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DeMoivre's Theorem
Main formulas:
- If the complex number z is written in polar form z = reiθ
, then we can find n-th powers as follows:
zn = (reiθ )n = [r(cosθ + i sinθ )]n = rn(cosn θ + i sinn θ ) = rn ei nθ - Every nonzero complex number has exactly n n-th roots.
- We can find n-th roots as follows:
where k = 0,1,2,..., n− 1.n√z= z[1/(n)] = (reiθ )[1/(n)] = [r(cosθ + i sinθ )][1/(n)] = r[1/(n)] ( cos θ + 2kπ n+ i sin θ + 2kπ n) = r[1/(n)] ei [(θ + 2kπ)/n]
Example 1:
Convert the complex number z = − √ 3 + i into polar form and then use DeMoivre's Theorem to calculate z7.Example 2:
Find all complex eighth roots of 16.Example 3:
Find all complex cube roots of − 1.Example 4:
Convert the complex number z = 2√ 2 − 2√ 2i into polar form and then use DeMoivre's Theorem to calculate z5.Example 5:
Find all complex fourth roots of z = − 2 − 2√ 3i.Give me another blank page here.
Convert the complex number w = − 1 + √3 i into polar form and then use DeMoivre's Theorem to calculate w5
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w5 = (2e[(2π)/3]i)5 = 25e[(10π)/3]i = 32e[(4π)/3]i
- 32(cos[(4π)/3] + isin[(4π)/3]) = 32[ ( − [1/2]) + i([( − √3 )/2]) ]
− 16 − 16√{3 i}
Convert the complex number w = − 4√3 + 4i into polar form and then use DeMoivre's Theorem to calculate w6
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w6 = (8e[(5π)/6]i)6 = 86e[(30π)/6]i = 262144e5πi = 262144eπi
- 262144(cosπ+ isinπ) = 262144[ ( − 1) + i(0) ]
− 262144
Convert the complex number w = − 2 + 2√{3 i} into polar form and then use DeMoivre's Theorem to calculate w7
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w7 = (4e[(2π)/3]i)7 = 47e[(14π)/3]i = 16384e[(2π)/3]i
- 16384(cos[(2π)/3] + isin[(2π)/3]) = 16384[ ( − [1/2]) + i([(√3 )/2]) ]
− 8192 + 8192√{3i}
Convert the complex number w = − 3√2 − 3√{2i} into polar form and then use DeMoivre's Theorem to calculate w4
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w4 = (6e[(5π)/4]i)4 = 64e5πi = 1296e5πi = 1296eπi
- 1296(cosπ+ isinπ) = 1296[ ( − 1) + i(0) ]
− 1296
Convert the complex number w = 4 + 4i into polar form and then use DeMoivre's Theorem to calculate w3
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w3 = (4√2 e[(π)/4]i)4 = (4√2 )3e[(3π)/4]i = 128√2 e[(3π)/4]i
- 128√2 (cos[3p/4] + isin[3p/4]) = 16384[ ( − [(√2 )/2]) + i([(√2 )/2]) ]
− 128 + 128i
Convert the complex number w = 2 − 2√{3 i} into polar form and then use DeMoivre's Theorem to calculate w4
- DeMoivre's Theorem: zn = (reiθ)n = [ r(cosθ+ isinθ) ]n = rn(cosnθ+ isinnθ) = rneinθ
- r = √{x2 + y2}
- 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]
- w4 = (4e[(5π)/3]i)4 = 44e[(20π)/3]i = 256e[(2π)/3]i
- 256(cos[(2π)/3] + isin[(2π)/3]) = 256[ ( − [1/2]) + i([(√3 )/2]) ]
− 128 + 128√{3i}
Find all sixth roots of 1
- There should be six different answers
- DeMoivre's Theorem: z[1/n] = (reiθ)[1/n] = [ r(cosθ+ isinθ) ][1/n] = r[1/n](cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r[1/n]ei[(θ+ 2kπ)/n]
- r = √{x2 + y2}
- 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 = 1ei0 ⇒ 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 |
Find all fourth roots of - 4
- There should be four different answers
- DeMoivre's Theorem: z[1/n] = (reiθ)[1/n] = [ r(cosθ+ isinθ) ][1/n] = r[1/n](cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r[1/n]ei[(θ+ 2kπ)/n]
- r = √{x2 + y2}
- 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 = 4eiπ ⇒ 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 |
Find all cube roots of 8
- There should be three different answers
- DeMoivre's Theorem: z[1/n] = (reiθ)[1/n] = [ r(cosθ+ isinθ) ][1/n] = r[1/n](cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r[1/n]ei[(θ+ 2kπ)/n]
- r = √{x2 + y2}
- 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 = 8ei0 ⇒ 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 |
Find all cube roots of -125
- There should be three different answers
- DeMoivre's Theorem: z[1/n] = (reiθ)[1/n] = [ r(cosθ+ isinθ) ][1/n] = r[1/n](cos[(θ+ 2kπ)/n] + isin[(θ+ 2kπ)/n]) = r[1/n]ei[(θ+ 2kπ)/n]
- r = √{x2 + y2}
- 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 = 125eiπ ⇒ 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































0 answers
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