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 ├── 2-1 (b) Modulus 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 │ └── Expressing sine and cosine with exponentials ├── 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 │ ├── Square roots of a complex number in algebraic form │ └── Quadratic equations with complex coefficients └── 2-3 The cube roots of unity ├── Finding them ├── Properties of the cube roots of unity └── Applications
Polar and Cartesian coordinates Comparison aspect Main idea Keywords • 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 ) . point distance origin radius Polar axis the positive part of the x -axis, from which the angle θ is measured. positive part angle measured Polar to Cartesian rcos rsin Cartesian to polar • So the Cartesian point can be written as ( r cos θ , r sin θ ) . Cartesian point written rcos
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 + y i 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 z ˉ are symmetric about the real axis (x -axis). The point representing z 1 + z 2 is the fourth vertex of the parallelogram built on the vectors of z 1 and z 2 . 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.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 ∣ = x 2 + y 2
Amplitude (argument) θ of z : the angle from the positive real axis to the vector of z , with tan θ = y x .
Principal amplitude the value of the amplitude lying in the interval ] − π , π ] .
Comparison aspect Main idea Keywords • First find the quadrant in which the point ( x , y ) lies, then use α = tan − 1 ∣ y x ∣ as the acute reference angle. First find quadrant point 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 positive real axis negative • Radians convert to degrees with degrees = radians × 18 0 ∘ π . Radians convert degrees
✍️ Worked examples Example 1 Find the modulus and the principal amplitude of z 1 = − 3 + i and z 2 = − 1 − i .Solution: For z1 x=−3, y=1, so z1 lies in the second quadrant.r = 3 + 1 = 2 .Reference angle α = tan − 1 1 3 = π 6 . θ = π − π 6 = 5 π 6 .For z2 x=−1, y=−1, so z2 lies in the third quadrant.r = 1 + 1 = 2 .Reference angle α = tan − 1 ( 1 ) = π 4 . θ = − π + π 4 = − 3 π 4 .
Properties of the modulus and amplitude ∣ z ∣ ≥ 0 for every complex number z .θ+2πn where n∈Z. ∣ z ∣ = ∣ z ˉ ∣ = ∣ − z ∣ .z z ˉ = ∣ z ∣ 2 .If the principal amplitude of z is θ , then that of z ˉ is − θ .
The form Comparison aspect Main idea Keywords • If z = x + y i has modulus r and principal amplitude θ , then x = r cos θ and y = r sin θ . modulus principal amplitude then The trigonometric form (polar form) is z=r(cosθ+isinθ) isin The conversion is done in three steps find r find the quadrant find θ find quadrant
1 = cos 0 + i sin 0 , − 1 = cos π + i sin π .i = cos π 2 + i sin π 2 , − i = cos ( − π 2 ) + i sin ( − π 2 ) .cos θ − i sin θ = cos ( − θ ) + i sin ( − θ ) .✍️ Worked examples Example 2 Write z = 2 − 2 3 i in trigonometric form.Solution: x = 2 , y = − 2 3 , so z lies in the fourth quadrant.r = 4 + 12 = 4 .θ = tan − 1 ( − 2 3 2 ) = tan − 1 ( − 3 ) = − π 3 .z = 4 [ cos ( − π 3 ) + i sin ( − π 3 ) ] .Example 3 Write z = − 4 i and z = − 3 in trigonometric form.Solution: −4i lies on the negative imaginary axis r=4, θ=−π2.− 4 i = 4 [ cos ( − π 2 ) + i sin ( − π 2 ) ] .−3 lies on the negative real axis r=3, θ=π.− 3 = 3 ( cos π + i sin π ) .A form such as − 8 ( cos 4 5 ∘ + i sin 4 5 ∘ ) is not a trigonometric form because the coefficient is negative. Use reduction formulas to turn it into r ( cos θ + i sin θ ) with r > 0 . Reduction rules sin(π2+θ)=cosθ, cos(π2+θ)=−sinθ, sin(π+θ)=−sinθ, cos(π+θ)=−cosθ.✍️ Worked examples Example 4 Find the modulus and principal amplitude of z = − 8 ( cos 4 5 ∘ + i sin 4 5 ∘ ) .Solution: − cos 4 5 ∘ = cos 22 5 ∘ and − sin 4 5 ∘ = sin 22 5 ∘ .z = 8 ( cos 22 5 ∘ + i sin 22 5 ∘ ) .225∘ is not in ]−180∘,180∘], so subtract 360∘ 225∘−360∘=−135∘.∣ z ∣ = 8 and the principal amplitude is − 13 5 ∘ = − 3 π 4 .Example 5 Find the modulus and principal amplitude of z = 2 ( sin 5 π 6 − i cos 5 π 6 ) .Solution: Write the sine and cosine with π2 sin5π6=cosπ3 and −cos5π6=sinπ3.z = 2 ( cos π 3 + i sin π 3 ) .∣ z ∣ = 2 and the principal amplitude is π 3 .
Rules Let z 1 = r 1 ( cos θ 1 + i sin θ 1 ) and z 2 = r 2 ( cos θ 2 + i sin θ 2 ) . 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 ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ , ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ , arg ( z 1 z 2 ) = θ 1 + θ 2 , arg z 1 z 2 = θ 1 − θ 2 . For z=r(cosθ+isinθ) 1z=1r[cos(−θ)+isin(−θ)],z2=r2(cos2θ+isin2θ)
For a product of several numbers zk=rk(cosθk+isinθk) z1z2⋯zn=r1r2⋯rn[cos(θ1+⋯+θn)+isin(θ1+⋯+θn)]
When all the numbers are equal to z=r(cosθ+isinθ) zn=rn(cosnθ+isinnθ)
✍️ Worked examples Example 6 Write 3 ( cos π 3 + i sin π 3 ) × 4 ( cos π 6 + i sin π 6 ) in the form x + y i .Solution: Multiply the moduli 3×4=12.Add the amplitudes π3+π6=π2.The product is 12 ( cos π 2 + i sin π 2 ) . = 12 ( 0 + i ) = 12 i .Example 7 Find 6 ( cos 10 0 ∘ + i sin 10 0 ∘ ) 3 ( cos 4 0 ∘ + i sin 4 0 ∘ ) in the form x + y i .Solution: Divide the moduli 6÷3=2.Subtract the amplitudes 100∘−40∘=60∘.The quotient is 2 ( cos 6 0 ∘ + i sin 6 0 ∘ ) . = 2 ( 1 2 + 3 2 i ) = 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 ( − π 4 ) + i sin ( − π 4 ) ] .( 1 − i ) 8 = ( 2 ) 8 [ cos ( − 2 π ) + i sin ( − 2 π ) ] .= 16 ( 1 + 0 i ) = 16 .
Series that lead to Euler’s formula A function can be written as a Taylor series in powers of x . sin x = x − x 3 3 ! + x 5 5 ! − ⋯ (an odd function, so only odd powers).cos x = 1 − x 2 2 ! + x 4 4 ! − ⋯ (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 e i θ = cos θ + i sin 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 e i θ It comes from z = r ( cos θ + i sin 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 × π 18 0 ∘ . ✍️ Worked examples Example 9 Write in exponential form: (a) 1 + i (b) − 1 + 3 i (c) − 2 i (d) 5 .Solution: (a) r = 2 , first quadrant, θ = tan − 1 1 = π 4 , so 1 + i = 2 e π 4 i . (b) r = 1 + 3 = 2 , second quadrant, θ = π − π 3 = 2 π 3 , so − 1 + 3 i = 2 e 2 π 3 i . (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 2 5 ∘ + i sin 2 5 ∘ ) × 3 ( cos 6 8 ∘ + i sin 6 8 ∘ ) 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 e 31 π 60 i . Example 11 Find 1 + i 1 − i in exponential form.Solution: 1 + i = 2 ( cos 4 5 ∘ + i sin 4 5 ∘ ) and 1 − i = 2 ( cos ( − 4 5 ∘ ) + i sin ( − 4 5 ∘ ) ) .Divide the moduli 22=1.Subtract the amplitudes 45∘−(−45∘)=90∘=π2.The result is e π 2 i (which equals i ). Example 12 Write 2 e 3 π 4 i in the algebraic form x + y i .Solution: 2 ( cos 3 π 4 + i sin 3 π 4 ) .cos 3 π 4 = − 2 2 and sin 3 π 4 = 2 2 .= 2 ( − 2 2 + 2 2 i ) = − 1 + i .
Expressing sine and cosine with exponentials From eiθ=cosθ+isinθ and e−iθ=cosθ−isinθ → cosθ=eiθ+e−iθ2,sinθ=eiθ−e−iθ2i
Statement Comparison aspect Main idea Keywords For every positive integer n (cosθ+isinθ)n=cosnθ+isinnθ isin • Consequently [ r ( cos θ + i sin θ ) ] n = r n ( cos n θ + i sin n θ ) . Consequently isin In exponential form (reiθ)n=rneinθ reiθ rneinθ
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 + i sin θ ) n with the binomial theorem. Equate it with cos n θ + i sin n θ . Equate the real parts to get cos n θ and the imaginary parts to get sin n θ . Use sin 2 θ = 1 − cos 2 θ (or the reverse) to write the result in a single function. ✍️ Worked examples Comparison aspect Main idea Keywords Example 13 Express cos 3 θ in terms of cos θ . Express terms • Solution Solution By De Moivre (cosθ+isinθ)3=cos3θ+isin3θ isin By the binomial theorem (cosθ+isinθ)3=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ isin 3icos 2θsin 3cos Equate the real parts cos3θ=cos3θ−3cosθsin2θ 3cos θsin Replace sin2θ=1−cos2θ cos3θ=cos3θ−3cosθ+3cos3θ 3cos • cos 3 θ = 4 cos 3 θ − 3 cos θ .4cos 3cos • In the same way, equating the imaginary parts gives sin 3 θ = 3 sin θ − 4 sin 3 θ . same equating imaginary parts
Rational power Since cos θ + i sin θ = cos ( θ + 2 π r ) + i sin ( θ + 2 π r ) for any integer r , a root has several values. For a positive integer k (cosθ+isinθ)1k=cosθ+2πrk+isinθ+2πrkThis 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 z n = 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 2 π n , so the roots are the vertices of a regular polygon with n sides. ✍️ Worked examples Example 14 Find in C the roots of z 4 = 8 ( 1 − 3 i ) in exponential form.Solution: x=8, y=−83 fourth quadrant, r=64+192=16, θ=tan−1(−3)=−π3.z 4 = 16 [ cos ( − π 3 ) + i sin ( − π 3 ) ] .z = 2 [ cos − π 3 + 2 π r 4 + i sin − π 3 + 2 π r 4 ] .r=0 z1=2e−π12i.r=1 z2=2e5π12i.r=−1 z3=2e−7π12i.r=2 z4=2e11π12i.Example 15 Find the roots of z 3 = 1 and represent them on the Argand plane.Solution: z 3 = cos 0 ∘ + i sin 0 ∘ , so z = cos 2 π r 3 + i sin 2 π r 3 .r=0 z1=cos0∘+isin0∘=1.r=1 z2=cos120∘+isin120∘.r=−1 z3=cos(−120∘)+isin(−120∘).The three points lie on the unit circle and divide it into three equal arcs of 12 0 ∘ . They are the vertices of an equilateral triangle. Example 16 Find the fifth roots of − 32 and represent them on the Argand plane.Solution: z 5 = − 32 = 32 ( cos π + i sin π ) .z = 2 [ cos π + 2 π r 5 + i sin π + 2 π r 5 ] .For r = 0 the amplitude is π 5 = 3 6 ∘ . The angle between successive roots is 36 0 ∘ 5 = 7 2 ∘ . z 1 = 2 ( cos 3 6 ∘ + i sin 3 6 ∘ ) .z 2 = 2 ( cos 10 8 ∘ + i sin 10 8 ∘ ) .z 3 = 2 ( cos 18 0 ∘ + i sin 18 0 ∘ ) = − 2 .z 4 = 2 ( cos 25 2 ∘ + i sin 25 2 ∘ ) = 2 ( cos ( − 10 8 ∘ ) + i sin ( − 10 8 ∘ ) ) .z 5 = 2 ( cos 32 4 ∘ + i sin 32 4 ∘ ) = 2 ( cos ( − 3 6 ∘ ) + i sin ( − 3 6 ∘ ) ) .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 ( x 2 + y 2 ) 2 = ( x 2 − y 2 ) 2 + ( 2 x y ) 2 to find x 2 + y 2 = a 2 + b 2 . Solve the two equations in x 2 and y 2 ; the signs of x and y must satisfy 2 x y = b . ✍️ Worked examples Example 17 Find the square roots of 3 + 4 i .Solution: Let 3 + 4 i = x + y i , then x 2 − y 2 = 3 (1) and 2 x y = 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 ± b 2 − 4 a c 2 a ; the square root of the discriminant is found by the method above. ✍️ Worked examples Comparison aspect Main idea Keywords Example 18 Solve ( 1 − i ) x 2 − ( 6 − 4 i ) x + 9 − 7 i = 0 in C . Solve • Solution Solution Divide by 1−i 6−4i1−i=(6−4i)(1+i)2=5+i and 9−7i1−i=(9−7i)(1+i)2=8+i • The equation becomes x 2 − ( 5 + i ) x + 8 + i = 0 . equation becomes 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 + 3 i ) 2 , so x = 3 + 2 i or x = 2 − i .Solution set
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 , ω 2 where:
ω = − 1 2 + 3 2 i , ω 2 = − 1 2 − 3 2 i On the Argand plane the roots lie on the unit circle, at angles 0 ∘ , 12 0 ∘ and − 12 0 ∘ , and form an equilateral triangle. Properties of the cube roots of unity Comparison aspect Main idea Keywords The sum of the roots is zero 1+ω+ω2=0 • Hence 1 + ω = − ω 2 , 1 + ω 2 = − ω , ω + ω 2 = − 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 roots ω−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. simplify power divide exponent
✍️ Worked examples Example 19 Find the value of 5 + 5 ω + 5 ω 2 and of ( 1 − 2 ω − 2 ω 2 ) ( 3 + 5 ω + 5 ω 2 ) .Solution: 5 + 5 ω + 5 ω 2 = 5 ( 1 + ω + ω 2 ) = 5 × 0 = 0 .Use 1 ω = ω 2 and 1 ω 2 = 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 , ω 2 are the cube roots of unity, prove that ( 1 − ω + ω 2 ) ( 1 + ω − ω 2 ) = 4 .Solution: 1 + ω 2 = − ω , so 1 − ω + ω 2 = − ω − ω = − 2 ω .1 + ω = − ω 2 , so 1 + ω − ω 2 = − ω 2 − ω 2 = − 2 ω 2 .The product is ( − 2 ω ) ( − 2 ω 2 ) = 4 ω 3 = 4 . Example 21 Prove that x = − 1 + 3 i 2 is a root of x 10 + x 5 + 1 = 0 .Solution: x = − 1 2 + 3 2 i = ω , one of the complex cube roots of unity.x 10 = ω 10 = ω 9 ⋅ ω = ω .x 5 = ω 5 = ω 3 ⋅ ω 2 = ω 2 .x 10 + x 5 + 1 = ω + ω 2 + 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 + ω 2 = 0 and ω 3 = 1 , then write x 2 − ( 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.
🔍 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 Form z = x + i y (Modulus r = √(x2 + y2 ), Argument tan θ = y/x)
│ ├── Trigonometric Form z = r · (cos θ + i sin θ)
│ └── Exponential (Euler) Form z = 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 1 Principal amplitude (argument) of a complex number must lie in the interval: -π < θ ≤ π.
RULE 2 Cube roots of unity fundamental property: 1 + ω + ω2 = 0, which implies 1 + ω = -ω2 , 1 + ω2 = -ω, and ω + ω2 = -1.
RULE 3 Multiplication of complex numbers: Moduli multiply and amplitudes add (z1 · z2 = r1 r2 [cos(θ1 + θ2) + i sin(θ1 + θ2)]).
📊 Matrix Properties 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