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Mathematics Dictionary
Dr. K. G. Shih
Complex Number
Questions

    Read Symbol defintion

  • Q01 | - Imaginary number
  • Q02 | - Complex number and cojugate complex
  • Q03 | - Complex number in polar form and in rectangular form
  • Q04 | - e^(i*A) = cos(A) + i*sin(A)
  • Q05 | - (cos(A) + i*sin(A))*(cos(B) + i*sin(B)) = cos(A+B) + i*sin(A+B)
  • Q06 | - DeMoivre's theorem.
  • Q07 | - What is z = cis(A) ?
  • Q08 | - Example : Solve z^4 = 8 + 8*Sqr(3)*i
  • Q09 | - Solve x^3 + 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 numbers
  • Q15 | - Answer to quiz
Answers


Q01. Imaginary number

Defintion
  • A number has no position in real number system
  • The symbol i is used to define the imaginary number. That is i = Sqr(-1).
  • Properties of i
    • i^1 = +i. It is in A = 000 direction on xy coordinate system.
    • i^2 = -1. It is in A = 090 direction on xy coordinate system.
    • i^3 = -i. It is in A = 180 direction on xy coordinate system.
    • i^4 = +1. It is in A = 270 direction on xy coordinate system.
    • i^5 = +i. It is in A = 360 direction on xy coordinate system
  • i^n = ?
    • Let n divide 4.
    • If the remainder is 1 then it is +i.
    • If the remainder is 2 then it is -1.
    • If the remainder is 3 then it is -i.
    • If the remainder is 0 then it is +1.
  • The symbol i was introduced into mathematics by Euler.
  • Example : Find i^100.
    • Since 100 / 4 = 25 and remainder is 0.
    • Hence i^100 = 1.
    • 2nd method : Using DeMoivre's theorem in Q05.
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Q02. Complex number

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Q03. Complex number in polar form

Defintion
  • Let a + b*i = R*(coa(A) + i*sin(A)) be the polar form.
  • Since a = R*cos(A) and b = R*sin(A).
  • Hence R^2 = a^2 + b^2.
  • A = arctan(y/x).
    • If x = + and y = +, then A is in 1st quadrant.
    • If x = - and y = +, then A is in 2nd quadrant.
    • If x = - and y = -, then A is in 3rd quadrant.
    • If x = + and y = -, then A is in 4th quadrant.
Example : Convert z = 1 - i to polar form
  • x = 1 and y = -1. Hence angle A in 4th quadrant.
  • A = arctan(-1/1) = 315 degrees.
  • R = Sqr(1^2+(-1)^2) = Sqr(2)
  • Hence Z = 1 - i = Sqr(2)*(cos(315) + i*sin(315)).
Example : Convert Z = 4*(cos(135)+i*sin(135) to rectangular form
  • Let z = x + i*y.
  • Hence x = R*cos(A) = 4*cos(135) = 4*(-Sqr(2)/2) = -Sqr(2).
  • Hence y = R*sin(A) = 4*sin(135) = 4*(+Sqr(2)/2) = +Sqr(2).
  • Hence z = Sqr(2)*(-1 + i)
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Q04. z = e^(i*A) = cos(A) + i*sin(A)
  • Find e^(i*pi)
    • e^(i*pi) = cos(pi) + isin(pi).
    • cos(pi) = -1 and sin(pi) = 0.
    • e^(i*pi) = -1.
    • These include the most significant symbols e, i, pi, -, and 1.
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Q05. (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)*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)
Example : (cos(A)+i*sin(A))^2 = cos(2*A) + i*sin(2*A)
  • Let A = B in above formula, we have the proof
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)
Example : (cos(A)+i*sin(A))^n = cos(n*A) + i*sin(n*A)
  • From above examples we can get this conclusion.
  • This is call DeMoivre's theorem.
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Q06. DeMoivre's theorem.
  • Theorem 1 : (cos(A)+i*sin(A))^n = cos(n*A) + i*sin(n*A)
    • Example : Find i^100.
      • Let i = x + i*y = R*(cos(A) + i*sin(A)).
      • Hence x = 0 and y = 1.
      • Hence A = 90 degrees and R = 1.
      • Hence i^100 = (cos(90) + i*sin(90))^100 = cos(9000) + i*sin(9000)
      • Sine 9000/360 = 50*pi
      • i^100 = 1
  • Theorem 2 : (cos(A)+i*sin(A))^n = cos((2*k*pi+A)/n) + i*sin((2*k*pi+A)/n).
    Example : Solve z^6 = i
    • Convert i to polar form.
    • i = cos(A) + i*sin(A) = x + iy and x = 0 and y = 1.
    • Hence A = arctan(1/0) = 90 degrees = pi/2.
    • Hence z^6 = (cos(90) + i*sin(90)).
    • Hence z = (cos(90) + i*sin(90))^(1/6).
    • Hence z = cos((2*k*180+90)/6) + i*sin((2*k*180+90)/6).
    • If k= 0 then z=cos(90/6) + i*sin(90/6) = cos(15) + i*sin(15).
    • If k= 1 then z=cos((360+90)/6)+ i*sin((360+90)/6) = cos(075) + i*sin(075).
    • If k= 2 then z=cos((720+90)/6)+ i*sin((720+90)/6) = cos(135) + i*sin(135).
    • if k= 3 then z=cos((1080+90)/6)+ i*sin((1080+90)/6) = cos(195) + i*sin(195).
    • if k= 4 then z=cos((1440+90)/6)+ i*sin((1440+90)/6) = cos(255) + i*sin(255).
    • if k= 5 then z=cos((1800+90)/6)+ i*sin((1080+90)/6) = cos(315) + i*sin(315).
    • If k= 6 then z=value when k = 0. Hence k is from 0 to 5.
    Second method
    • z = cos(A) + i*sin(A).
    • Find the angle for k = 0.
        Since z^6 = cos(90) + i*sin(90).
      • First angle is 90/6. That is ist angle is A = 15 degrees.
      • The other angles = previous angle + 360/6
      • Hence 2nd angle = 15 + 60, etc.
    • For first angle A is 15 and hence z = cos(15) + i*sin(15).
    • For next angle is 15 + 60 = 75. Hence z = cos(75) + i*sin(75).
    • For next angle is 75 + 60 =135. Hence z = cos(135) + i*sin(135).
    • For next angle is 135+ 60 =195. Hence z = cos(195) + i*sin(195).
    • For next angle is 195+ 60 =255. Hence z = cos(255) + i*sin(255).
    • For next angle is 225+ 60 =315. Hence z = cos(315) + i*sin(315).
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Q07. What is z = cis(A) ?
  • cis(A) = cos(A) + i*sin(A).
  • Example : convert z = cis(270) to rectangular form
    • z = cis(270) = cos(270) + i*sin(270) = x + i*y.
    • Hence x = cos(270) = 0 and y = sin(270) = -1.
    • Hence z = -i.
  • Cis(A)*cis(B) = Cis(A+B). Proof : See Q5.
  • Cis(A)/cis(B) = Cis(A-B). Proof : See Q13.
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Q08. Example : Solve z^4 = 8 + 8*Sqr(3)*i
  • Let z^4 = R*(cos(A) + i*sin(A))
  • Hence R = Sqr(8^2 + 8*Sqr(3)^2) = Sqr(64 + 192) = 16.
  • Hence A = arctan(y/x) = arctan(8*Sqr(3)/8) = arctan(Sqr(3)) = 60.
  • First angle is 60/4 = 15 and secon angle is 15 + 360/4 etc.
  • Hence z0 = R^(1/4)*(cos(15) + i*sin(15))
  • Hence z1 = R^(1/4)*(cos(105) + i*sin(105))
  • Hence z2 = R^(1/4)*(cos(109) + i*sin(195))
  • Hence z3 = R^(1/4)*(cos(285) + i*sin(285))
  • Hence z4 = R^(1/4)*(cos(375) + i*sin(375)) = R^(1/4)*(cos(15)+i*sin(15) = z0.
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Q09. 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
    • 1st angle is 180.
      • Hence x1 = cos(180) + i*sin(180) = -1.
    • 2nd angle is 180 + 1*360/3 = 300.
      • Hence x2 = cos(300) + i*sin(300).
      • x2 = cos(60) - i*sin(60)
      • x2 = 1/2 - i*Sqr(3)/2
    • 3rd angle is 180 + 2*360/3 = 420.
      • Hence 3x = cos(420) + i*sin(420).
      • x3 = cos(60) + i sin(60)
      • x3 = 1/2 + i*Sqr(3)/2
    • Note : x2 and x3 are conjugate.
Method 3 : DeMoivre's Theorem
  • x^3 = -1 = cos(180) + i*sin(180).
  • x = (cos(180) + i*sin(180))^(1/3).
  • x = cos((2*k*180+180)/3) + i*sin((2*k*180+180)/3)
  • For k = 0,
      x1 = cos(60) + i*sin(60).
    • = 1/2 - i*Sqr(3)/2.
  • For k = 1,
    • x2 = cos(540/3) + i*sin(540/3)
    • = cos(180) + i*sin(180)
    • = -1.
  • For k = 2,
    • x3 = cos(900/3) + i*sin(900/3)
    • = cos(300) + i*sin(300)
    • = cos(60) - i* sin(60)
    • = 1/2 - i*Sqr(3)/2.
Method 4 : Construction method
  • Draw a circle with radius = 1 inch and center at C(0,0).
  • Draw axis ox and axis oy.
  • Draw a point P(x,y) on circle and make angle PCQ = 60.
  • Draw a point Q on x-axis so that PQ perpendicular to x-axis.
  • Hence x1 = CQ + PQ*i = 1/2 + i*Sqr(3)/2.
  • Similarly draw angle PCQ = 180 and x2 = -1.
  • Similarly draw angle PCQ = 300 and x3 = 1/2 - i*Sqr(3)/2
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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
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Q11. Reference
  • On PC computer : Use MD2002 Chapter 8.
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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)).
  • Complex Number : 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 (cos(A) + i*sin(A))^n = cos(n*A) + i*sin(n*A).
  • DeMoivre's Theorem (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).
  • Sqr(cis(A)) = cis(A/2) and -cis(A/2).
  • Cis(A)*cis(B) = Cis(A+B).
  • Cis(A)/cis(B) = Cis(A-B).
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Q13. (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).
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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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