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Mathematics Dictionary
Dr. K. G. Shih
DeMovire's Theorem


  • Q01 | - D002 - DeMovire's Theorem : Rule 2
  • Q02 | - Solve x^2 - 1 = 0
  • Q03 | - Solve x^3 - 1 = 0
  • Q04 | - Solve x^5 - 1 = 0
  • Q05 | - Solve x^4 + x^3 + x^2 + x + 1 = 0
  • Q06 | - References

    • Q01. DeMovire's Theorem

    Rule 2 Definition
    • Expression : x + i*y = r*(cos(A) + i*sin(A))
      • r = Sqr(x^2 + y^2)
      • A = arctan(y/x)
    • z = x + i*y is rectangular form
    • z = r*(cos(A) + i*sin(A)) is polor form
    • cis(A) = cos(A) + i*sin(A)
    • Rule 2
      • (cos(A) + i*sin(A))^(1/n) = cos((2*k*pi + A)/n) + i*sin((2*k*pi + A)/n)
      • Where k = 0, 1, 2, .... (n - 1)
    Example : Express z = 1 + i in polar form
    • r = Sqr(1^2 + 1^2) = Sqr(2)
    • A = arctan(1/1) = arctan(1) = 45 degrees
    • Hence Z = Sqr(2)*(cos(45) + i*sin(45) = Sqr(2)*cis(45)

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    Q02. Application : Solve x^2 - 1 = 0

    Method 1 : Factor method
    • x^2 - 1 = 0
    • (x - 1)*(x + 1) = 0
    • Hence x - 1 = 0 or x = 1
    • Hence x + 1 = 0 or x = -1
    Method 2 : Use DeMovire's theory 2
    • x^2 = 1
    • x^2 = 1*(cos(0) + i*sin(0))
    • x
      • = (cos(0) + i*sin(0))^(1/2)
      • = cos((2*k*pi + 0)/2) + i*sin((2*k*pi + 0)/2)
    • k = 0
      • x = (cos(0) + i*sin(0)) = 1
    • k = 1
      • x = cos((2*1*pi+0)/2) + i*sin((2*1*pi+0)/2)
      • x = cos(180) + i*sin(180) = -1

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    Q03. Application : Solve x^3 - 1 = 0

    Method 1 : Factor method
    • x^3 - 1 = 0
    • (x - 1)*(x^2 + x + 1) = 0
    • Hence x - 1 = 0 or x0 = 1
    • Hence x^2 + x + 1 = 0
      • x1 = (-1 + Sqr(1^2 - 4*1*1))/2 = (-1 + i*Sqr(3))/2
      • x2 = (-1 - Sqr(1^2 - 4*1*1))/2 = (-1 - i*Sqr(3))/2
    Method 2 : Use DeMovire's theory 2
    • x^3 = 1
    • x^3 = 1*(cos(0) + i*sin(0))
    • Hence
      • x = (cos(0) + i*sin(0))^(1/3)
      • x = cos((3*k*pi + 0)/3) + i*sin((3*k*pi + 0)/3)
    • k = 0
      • x0 = (cos(0) + i*sin(0)) = 1
    • k = 1
      • x1 = cos((2*1*pi+0)/3) + i*sin((2*1*pi+0)/3)
      • x1 = cos(120) + i*sin(120) = -1/2 + i*Sqr(3)/2
    • k = 2
      • x2 = cos((2*2*pi+0)/3) + i*sin((2*2*pi+0)/3)
      • x2 = cos(240) + i*sin(240) = -1/2 - i*Sqr(3)/2
    Notes
    • Conjugate complex
      • Sum of two complex roots is real
      • Product of two complex roots is also real
    • Three angles for x^3 - 1 = 0
      • Angle A0 = A = 0
      • Angle A1 = (2*pi + A)/3 = 120
      • Angle A2 = (4*pi + A)/3 = 240

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    Q04. Application : Solve x^5 - 1 = 0

    Method 1 : Factor method
    • x^5 - 1 = 0
    • (x - 1)*(x^4 + x^3 + x^2 + x + 1) = 0
    • Hence x - 1 = 0 or x0 = 1
    • Hence x^4 + x^3 + x^2 + x + 1 = 0
      • Formula is not easy in use
    Method 2 : Use DeMovire's theory 2
    • x^5 = 1
    • x^5 = 1*(cos(0) + i*sin(0))
    • Hence
      • x = (cos(0) + i*sin(0))^(1/5)
      • x = cos((5*k*pi + 0)/5) + i*sin((5*k*pi + 0)/5)
    • k = 0
      • x0 = (cos(0) + i*sin(0)) = 1
    • k = 1
      • x1 = cos((2*1*pi+0)/5) + i*sin((2*1*pi+0)/5)
      • x1 = cos(72) + i*sin(72) = ?
    • k = 2
      • x2 = cos((2*2*pi+0)/5) + i*sin((2*2*pi+0)/5)
      • x2 = cos(144) + i*sin(144) = ?
    • k = 3
      • x3 = cos((2*3*pi+0)/5) + i*sin((2*3*pi+0)/5)
      • x3 = cos(216) + i*sin(216) = ?
    • k = 4
      • x4 = cos((2*4*pi+0)/5) + i*sin((2*4*pi+0)/5)
      • x4 = cos(288) + i*sin(288) = ?
    Notes
    • Conjugate complex
      • Sum of two complex roots is real
      • Product of two complex roots is also real
    • Five angles angle for x^5 - 1 = 0
      • Angle A0 = A = 0
      • Angle A1 = (2*pi + A)/5 = 072
      • Angle A2 = (4*pi + A)/5 = 144
      • Angle A3 = (2*pi + A)/5 = 216
      • Angle A4 = (4*pi + A)/5 = 288

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    Q05. Application : Solve x^4 + x^3 + x^2 + x + 1 = 0

    DeMovire's theory method
    • Use solution of x^5 - 1 = 0 in Q04
    • Hense x1, x2, x3, x4 are the solutions
    • The angles are 72, 144, 216 and 288
    Construction method
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    Q6. References

    References :

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