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Mathematics Dictionary
Dr. K. G. Shih
Demoivre's Theory
Subjects

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  • Q01 | - Imaginary number and complex numbers
  • Q02 | - Ptove that cis(A)*cis(B) = cis(A+B)
  • Q03 | - Prove that cis(A)*cis(B)*cis(C) = cis(A+B+C)
  • Q04 | - Prove that DeMoivre Theory cis(A)^n = cis(n*A)
  • Q05 | - Solve x^3 - 1 = 0 by using DeMoivre's theory
  • Q06 | - Solve x^5 - 1 = 0 by using DeMoivre's theory
  • Q07 | - Solve x^3 + 1 = 0 by using DeMoivre's theory
  • Q08 | - Solve x^4 + i = 0 by using DeMoivre's theory
  • Q09 | - Solve x^4 - x^3 + x^2 - x + 1 = 0. using x^5 + 1 = 0
  • Q10 | - Solve x^4 + x^3 + x^2 + x + 1 = 0. using x^5 - 1 = 0
  • Q11 | - Reference
  • Q12 | - Summary and formula
  • Q13 | - Prove that (cos(A)+i*sin(A))/(cos(B)+i*sin(B))=cos(A-B)+i*sin(A-B).
  • Q14 | - Quiz for complex numbersNew
  • Q15 | - Answer to quiz
Answers


Q01. Imaginary number and complex numbers

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Q02. Identities

Prove that (cos(A)+i*sin(A))*(cos(B)+i*sin(B)) = cos(A+B) + i*sin(A+B)
  • cis(A)*cis(B)
  • = (cos(A) + i*sin(A))*(cos(B) + i*sin(B))
  • = cos(A)*cos(B) + i*cos(A)*sin(B) + i*sin(A)*cos(B) + (i^2)*sin(A)*sin(B).
  • = cos(A)*cos(B) - sin(A)*sin(B) + i*(cos(A)*sin(B) + sin(A)*cos(B)).
  • = cos(A+B) + i*sin(A+B)
  • = cis(A+B).
Prive that (cos(A)+i*sin(A))/(cos(B)+i*sin(B)) = cos(A-B) + i*sin(A-B)
  • cis(A)/cis(B)
  • = (cos(A) + i*sin(A))/(cos(B) + i*sin(B))
  • = (cos(A) + i*sin(A))*(cos(A) - i*sin(B))/(cos(B)^2 - (i*sin(B)^2)
  • = (cos(A) + i*sin(A))*(cos(B) - i*sin(B))
  • = cos(A)*cos(B) - i*cos(A)*sin(B) + i*sin(A)*cos(B) - (i^2)*sin(A)*sin(B).
  • = cos(A)*cos(B) + sin(A)*sin(B) + i*(cos(A)*sin(B) + sin(A)*cos(B)).
  • = cos(A-B) + i*sin(A-B)
  • = cis(A-B).
Other form :
  • If cis(A) = cos(A) + i*sin(A) then cis(A)*cis(B) = cis(A+B).
  • If cis(A) = cos(A) + i*sin(A) then cis(A)/cis(B) = cis(A-B).
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Q03. Prove that cis(A)*cis(B)*cis(C) = cis(A+B+C)

  • In Q02 we have cis(A)*cis(B) = cis(A+B).
  • Hence we can have cis(A+B)*cis(C) = cis(A+B+C).
  • If A = B = C then we have cis(A)^3 = cis(3*A).
  • Similarly cis(A)^4 = cis(4*A).
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Q04. DeMoivre's theorem

  • 1. cis(A)^n = cis(n*A)
  • 2. cis(A)^(1/n) = cis((2*k*pi+A)/n)
Example : (cos(A)+i*sin(A))^2 = cos(2*A) + i*sin(2*A)
  • Let A = B in Q01, we have cis(A)^2 = cis(2*A).
Example : (cos(A)+i*sin(A))^3 = cos(3*A) + i*sin(3*A)
  • (cos(A)+i*sin(A))^3 = (cos(A)+i*sin(A))*(cos(A)+i*sin(A))^2
  • = (cos(A)+i*sin(A))*(cos(2*A)+i*sin(2*A)
  • = cos(3*A) + i*sin(3*A)
  • Hence cis(A)^3 = cis(3*A)
Example : (cos(A)+i*sin(A))^n = cos(n*A) + i*sin(n*A)
  • From above examples we can get this conclusion cis(A)^n = cis(n*A).
  • This is call DeMoivre's theorem.
Example : if z^n = cis(A), then z = cis(A)^(1/n) = cis((2*k*pi+A)/n)
  • k = 0, z1 = cis(A/n).
  • k = 1, z2 = cis((2*pi+A)/n).
  • k = 2, z3 = cis((4*pi+A)/n).
  • k = 3, z2 = cis((6*pi+A)/n).
  • k = n, z2 = cis((2*n*pi+A)/n) = cis(A/n).
  • Hence k = 1, 2, 3, ...... (n-1).
Example : Solve x^5 - 1 = 0
  • Since x^5 = 1 = cos(0) + i*sin(0).
  • The solutions
    • k = 0, x0 = cis((0*360+0)/5) = cos(000) + i*sin(000).
    • k = 1, x1 = cis((1*360+0)/5) = cos(072) + i*sin(072).
    • k = 2, x2 = cis((2*360+0)/5) = cos(144) + i*sin(144).
    • k = 3, x3 = cis((3*360+0)/5) = cos(216) + i*sin(216).
    • k = 4, x4 = cis((4*360+0)/5) = cos(288) + i*sin(288).
  • Five angles of (2*k*pi+A)/5 : 0, 72, 2*72, 3*72, 4*72.
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Q05. Solve x^3 - 1 = 0

  • x^3 = 1 = cos(0) + i*sin(0)
  • The angles
    • A0 = (0*360+0)/3 = 0.
    • A1 = (1*360+0)/3 = 120.
    • A2 = (2*360+0)/3 = 240.
  • The solutions
    • x0 = cos(000) + i*sin(000) = 1.
    • x1 = cos(120) + i*sin(120) = -1/2 + i*Sqr(3)/2.
    • x2 = cos(240) + i*sin(240) = -1/2 - i*Sqr(3)/2.
    • Note : x1 and x2 are conjugate.
Example : Solve x^2 + x + 1 = 0
Method 1 : Using DeMoivre theory
  • Since x^2 + x + 1 is a factor of (x^3 - 1).
  • Hence we can use the solutions of x^3 - 1 = 0.
  • x^3 = 1 = cos(0) + i*sin(0).
  • The angles are 0, 120, 240.
  • The answer of x^3 -1 = 0 are.
    • x0 = cis((0*360+0)/3) = cis(0) = 1.
    • x1 = cis((1*360+0)/3) = cis(120) = -1/2 + i*Sqr(3)/2 .
    • x2 = cis((2*360+0)/3) = cis(240) = -1/2 - i*Sqr(3)/2 .
  • Hence x1 and x2 are the answer of x^2 + x + 1 = 0.
  • Note : x1 and x2 are conjugate.
Method 2 : Using quadratic formula
  • 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 3 : Construction method from results of DeMoivre's theory
  • Draw a circle with 1 unit.
  • Draw OX where O is center and X is x-axis.
  • The angles are 0, 120, 240.
  • Draw angle POX = 120 where P(x,y) on circle.
  • Hence solution is x1 = x + i*y.
  • Measure x and y we get first solution.
  • Find 2nd solution.
    • Same as above, let angle POX = 240 and measure x and y.
    • Or use the conjugate property ti find second solution.
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Q06. Solve x^5 - 1 = 0.

  • x^5 = 1 = cos(0) + i*sin(0)
  • The angles : 0, 72, 2*72, 3*72, 4*72 (360/5 = 72)
    • A0 = (0*360+0)/5 = 0.
    • A1 = (1*360+0)/5 = 1*72 = 072.
    • A2 = (2*360+0)/5 = 2*72 = 144.
    • A3 = (2*360+0)/5 = 3*72 = 216.
    • A4 = (2*360+0)/5 = 4*72 = 288.
  • The solutions
    • x0 = cis(000) = 1.
    • x1 = cis(072).
    • x2 = cis(144).
    • x3 = cis(216).
    • x4 = cis(288).
  • Note 1 : x1 and x4 are conjugate.
  • Note 2 : x2 and x3 are conjugate.
Example : Solve x^4 + x^3 + x^2 + x + 1 = 0
Method 1 : Using DeMoivre theory
  • Since x^4 + x^3 + x^2 + x + 1 is a factor of (x^5 - 1).
  • Hence we can use the solutions of x^5 - 1 = 0.
  • The answer are.
    • x1 = cis(072).
    • x2 = cis(144).
    • x3 = cis(216).
    • x4 = cis(288).
Method 2 : Construction method from results of DeMoivre's theory
  • Draw a circle with 1 unit.
  • Draw OX where O is center and X is x-axis.
  • The angles are 0, 72, 144, 216, 288.
  • Draw angle POX = 72 where P(x,y) on circle.
  • Hence solution is x1 = x + i*y.
  • Measure x and y we get first solution.
  • Similarly Find 2nd, 3rd, 4th solution.
Example : Using solutions of x^4 + x^3 + x^2 + x + 1 = 0 prove that
  • 1. cos(72) + cos(144) + cos(216) + cos(288) = -1.
  • 2. sin(72) + sin(144) + sin(216) + sin(288) = 0.
  • 3. cis(72)*cis(144)*cis(216)*cis(288) = 1.
Proof
  • Using sum of roots = -(coeff of X^3)/(coeff of x^4).
    • cis(72)+cis(144)+cis(216)+cis(288) = -1
    • cos(72)+cos(144)+cos(216)+cos(288) + i*(sin(72)+sin(144)+sin(216)+sin(288)) = -1
    • Hence cos(72)+cos(144)+cos(216)+cos(288) = -1.
    • Hnece sin(72)+sin(144)+sin(216)+sin(288) = 0.
  • Using product of roots = +(coeff of constant)/(coeff of x^4).
    • Hence cis(72)*cis(144)*cis(216)*cis(288) = 1.
  • Other method : cos(72)+cos(144)+cos(216)+cos(288) = -1.
    • cos(72)+cos(144)+cos(216)+cos(288)
    • = cos(72) + cos(180-36) + cos(180+36) + cos(360-72)
    • = 2*cos(72) - 2*cos(36)
    • = -1.
  • Other method : sin(72)+sin(144)+sin(216)+sin(288) = 0.
    • sin(72)+sin(144)+sin(216)+sin(288)
    • = sin(72) + sin(180-36) + sin(180+36) + sin(360-72)
    • = sin(72) + sin(36) - sin(36) - sin(72)
    • = 0.
  • Other method : cis(72)*cos(144)*cis(216)*cis(288) = 1.
    • (cos(72)+i*sin(72))*(cos(144)+i*sin(144))*(cos(216)+i*sin(216))*(cos(288) +i*sin(288)
    • = (cos(72)+i*sin(72))*(cos(72)-i*sin(72)*(-cos(36)+i*sin(36))*(-cos(36)-i*sin(36)
    • = (cos(72)^2 + sin(72)^2)*(cos(36)^2 + sin(36)^2)
    • = 1*1
    • = 1
    • where we use
      • cos(288)+i*sin(288) = cos(72) - i*sin(72)
      • cos(144)+i*sin(144) = -cos(36) + i*sin(36)
      • cos(216)+i*sin(216) = -cos(36) - i*sin(36)
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Q07. Solve x^3 + 1 = 0


Method 1 : Factor method with quadratic formula
  • Since X^3 + 1 = (x + 1)*(x^2 - x + 1) = 0
  • Hence X + 1 = 0 and x = -1.
  • For x^2 - x + 1 = 0 we can use quadratic formula.
    • Hence x = (-(-1) + Sqr((-1)^2 - 4*1*1))/2 = (1 + Sqr(3)*i)/2.
    • Hence x = (-(-1) - Sqr((-1)^2 - 4*1*1))/2 = (1 - Sqr(3)*i)/2.
    • Noe : these two roots are conjugate.
Method 2 : DeMoivre's Theorem by using angles
  • x^3 = - 1 = cos(180) + i*sin(180).
  • The angles in Demoivre's solutions are
    • (0*360+180)/3 = 60.
    • (1*360+180)/3 = 180.
    • (2*360+180)/3 = 300.
  • The solutions are
    • x0 = cis(60) = 1/2 - i*Sqr(3)/2.
    • x1 = cis(180) = -1.
    • x2 = cis(300) = 1/2 + i*Sqr(3)/2.
    • Note x0 and x2 are conjugate.
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Q08. Solve x^4 + i = 0

  • x^4 = -i = cos(270) + i*sin(270) = cis(270).
  • x = cis(270)^4 = cis((2*k*pi+270)/4).
  • The solutions are
    • x0 = cis(270/4) = cis(067.5).
    • x1 = cis(2*180+270/4) = cis(090+067.5).
    • x2 = cis(4*180+270/4) = cis(180+067.5).
    • x3 = cis(6*180+270/4) = cis(270+067.5).
Solve x^4 - i = 0

  • x^4 = i = cos(90) + i*sin(90) = cis(90).
  • x = cis(90)^4 = cis((2*k*pi+90)/4).
  • The solutions are
    • x0 = cis(0*180+90/4) = cis(22.5).
    • x1 = cis(2*180+90/4) = cis(090+22.5).
    • x2 = cis(4*180+90/4) = cis(180+22.5).
    • x3 = cis(6*180+90/4) = cis(270+22.5).
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Q09. Solve x^4 - x^3 + x^2 - x + 1 = 0. using x^5 + 1 = 0

  • Since x^4 - x^3 + x^2 - x + 1 is a factor of (x^5 + 1).
  • Hence we can use the solutions of x^5 + 1 = 0.
  • The answers for x^5 + 1 = 0 are
    • Since x^5 = -1 = cos(180) + i*sin(180)
    • x0 = cis((0*360+180)/5) = cis(0*72+36) = cis(36).
    • x1 = cis((1*360+180)/5) = cis(1*72+36) = cis(108).
    • x2 = cis((2*360+180)/5) = cis(2*72+36) = cis(180).
    • x3 = cis((3*360+180)/5) = cis(3*72+36) = cis(252).
    • x4 = cis((4*360+180)/5) = cis(4*72+36) = cis(324).
  • Hence x0,x2,x3,x4 are the required solution.
  • Where x0 and x4 are conjugate and also x1 and x3 are conjugate.
Appliction : Prove that cis(36)*cis(108)*cis(252)*cis(324) = 1.
  • Method 1 : Use product of roots = + (coeff of constant)/(coeff of x^4).
  • Method 2 : Change all angles to 1st quadrant angles.
Appliction : Prove that cos(36)+cos(108)+cos(252)+cos(324) = 1.
  • Method 1 : Use sum of roots = + (coeff of x^3)/(coeff of x^4).
  • Method 2 : Change all angles to 1st quadrant angles (see Q06).
Appliction : Prove that sin(36)+sin(108)+sin(252)+sin(324) = 1.
  • Method 1 : Use sum of roots = + (coeff of x^3)/(coeff of x^4).
  • Method 2 : Change all angles to 1st quadrant angles (see Q06).
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Q10. Solve x^4 + x^3 + x^2 + x + 1 = 0. Then using DeMoivre's theorem prove that

  • cos(72) + cos(144) + cos(216) + cos(288) = -1.
  • sin(72) + sin(144) + sin(216) + sin(288) = 0.
Solution (See Q06)
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Q11. Reference
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Q12. Summary and formula
  • Imaginary number i = Sqr(-1) or i^2 = -1.
  • Complex Number
    • Polar form z = R*(cos(A) + i*sin(A)).
    • Polar form z = cis(A).
    • Rectangular form z = x + i*y.
  • Relation between rectangular and polar
    • x = R*cos(A).
    • y = R*sin(A).
    • R = Sqr(x^2 + y^2).
    • A = arctan(y/x).
  • e^(i*A) = cos(A) + i*sin(A)
  • (cos(A)+i*sin(A))*(cos(B)+i*sin(B)) = cos(A+B) + i*sin(A+B).
  • (cos(A)+i*sin(A))/(cos(B)+i*sin(B)) = cos(A-B) + i*sin(A-B).
  • DeMoivre's Theorem
    • cis(A)^n = (cos(A) + i*sin(A))^n = cos(n*A) + i*sin(n*A) = cis(n*A).
    • cis(A)^(1/n) = (cos(A) + i*sin(A))^(1/n).
    • = cos((2*k*pi+A)/n) + i*sin((2*k*pi+A)/n) where k = 1,2,3,....(n-1).
    • = cis((2*k*pi+A)/n)
  • Sqr(cis(A)) = cis(A/2) and -cis(A/2).
  • Cis(A)*cis(B) = Cis(A+B).
  • Cis(A)/cis(B) = Cis(A-B).
  • Solutions of x^4 + x^3 + x^2 + x + 1 = 0 are cis(72), cis(144), cis(216) and cis(288)
    • then cos(72) + cos(144) + cos(216) + cos(288) = -1.
    • then sin(72) + sin(144) + sin(216) + sin(288) = 0.
    • then cis(72)*cis(144)*cis(216)*cis(288) = 1.
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Q13. Prove that (cos(A)+i*sin(A))/(cos(B)+i*sin(B)) = cos(A-B) + i*sin(A-B).
  • Multiply numerator and denominator by (cos(B) - i*sin(B)).
  • Hence denominator is cos(B)^2 - (i*sin(B))^2 = cos(B)^2 + sin(B)^2 = 1.
  • (cos(A)+i*sin(A))/(cos(B)+i*sin(B)) = (cos(A)+i*sin(A))*(cos(B)-i*sin(B)).
  • = (cos(A)*cos(B)+sin(A)*(sin(B)) + i*(sin(A)*cos(B)-cos(A)*cos(B))
  • = cos(A-B) + i*sin(A-B).
  • Or cis(A)/cis(B) = cis(A-B).
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Q14. Quiz
  • 1. Rationalize z = 1/(1/2 - i*Sqr(3)/2).
  • 2. What are conjugate complex numbers.
  • 3. Express z = 1 + i in polar form.
  • 4. Express z = cis(270) as z = x + i*y.
  • 5. Express z = 1/(1-i) in polar form.
  • 6. Prove that (cos(A)+i*sin(A))*(cos(B)+i*sin(B)) = cos(A+B)+i*sin(A+B).
  • 7. Solve x^4 - x^3 + x^2 - x + 1 = 0 by solving (x^5 + 1) = 0.
  • 8. If z = 1 + i, find z^10.
  • 9. If z = cis(pi/3), find summmation Sum[z^n] for n = 1 to 6.
  • 10 Solve z^4 = -8 +i*8*Sqr(3).
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Q15. Quiz
  • 1. Rationalize z = 1/(1/2 - i*Sqr(3)/2).
    • Multiply numeritor and denominator by (1/2 + i*Sqr(3)/2).
    • z = (1/2+i*Sqr(3)/2)/((1/2-i*Sqr(3)/2)*(1/2+i*Sqr(3)/2)).
    • z = (1/2+i*Sqr(3)/2)/(1/4 - (i^2)*3/4).
    • z = (1/2+i*Sqr(3)/2)/(1/4 +3/4).
    • z = (1/2-i*Sqr(3)/2).
  • 2. What are conjugate complex numbers.
    • Sum of complex numbers = real.
    • Product of complex numbers = real.
  • 3. Express z = 1 + i in polar form.
    • z = 1 + i = R*(cos(A) + i*sin(A)).
    • R = Sqr(1^2 + 1^2) = Sqr(2).
    • A = arctan(1/1) = 45 degrees.
    • z = 1 + i = Sqr(2)*(cos(45) + i*sin(45)/2.
  • 4. Express z = cis(270) as z = x + i*y.
    • z = cis 270 = cos(270) + i*sin(270) = -i.
  • 5. Express z = 1/(1-i) in polar form.
    • z = (1+i)/((1-i)*(1+i)).
    • z = (1+i)/2 = R*(sin(A) + i*sin(A)).
    • R = 1/2 and A = 45 degrees.
    • Hence z = 1/(1-i) = (cos(45) + i*sin(45))/2
  • 6. Prove that (cos(A)+i*sin(A))*(cos(B)+i*sin(B)) = cos(A+B)+i*sin(A+B).
    • Use cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B).
    • Use sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B).
    • Then we can get the prove. (See Q5)
  • 7. Solve x^4 - x^3 + x^2 - x + 1 = 0 by solving (x^5 + 1) = 0.
    • Solve x^5 + 1 = 0
    • The five angles are 180, 180+72, 180+2*72, 180+3*72, 180+4*72.
    • x1 = cos(180) + i*sin(180) = -1
    • x2 = cos(252) + i*sin(252) = -cos(72) - i*sin(72) = ?
    • x3 = cos(324) + i*sin(324) = +cos(36) - i*sin(36) = ?
    • x4 = cos(396) + i*sin(396) = +cos(36) + i*sin(36) = ?
    • x5 = cos(468) + i*sin(468) = -cos(72) + i*sin(72) = ?
    • Hence x2, x3, x4, x5 are the required solutions.
    • Note : x2 and x5 are conjugate and x3 and x4 are also conjugate.
  • 8. If z = 1 + i, find z^10.
    • z = (1 + i) = (cos(45) + i*sin(45))/Sqr(2).
    • z^10 = ((1/Sqr(2))^10)*(cos(45) + i*sin(45))^10.
    • z^10 = (1/32)*(cos(450) + i*sin(450))
    • z^10 = (cos(90) + i*sin(90))/32
    • z^10 = i/32.
  • 9. If z = cis(pi/3), find summmation Sum[z^n] for n = 1 to 6.
    • z^1 = cos(060) + i*sin(060) = +cos(60) + i*sin(60).
    • z^2 = cos(120) + i*sin(120) = -cos(60) + i*sin(60).
    • z^3 = cos(180) + i*sin(180) = -1.
    • z^4 = cos(240) + i*sin(240) = -cos(60) - i*sin(60).
    • z^5 = cos(300) + i*sin(300) = +cos(60) - i*sin(60).
    • z^6 = cos(360) + i*sin(360) = +1.
    • Hence z^1 + z^2 + z^3 + z^4 + z^5 + z^6 = 0.
  • 10 Solve z^4 = -8 +i*8*Sqr(3).
    • Z^4 = 8*(-1 + i*Sqr(3)) = R*(cos(A) + i*sin(A))
    • Hence R = 16 and A = arctan(Sqr(3)/(-1)) = 120 degrees.
    • z^4 = (16*(cos(120) + i*sin(120)).
    • z1 = 2*(cos(120/4) + i*sin(120/4)) = 2*(+cos(30) + i*sin(30)).
    • z2 = 2*(cos(480/4) + i*sin(480/4)) = 2*(-cos(60) - i*sin(60)).
    • z3 = 2*(cos(840/4) + i*sin(840/4)) = 2*(-cos(30) - i*sin(30)).
    • z4 = 2*(cos(1200/4) + i*sin(120/4))= 2*(+cos(60) - i*sin(60)).
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