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Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem

Mathematics

Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem

Overview

Overview

  • Represent a complex number and its conjugate on the Argand plane and find its modulus and (principal) amplitude.
  • Write a complex number in trigonometric and exponential (Euler) form, and multiply and divide in these forms.
  • Use De Moivre’s theorem for powers, identities and the n-th roots of a complex number.
  • Know the cube roots of unity, their properties and their applications.

🌳 Concept tree

  • Lesson 1 Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem
  • ├── 2-1 (a) Polar coordinates and the Argand plane

│ ├── Polar and Cartesian coordinates

│ ├── Argand plane

│ └── Symmetry on the Argand plane

  1. ├── 2-1 (b) Modulus
  2. amplitude and principal amplitude

│ ├── Definitions

  • │ ├── Finding the principal amplitude from the quadrant of z

│ └── Properties of the modulus and amplitude

  • ├── 2-1 (c) The trigonometric (polar) form of a complex number

│ ├── The form

│ ├── Useful special values

│ └── Numbers that are not in standard form

  • ├── 2-1 (d) Product and quotient in trigonometric form

│ ├── Rules

│ └── Consequences

├── 2-1 (e) The exponential (Euler) form

│ ├── Series that lead to Euler's formula

│ ├── Euler's formula

│ ├── Product and quotient in exponential form

  1. │ └── Expressing sine and cosine with exponentials
  2. ├── 2-2 (a) De Moivre's theorem for a positive integer power

│ ├── Statement

  • │ └── Using the theorem to prove trigonometric identities
  • ├── 2-2 (b) De Moivre's theorem for a rational power and the roots of a complex number

│ ├── Rational power

│ ├── The n-th roots of a complex number

  1. │ ├── Square roots of a complex number in algebraic form
  2. │ └── Quadratic equations with complex coefficients

└── 2-3 The cube roots of unity

├── Finding them

├── Properties of the cube roots of unity

└── Applications

Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 1 / 10
2-1 (a) Polar coordinates and the Argand plane

2-1 (a) Polar coordinates and the Argand plane

Polar and Cartesian coordinates

Comparison aspectMain ideaKeywords
•A point A(x,y) at distance r from the origin, with the radius making angle θ with the positive x-axis, has polar coordinates (r,θ).pointdistanceoriginradius
Polar axisthe positive part of the x-axis, from which the angle θ is measured.positivepartanglemeasured
Polar to Cartesian
  • x=rcos⁡θ
  • y=rsin⁡θ
rcosrsin
Cartesian to polar
  • r=x2+y2
  • tan⁡θ=yx
•So the Cartesian point can be written as (rcos⁡θ, rsin⁡θ).Cartesianpointwrittenrcos
  • Argand plane (Argand diagram): a Cartesian plane in which the x-axis is the real axis and the y-axis is the imaginary axis.
  • The complex number z=x+yi is represented by the point (x,y) (or by the vector from the origin to that point).
  • Jean-Robert Argand was the first to study complex numbers in detail; he used them to show that every algebraic equation has roots, real or imaginary.
  • The points representing z and −z are symmetric about the origin.
  • The points representing z and its conjugate zˉ are symmetric about the real axis (x-axis).
  • The point representing z1+z2 is the fourth vertex of the parallelogram built on the vectors of z1 and z2.
  • Example z1=2+i and z2=−2−i are symmetric about O; z1=2+i and zˉ1=2−i are symmetric about the real axis.
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 2 / 10
2-1 (b) Modulus, amplitude and principal amplitude

2-1 (b) Modulus, amplitude and principal amplitude

Definitions

Modulus∣z∣=r: the distance of the point z(x,y) from the origin.
r=∣z∣=x2+y2
Amplitude (argument)θ of z: the angle from the positive real axis to the vector of z, with tan⁡θ=yx.
Principal amplitudethe value of the amplitude lying in the interval ]−π,π].
Comparison aspectMain ideaKeywords
•First find the quadrant in which the point (x,y) lies, then use α=tan⁡−1∣yx∣ as the acute reference angle.Firstfindquadrantpoint
First quadrant (x>0, y>0)θ=α
Second quadrant (x<0, y>0)θ=π−α
Third quadrant (x<0, y<0)θ=−π+α
Fourth quadrant (x>0, y<0)θ=−α
On the axes
  • positive real axis θ=0
  • negative real axis θ=π
  • positive imaginary axis θ=π2
  • negative imaginary axis θ=−π2
positiverealaxisnegative
•Radians convert to degrees with degrees=radians×180∘π.Radiansconvertdegrees

✍️ Worked examples

  • Example 1 Find the modulus and the principal amplitude of z1=−3+i and z2=−1−i.
  • Solution:
  • For z1 x=−3, y=1, so z1 lies in the second quadrant.
  • r=3+1=2.
  • Reference angle α=tan⁡−113=π6.
  • θ=π−π6=5π6.
  • For z2 x=−1, y=−1, so z2 lies in the third quadrant.
  • r=1+1=2.
  • Reference angle α=tan⁡−1(1)=π4.
  • θ=−π+π4=−3π4.

Properties of the modulus and amplitude

  1. ∣z∣≥0 for every complex number z.
  2. θ+2πn where n∈Z.
  3. ∣z∣=∣zˉ∣=∣−z∣.
  4. z zˉ=∣z∣2.
  5. If the principal amplitude of z is θ, then that of zˉ is −θ.
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 3 / 10
2-1 (c) The trigonometric (polar) form of a complex number

2-1 (c) The trigonometric (polar) form of a complex number

The form

Comparison aspectMain ideaKeywords
•If z=x+yi has modulus r and principal amplitude θ, then x=rcos⁡θ and y=rsin⁡θ.modulusprincipalamplitudethen
The trigonometric form (polar form) isz=r(cos⁡θ+isin⁡θ)isin
The conversion is done in three steps
  • find r
  • find the quadrant
  • find θ
findquadrant
  1. 1=cos⁡0+isin⁡0, −1=cos⁡π+isin⁡π.
  2. i=cos⁡π2+isin⁡π2, −i=cos⁡(−π2)+isin⁡(−π2).
  3. cos⁡θ−isin⁡θ=cos⁡(−θ)+isin⁡(−θ).

✍️ Worked examples

  • Example 2 Write z=2−23 i in trigonometric form.
  • Solution:
  • x=2, y=−23, so z lies in the fourth quadrant.
  • r=4+12=4.
  • θ=tan⁡−1(−232)=tan⁡−1(−3)=−π3.
  • z=4[cos⁡(−π3)+isin⁡(−π3)].
  • Example 3 Write z=−4i and z=−3 in trigonometric form.
  • Solution:
  • −4i lies on the negative imaginary axis r=4, θ=−π2.
  • −4i=4[cos⁡(−π2)+isin⁡(−π2)].
  • −3 lies on the negative real axis r=3, θ=π.
  • −3=3(cos⁡π+isin⁡π).
  • A form such as −8(cos⁡45∘+isin⁡45∘) is not a trigonometric form because the coefficient is negative.
  • Use reduction formulas to turn it into r(cos⁡θ+isin⁡θ) with r>0.
  • Reduction rules sin⁡(π2+θ)=cos⁡θ, cos⁡(π2+θ)=−sin⁡θ, sin⁡(π+θ)=−sin⁡θ, cos⁡(π+θ)=−cos⁡θ.

✍️ Worked examples

  • Example 4 Find the modulus and principal amplitude of z=−8(cos⁡45∘+isin⁡45∘).
  • Solution:
  • −cos⁡45∘=cos⁡225∘ and −sin⁡45∘=sin⁡225∘.
  • z=8(cos⁡225∘+isin⁡225∘).
  • 225∘ is not in ]−180∘,180∘], so subtract 360∘ 225∘−360∘=−135∘.
  • ∣z∣=8 and the principal amplitude is −135∘=−3π4.
  • Example 5 Find the modulus and principal amplitude of z=2(sin⁡5π6−icos⁡5π6).
  • Solution:
  • Write the sine and cosine with π2 sin⁡5π6=cos⁡π3 and −cos⁡5π6=sin⁡π3.
  • z=2(cos⁡π3+isin⁡π3).
  • ∣z∣=2 and the principal amplitude is π3.
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 4 / 10
2-1 (d) Product and quotient in trigonometric form

2-1 (d) Product and quotient in trigonometric form

Rules

  • Let z1=r1(cos⁡θ1+isin⁡θ1) and z2=r2(cos⁡θ2+isin⁡θ2).
  • Multiplication moduli multiply, amplitudes add. z1z2=r1r2[cos⁡(θ1+θ2)+isin⁡(θ1+θ2)]
  • Division moduli divide, amplitudes subtract. z1z2=r1r2[cos⁡(θ1−θ2)+isin⁡(θ1−θ2)]
  • Therefore ∣z1z2∣=∣z1∣ ∣z2∣,  ∣z1z2∣=∣z1∣∣z2∣,  arg⁡(z1z2)=θ1+θ2,  arg⁡z1z2=θ1−θ2.
For z=r(cos⁡θ+isin⁡θ)1z=1r[cos⁡(−θ)+isin⁡(−θ)],z2=r2(cos⁡2θ+isin⁡2θ)
For a product of several numbers zk=rk(cos⁡θk+isin⁡θk)z1z2⋯zn=r1r2⋯rn[cos⁡(θ1+⋯+θn)+isin⁡(θ1+⋯+θn)]
When all the numbers are equal to z=r(cos⁡θ+isin⁡θ)zn=rn(cos⁡nθ+isin⁡nθ)

✍️ Worked examples

  • Example 6 Write 3(cos⁡π3+isin⁡π3)×4(cos⁡π6+isin⁡π6) in the form x+yi.
  • Solution:
  • Multiply the moduli 3×4=12.
  • Add the amplitudes π3+π6=π2.
  • The product is 12(cos⁡π2+isin⁡π2).
  • =12(0+i)=12i.
  • Example 7 Find 6(cos⁡100∘+isin⁡100∘)3(cos⁡40∘+isin⁡40∘) in the form x+yi.
  • Solution:
  • Divide the moduli 6÷3=2.
  • Subtract the amplitudes 100∘−40∘=60∘.
  • The quotient is 2(cos⁡60∘+isin⁡60∘).
  • =2(12+32i)=1+3 i.
  • Example 8 Write 1−i in trigonometric form, then find (1−i)8.
  • Solution:
  • x=1, y=−1 fourth quadrant, r=2, θ=−π4.
  • 1−i=2[cos⁡(−π4)+isin⁡(−π4)].
  • (1−i)8=(2)8[cos⁡(−2π)+isin⁡(−2π)].
  • =16(1+0i)=16.
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 5 / 10
2-1 (e) The exponential (Euler) form

2-1 (e) The exponential (Euler) form

Series that lead to Euler’s formula

  1. A function can be written as a Taylor series in powers of x.
  2. sin⁡x=x−x33!+x55!−⋯ (an odd function, so only odd powers).
  3. cos⁡x=1−x22!+x44!−⋯ (an even function, so only even powers).

ex=1+x+x22!+x33!+⋯

Replace x by ix and use i2=−1, i3=−i, i4=1→the real terms form the cosine series and the imaginary terms form i times the sine series.

Euler’s formula

  • eiθ=cos⁡θ+isin⁡θ(θ in radians)
  • It links the most famous constants eiπ+1=0.
  • The exponential form (Euler form) of a complex number is
    z=r eiθ
  • It comes from z=r(cos⁡θ+isin⁡θ); the angle must be in radians.
  • If z1=r1eiθ1 and z2=r2eiθ2 z1z2=r1r2 ei(θ1+θ2),z1z2=r1r2 ei(θ1−θ2)
  • Convert degrees to radians with radians=degrees×π180∘.

✍️ Worked examples

  • Example 9 Write in exponential form: (a) 1+i (b) −1+3 i (c) −2i (d) 5.
  • Solution:
  • (a) r=2, first quadrant, θ=tan⁡−11=π4, so 1+i=2 eπ4i.
  • (b) r=1+3=2, second quadrant, θ=π−π3=2π3, so −1+3 i=2e2π3i.
  • (c) −2i lies on the negative imaginary axis r=2, θ=−π2, so −2i=2e−π2i.
  • (d) 5 lies on the positive real axis r=5, θ=0, so 5=5e0.
  • Example 10 Find 2(cos⁡25∘+isin⁡25∘)×3(cos⁡68∘+isin⁡68∘) in exponential form.
  • Solution:
  • Multiply the moduli 2×3=6.
  • Add the amplitudes 25∘+68∘=93∘.
  • Convert to radians 93∘×π180∘=31π60.
  • The result is 6 e31π60i.
  • Example 11 Find 1+i1−i in exponential form.
  • Solution:
  • 1+i=2(cos⁡45∘+isin⁡45∘) and 1−i=2(cos⁡(−45∘)+isin⁡(−45∘)).
  • Divide the moduli 22=1.
  • Subtract the amplitudes 45∘−(−45∘)=90∘=π2.
  • The result is eπ2i (which equals i).
  • Example 12 Write 2 e3π4i in the algebraic form x+yi.
  • Solution:
  • 2(cos⁡3π4+isin⁡3π4).
  • cos⁡3π4=−22 and sin⁡3π4=22.
  • =2(−22+22i)=−1+i.

Expressing sine and cosine with exponentials

From eiθ=cos⁡θ+isin⁡θ and e−iθ=cos⁡θ−isin⁡θ→cos⁡θ=eiθ+e−iθ2,sin⁡θ=eiθ−e−iθ2i
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 6 / 10
2-2 (a) De Moivre’s theorem for a positive integer power

2-2 (a) De Moivre’s theorem for a positive integer power

Statement

Comparison aspectMain ideaKeywords
For every positive integer n(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθisin
•Consequently [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ).Consequentlyisin
In exponential form(reiθ)n=rneinθreiθrneinθ
A turntable with six pegs shows the trigonometric form, De Moivres theorem and the sixth roots of unity every 60 degrees.
A turntable with six pegs shows the trigonometric form, De Moivres theorem and the sixth roots of unity every 60 degrees.

Using the theorem to prove trigonometric identities

  • Expand (cos⁡θ+isin⁡θ)n with the binomial theorem.
  • Equate it with cos⁡nθ+isin⁡nθ.
  • Equate the real parts to get cos⁡nθ and the imaginary parts to get sin⁡nθ.
  • Use sin⁡2θ=1−cos⁡2θ (or the reverse) to write the result in a single function.

✍️ Worked examples

Comparison aspectMain ideaKeywords
Example 13Express cos⁡3θ in terms of cos⁡θ.Expressterms
•SolutionSolution
By De Moivre(cos⁡θ+isin⁡θ)3=cos⁡3θ+isin⁡3θisin
By the binomial theorem(cos⁡θ+isin⁡θ)3=cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θisin3icos2θsin3cos
Equate the real partscos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ3cosθsin
Replace sin⁡2θ=1−cos⁡2θcos⁡3θ=cos⁡3θ−3cos⁡θ+3cos⁡3θ3cos
•cos⁡3θ=4cos⁡3θ−3cos⁡θ.4cos3cos
•In the same way, equating the imaginary parts gives sin⁡3θ=3sin⁡θ−4sin⁡3θ.sameequatingimaginaryparts
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 7 / 10
2-2 (b) De Moivre’s theorem for a rational power and the roots of a complex number

2-2 (b) De Moivre’s theorem for a rational power and the roots of a complex number

Rational power

  • Since cos⁡θ+isin⁡θ=cos⁡(θ+2πr)+isin⁡(θ+2πr) for any integer r, a root has several values.
  • For a positive integer k (cos⁡θ+isin⁡θ)1k=cos⁡θ+2πrk+isin⁡θ+2πrk
  • This gives exactly k different values, taking r as k consecutive integers (such as 0,1,2,…,k−1 or values that keep the angle within ]−π,π]).
  • The equation zn=a (where a is a complex number) has exactly n roots.
  • Method write a in trigonometric form, then apply the rational-power rule with r=0,1,…,n−1.
  • All the roots have the same modulus ∣a∣n, so they lie on a circle with centre O and radius ∣a∣n.
  • The angle between two successive roots is 2πn, so the roots are the vertices of a regular polygon with n sides.

✍️ Worked examples

  • Example 14 Find in C the roots of z4=8(1−3 i) in exponential form.
  • Solution:
  • x=8, y=−83 fourth quadrant, r=64+192=16, θ=tan⁡−1(−3)=−π3.
  • z4=16[cos⁡(−π3)+isin⁡(−π3)].
  • z=2[cos⁡−π3+2πr4+isin⁡−π3+2πr4].
  • r=0 z1=2e−π12i.
  • r=1 z2=2e5π12i.
  • r=−1 z3=2e−7π12i.
  • r=2 z4=2e11π12i.
  • Example 15 Find the roots of z3=1 and represent them on the Argand plane.
  • Solution:
  • z3=cos⁡0∘+isin⁡0∘, so z=cos⁡2πr3+isin⁡2πr3.
  • r=0 z1=cos⁡0∘+isin⁡0∘=1.
  • r=1 z2=cos⁡120∘+isin⁡120∘.
  • r=−1 z3=cos⁡(−120∘)+isin⁡(−120∘).
  • The three points lie on the unit circle and divide it into three equal arcs of 120∘.
  • They are the vertices of an equilateral triangle.
  • Example 16 Find the fifth roots of −32 and represent them on the Argand plane.
  • Solution:
  • z5=−32=32(cos⁡π+isin⁡π).
  • z=2[cos⁡π+2πr5+isin⁡π+2πr5].
  • For r=0 the amplitude is π5=36∘.
  • The angle between successive roots is 360∘5=72∘.
  • z1=2(cos⁡36∘+isin⁡36∘).
  • z2=2(cos⁡108∘+isin⁡108∘).
  • z3=2(cos⁡180∘+isin⁡180∘)=−2.
  • z4=2(cos⁡252∘+isin⁡252∘)=2(cos⁡(−108∘)+isin⁡(−108∘)).
  • z5=2(cos⁡324∘+isin⁡324∘)=2(cos⁡(−36∘)+isin⁡(−36∘)).
  • The five points lie on a circle of radius 2 and form a regular pentagon.

Square roots of a complex number in algebraic form

  • Put a+bi=x+yi and square both sides a+bi=x2−y2+2xyi.
  • Equate the real parts and the imaginary parts x2−y2=a and 2xy=b.
  • Use (x2+y2)2=(x2−y2)2+(2xy)2 to find x2+y2=a2+b2.
  • Solve the two equations in x2 and y2; the signs of x and y must satisfy 2xy=b.

✍️ Worked examples

  • Example 17 Find the square roots of 3+4i.
  • Solution:
  • Let 3+4i=x+yi, then x2−y2=3 (1) and 2xy=4 (2).
  • Square (1) and (2) and add (x2+y2)2=9+16=25, so x2+y2=5 (3).
  • Add (1) and (3) 2x2=8, so x=±2.
  • From (2) x=2⇒y=1 and x=−2⇒y=−1.
  • The square roots are 2+i and −2−i.
  • Use the general formula x=−b±b2−4ac2a; the square root of the discriminant is found by the method above.

✍️ Worked examples

Comparison aspectMain ideaKeywords
Example 18Solve (1−i)x2−(6−4i)x+9−7i=0 in C.Solve
•SolutionSolution
Divide by 1−i6−4i1−i=(6−4i)(1+i)2=5+i and 9−7i1−i=(9−7i)(1+i)2=8+i
•The equation becomes x2−(5+i)x+8+i=0.equationbecomes
Discriminant(5+i)2−4(8+i)=24+10i−32−4i=−8+6i
Let −8+6i=a+bi
  • a2−b2=−8
  • 2ab=6
  • a2+b2=10
  • so a=±1
  • b=±3 and −8+6i=±(1+3i)
•x=5+i±(1+3i)2, so x=3+2i or x=2−i.
Solution set
  • {3+2i
  • 2−i}
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 8 / 10
2-3 The cube roots of unity

2-3 The cube roots of unity

Finding them

  • Solve z3=1 with De Moivre’s theorem the roots are 1, −12+32i and −12−32i.
  • Algebraic check z3−1=(z−1)(z2+z+1)=0 gives z=1 or z=−1±3 i2.
  • The square of one complex root equals the other, so the roots are written as 1, ω, ω2 where:
    ω=−12+32i,ω2=−12−32i
  • On the Argand plane the roots lie on the unit circle, at angles 0∘, 120∘ and −120∘, and form an equilateral triangle.

Properties of the cube roots of unity

Comparison aspectMain ideaKeywords
The sum of the roots is zero1+ω+ω2=0
•Hence 1+ω=−ω2,  1+ω2=−ω,  ω+ω2=−1.Hence
The product is one
  • ω3=1
  • so ω3n=1
  • ω3n+1=ω and ω3n+2=ω2 for any integer n
integer
Reciprocals
  • 1ω=ω2 and 1ω2=ω
  • also ωˉ=ω2
The difference of the two complex rootsω−ω2=3 i and ω2−ω=−3 i
All three roots have modulus 1∣1∣=∣ω∣=∣ω2∣=1
•To simplify a power of ω, divide the exponent by 3 and keep the remainder.simplifypowerdivideexponent

✍️ Worked examples

  • Example 19 Find the value of 5+5ω+5ω2 and of (1−2ω−2ω2)(3+5ω+5ω2).
  • Solution:
  • 5+5ω+5ω2=5(1+ω+ω2)=5×0=0.
  • Use 1ω=ω2 and 1ω2=ω.
  • First bracket 1−2ω2−2ω=1−2(ω+ω2)=1+2=3.
  • Second bracket 3+5(ω+ω2)=3−5=−2.
  • The product is 3×(−2)=−6.
  • Example 20 If 1,ω,ω2 are the cube roots of unity, prove that (1−ω+ω2)(1+ω−ω2)=4.
  • Solution:
  • 1+ω2=−ω, so 1−ω+ω2=−ω−ω=−2ω.
  • 1+ω=−ω2, so 1+ω−ω2=−ω2−ω2=−2ω2.
  • The product is (−2ω)(−2ω2)=4ω3=4.
  • Example 21 Prove that x=−1+3 i2 is a root of x10+x5+1=0.
  • Solution:
  • x=−12+32i=ω, one of the complex cube roots of unity.
  • x10=ω10=ω9⋅ω=ω.
  • x5=ω5=ω3⋅ω2=ω2.
  • x10+x5+1=ω+ω2+1=0, so x is a root.
  • The same holds for x=ω2 (ω2)10=ω2 and (ω2)5=ω, giving again ω2+ω+1=0.

Applications

  • Forming a quadratic equation from two roots that are expressed in ω: find the sum and product of the roots using 1+ω+ω2=0 and ω3=1, then write x2−(sum)x+(product)=0.
  • Reducing expressions replace 1/ω by ω2, group terms into 1+ω+ω2 and reduce powers modulo 3.
  • Complex numbers are used in many fields, including electricity, dynamics, relativity and other physical applications.
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 9 / 10
Part 10

Part 10

  • 📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Mathematics Thanaweya Amma (2026/2027)

  • Lesson 1 – Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem
  • Complex numbers extend real numbers by introducing imaginary unit i = √(-1).
  • A complex number z = x + iy is represented on the Argand diagram and transformed into trigonometric form z = r(cos θ + i sin θ) and exponential form z = r·eiθ.
  • De Moivre's theorem evaluates powers and roots of unity.

🌳 Conceptual Hierarchy Tree

Complex Number Algebra

├── 1. Representations of Complex Numbers
│ ├── Algebraic Formz = x + i y (Modulus r = √(x2 + y2), Argument tan θ = y/x)
│ ├── Trigonometric Formz = r · (cos θ + i sin θ)
│ └── Exponential (Euler) Formz = r · ei θ (where θ is in radians)

├── 2. De Moivre's Theorem & Roots of Unity

  • │ ├── Power Theorem [r(cos θ + i sin θ)]n = rn · [cos(nθ) + i sin(nθ)]
  • │ └── Cube Roots of Unity (1, ω, ω2):

│ ├── 1 + ω + ω2 = 0

│ ├── ω3 = 1 and ω2 = 1/ω = ω̅

│ └── (ω – ω2) = ± i √3

📋 Master Teacher's Blackboard Secrets
RULE 1Principal amplitude (argument) of a complex number must lie in the interval: -π < θ ≤ π.
RULE 2Cube roots of unity fundamental property: 1 + ω + ω2 = 0, which implies 1 + ω = -ω2, 1 + ω2 = -ω, and ω + ω2 = -1.
RULE 3Multiplication of complex numbers: Moduli multiply and amplitudes add (z1 · z2 = r1 r2 [cos(θ1 + θ2) + i sin(θ1 + θ2)]).
📊 MatrixProperties of the Cube Roots of Unity (1, ω, ω2)
Property Mathematical Identity Application Shortcut
Sum of all 3 roots 1 + ω + ω2 = 0 Any two roots equal negative of third
Product of all 3 roots 1 · ω · ω2 = ω3 = 1 ω3n = 1 for any integer n
Difference between imaginary roots ω – ω2 = ± i √3 Squares yield: (ω – ω2)2 = -3
Modulus of each root |1| = |ω| = |ω2| = 1 All lie on unit circle centered at origin
Lesson 1: Complex Numbers in Trigonometric and Exponential Form and De Moivres Theorem · 10 / 10