - 256(cos[(2π)/3] + isin[(2π)/3]) = 256[ ( − [1/2]) + i([(√3 )/2]) ]

- There should be six different answers
- DeMoivre's Theorem: z
^{[1/n]}= (re^{iθ})^{[1/n]}= [ r(cosθ+ isinθ) ]^{[1/n]}= r^{[1/n]}(cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r^{[1/n]}e^{i[(θ+ 2kπ)/n]} - r = √{x
^{2}+ y^{2}} - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- 1 + 0i
- r = √{(1)
^{2}+ (0)^{2}} = 1 - θ = arctan([0/1]) = 0
- z = 1e
^{i0}⇒ z^{[1/6]}= 1^{[1/6]}(cos[(0 + 2kπ)/6] + isin[(0 + 2kπ)/6])

k | [(θ+ 2kπ)/n] = [(0 + 2kπ)/6] = [(kπ)/3] | cosα+i sinα | Six Answers |

0 | 0 | 1+0i | 1 |

1 | [(π)/3] | [1/2] + [(√3 )/2]i | [1/2] + [(√3 )/2]i |

2 | [(2π)/3] | − [1/2] + [(√3 )/2]i | − [1/2] + [(√3 )/2]i |

3 | π | − 1 + 0i | −1 |

4 | [(4π)/3] | − [1/2] − [(√3 )/2]i | − [1/2] − [(√3 )/2]i |

- There should be four different answers
- DeMoivre's Theorem: z
^{[1/n]}= (re^{iθ})^{[1/n]}= [ r(cosθ+ isinθ) ]^{[1/n]}= r^{[1/n]}(cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r^{[1/n]}e^{i[(θ+ 2kπ)/n]} - r = √{x
^{2}+ y^{2}} - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- − 4 + 0i
- r = √{( − 4)
^{2}+ (0)^{2}} = 4 - θ = arctan([0/( − 4)]) + π = π
- z = 4e
^{iπ}⇒ z^{[1/4]}= 1^{[1/4]}(cos[(π+ 2kπ)/4] + isin[(π+ 2kπ)/4])

k | [(θ+ 2kπ)/n] = [(π+ 2kπ)/4] | r^{[1/n]}(cosα+ isinα) | Four Answers |

0 | [(π)/4] | √2 ([(√2 )/2] + i[(√2 )/2]) | 1+i |

1 | [(3π)/4] | √2 ( − [(√2 )/2] + i[(√2 )/2]) | −1+i |

2 | [(5π)/4] | √2 ( − [(√2 )/2] + i( − [(√2 )/2]) ) | −1−i |

3 | [(7π)/4] | √2 ( [(√2 )/2] + i( − [(√2 )/2]) ) | 1−i |

- There should be three different answers
- DeMoivre's Theorem: z
^{[1/n]}= (re^{iθ})^{[1/n]}= [ r(cosθ+ isinθ) ]^{[1/n]}= r^{[1/n]}(cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r^{[1/n]}e^{i[(θ+ 2kπ)/n]} - r = √{x
^{2}+ y^{2}} - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- 8 + 0i
- r = √{(8)
^{2}+ (0)^{2}} = 1 - θ = arctan([0/8]) = 0
- z = 8e
^{i0}⇒ z^{[1/3]}= 8^{[1/3]}(cos[(0 + 2kπ)/3] + isin[(0 + 2kπ)/3])

k | [(θ+ 2kπ)/n] = [(0 + 2kπ)/3] | 2(cosα+ isinα) | Three Answers |

0 | 0 | 2(1 + 0i) | 2 |

1 | [(2π)/3] | 2( − [1/2] + i[(√3 )/2] ) | − 1 + √3 i |

2 | [(4π)/3] | 2( − [1/2] − i[(√3 )/2] ) | − 1 − √3 i |

- There should be three different answers
- DeMoivre's Theorem: z
^{[1/n]}= (re^{iθ})^{[1/n]}= [ r(cosθ+ isinθ) ]^{[1/n]}= r^{[1/n]}(cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r^{[1/n]}e^{i[(θ+ 2kπ)/n]} - r = √{x
^{2}+ y^{2}} - θ = arctan[y/x] if x is positive or θ = arctan[y/x] + π if x is negative
- − 125 + 0i
- r = √{( − 125)
^{2}+ (0)^{2}} = 125 - θ = arctan([0/( − 125)]) + π = π
- z = 125e
^{iπ}⇒ z^{[1/3]}= 125^{[1/3]}(cos[(π+ 2kπ)/3] + isin[(π+ 2kπ)/3])

k | [(θ+ 2kπ)/n] = [(π+ 2kπ)/3] | 5(cosα+ isinα) | Three Answers |

0 | [(π)/3] | 5( [1/2] + i[(√3 )/2] ) | [5/2] + [(5√3 )/2]i |

1 | π | 5( − 1 + i(0) ) | −5 |

2 | [(5π)/3] | 5( [1/2] + i( − [(√3 )/2]) ) | [5/2] − [(5√3 )/2]i |

*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

### DeMoivre's Theorem

Lecture Slides are screen-captured 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
- Introduction to DeMoivre's Theorem
- DeMoivre's Theorem: Finding nth Roots
- Example 1: Convert to Polar Form and Use DeMoivre's Theorem
- Example 2: Find Complex Eighth Roots
- Example 3: Find Complex Roots
- Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
- Extra Example 2: Find Complex Fourth Roots

- Intro 0:00
- Introduction to DeMoivre's Theorem 0:10
- n nth Roots
- DeMoivre's Theorem: Finding nth Roots 3:52
- Relation to Unit Circle
- One nth Root for Each Value of k
- Example 1: Convert to Polar Form and Use DeMoivre's Theorem 8:24
- Example 2: Find Complex Eighth Roots 15:27
- Example 3: Find Complex Roots 27:49
- Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
- Extra Example 2: Find Complex Fourth Roots

### 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 | |

Half-Angle 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: 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 z ^{5}.*0004

*Let us first convert that into polar form, remember my equations x ^{2} + y^{2}, 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= x ^{2} + y^{2}.*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 z ^{5} .*0151

*We can do that by z ^{n}=r^{n} x cos(n)( theta) + (i)sin(n)(theta).*0159

*That is how DeMoivre’s theorem applies to this one z ^{5} = r^{2}, that is 4^{5} 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 4 ^{5} 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 z ^{n} is r^{n}, 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, x ^{2} + y^{2} 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 -2 ^{2} is 4, 2 (root 3) ^{2} is 2^{2} 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 2^{2} 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}, y^{4} 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 z ^{n}, 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

*r ^{n}, that just gives you r^{n}.*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 r ^{n}×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 z ^{n}=r^{n}×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, z ^{n}=r^{n}×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

*z ^{n}=r^{n}×e^{niθ}, which was r^{n}×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/n}cos(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 z ^{7}.*0505

*Remember to convert it into polar form.*0518

*You do r equals the square root of x ^{2}+y^{2}, 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 x ^{2}+y^{2}, x^{2}=3, y^{2}=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

*z ^{7}, the whole point is we're going to use the polar form to find z^{7}, so this is 2e^{(5π/6)i7}.*0655

*That's 2 ^{7}.*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, 2 ^{7} 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 2^{7}, 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 x ^{2}+y^{2}.*0981

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

*My r is the square root of 16 ^{2}+0^{2}, that's just 16.*1001

*θ, my y=0, arctan(0)=0.*1007

*My z=16e ^{0i}.*1016

*I could have worked that out, certainly 16e ^{0}=16 just by itself because e^{0}=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=2^{4}, 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 w ^{8}=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=16e ^{0i}.*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 x ^{2}+y^{2}, 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 x ^{2}+y^{2}, 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 wr ^{3}=-1.*2106

*If you worked out wr ^{3} 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

1 answer

Last reply by: Dr. William Murray

Thu Oct 31, 2013 3:39 PM

Post by Heinz Krug on October 31, 2013

Thank you, Will. Great lecture! With age 57, I very much enjoyed learning trigonometry. Still too lazy to memorize all the common values of sin, cos, tan and using the calculator instead. Do you have a final recommendation for that?

1 answer

Last reply by: Dr. William Murray

Wed Aug 14, 2013 12:20 PM

Post by Ikze Cho on August 8, 2013

does your calculator have to be set to radians when doing these things?

1 answer

Last reply by: Dr. William Murray

Wed Aug 14, 2013 12:16 PM

Post by Tomas Hernandez on August 7, 2013

Example 1: Since arctan of minus squared root 3/3 can be 5pi/6 or 11pi/6, why did you choose -pi/6 (11pi/6) instead of 5pi/6?

1 answer

Last reply by: Dr. William Murray

Fri Jul 5, 2013 9:19 AM

Post by Emily Engle on July 1, 2013

On my calculator when I enter 16^(1/8)it gives me sqrt of 2. How do I get my calculator to give me the answer you got in ex.2 ?

1 answer

Last reply by: Dr. William Murray

Fri Jul 5, 2013 9:19 AM

Post by Emily Engle on June 27, 2013

In De Moivre's theorem why do we multiply theta by n and not cosine of theta by n?

1 answer

Last reply by: Dr. William Murray

Thu May 30, 2013 3:55 PM

Post by varsha sharma on June 9, 2011

I want to master your notes and use them in my classroom and i want to work as an instructor in school of math and science.

Thanks,

Varsha