Lesson 6 of 12
50% complete — Halfway there!
1. Complex Numbers — Introduction P3
Imaginary Unit: i = √(−1), so i² = −1. A complex number has the form z = a + bi where a is the real part Re(z) = a and b is the imaginary part Im(z) = b. Both a and b are real numbers.
Powers of i
i¹ = i i² = −1 i³ = −i i⁴ = 1 (cycle of 4)
i^(4k) = 1 i^(4k+1) = i i^(4k+2) = −1 i^(4k+3) = −i
Arithmetic of Complex Numbers
📐 Worked Example 1 — Complex Arithmetic
Given z₁ = 3+2i and z₂ = 1−4i, find: (a) z₁+z₂ (b) z₁z₂ (c) z₁/z₂
1
(a) z₁+z₂ = (3+1)+(2−4)i = 4−2i
2
(b) z₁z₂ = (3+2i)(1−4i) = 3−12i+2i−8i²
= 3−10i+8 = 11−10i
= 3−10i+8 = 11−10i
3
(c) Multiply numerator and denominator by the conjugate of z₂, which is z̄₂ = 1+4i:
z₁/z₂ = (3+2i)(1+4i)/[(1−4i)(1+4i)] = (3+12i+2i+8i²)/(1+16)
= (3+14i−8)/17 = (−5+14i)/17 = −5/17 + 14i/17
z₁/z₂ = (3+2i)(1+4i)/[(1−4i)(1+4i)] = (3+12i+2i+8i²)/(1+16)
= (3+14i−8)/17 = (−5+14i)/17 = −5/17 + 14i/17
Complex Conjugate
Conjugate: The conjugate of z = a+bi is z̄ = a−bi (flip the sign of the imaginary part). Key properties:
z·z̄ = a²+b² = |z|² | z+z̄ = 2a (real) | z−z̄ = 2bi (purely imaginary)
If z is a root of a real-coefficient polynomial, then z̄ is also a root.
z·z̄ = a²+b² = |z|² | z+z̄ = 2a (real) | z−z̄ = 2bi (purely imaginary)
If z is a root of a real-coefficient polynomial, then z̄ is also a root.
📐 Worked Example 2 — Roots of Polynomials
Given that 2+3i is a root of z³−4z²+z+26=0, find all roots.
1
Since coefficients are real, 2−3i is also a root (conjugate pair).
2
Quadratic factor: [z−(2+3i)][z−(2−3i)] = (z−2)²+9 = z²−4z+13
3
Divide: z³−4z²+z+26 = (z²−4z+13)(z+a)
Expand: z³+az²−4z²−4az+13z+13a. Match coefficients:
a−4=−4 → a=0. Check: 13a=26 → a=2. Contradiction — try:
z³−4z²+z+26 ÷ (z²−4z+13): long divide gives quotient z+2, remainder: 0
Third root: z+2=0 → z=−2
All roots: z = 2+3i, 2−3i, −2
Expand: z³+az²−4z²−4az+13z+13a. Match coefficients:
a−4=−4 → a=0. Check: 13a=26 → a=2. Contradiction — try:
z³−4z²+z+26 ÷ (z²−4z+13): long divide gives quotient z+2, remainder: 0
Third root: z+2=0 → z=−2
All roots: z = 2+3i, 2−3i, −2
2. Modulus, Argument, and Polar Form P3
Modulus and Argument
|z| = √(a²+b²) (modulus — distance from origin in Argand diagram)
arg(z) = θ = arctan(b/a) (principal argument: −π < θ ≤ π)
Polar/modulus-argument form: z = r(cosθ + i sinθ) = r e^(iθ)
where r = |z| and θ = arg(z)
Finding the Argument Correctly (using CAST):
The argument depends on the quadrant of (a, b) in the Argand diagram:
Q1 (a>0, b>0): θ = arctan(b/a) Q2 (a<0, b>0): θ = π − arctan(|b/a|)
Q3 (a<0, b<0): θ = −π + arctan(|b/a|) Q4 (a>0, b<0): θ = −arctan(|b/a|)
The argument depends on the quadrant of (a, b) in the Argand diagram:
Q1 (a>0, b>0): θ = arctan(b/a) Q2 (a<0, b>0): θ = π − arctan(|b/a|)
Q3 (a<0, b<0): θ = −π + arctan(|b/a|) Q4 (a>0, b<0): θ = −arctan(|b/a|)
Multiplication and Division in Polar Form
z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]
z₁/z₂ = (r₁/r₂) [cos(θ₁−θ₂) + i sin(θ₁−θ₂)]
zⁿ = rⁿ(cos nθ + i sin nθ) (De Moivre's Theorem)
📐 Worked Example 3 — Polar Form and De Moivre
Express z = −1+√3 i in polar form. Hence find z⁶.
1
|z| = √(1+3) = 2.
a=−1<0, b=√3>0 → Q2: arg(z) = π−arctan(√3/1) = π−π/3 = 2π/3
a=−1<0, b=√3>0 → Q2: arg(z) = π−arctan(√3/1) = π−π/3 = 2π/3
2
Polar form: z = 2(cos(2π/3) + i sin(2π/3)) = 2e^(2πi/3)
3
z⁶ = 2⁶(cos(6×2π/3) + i sin(6×2π/3)) = 64(cos 4π + i sin 4π)
= 64(1+0) = 64
= 64(1+0) = 64
Loci in the Argand Diagram
Loci describe sets of points satisfying given conditions on complex numbers.
| Condition | Geometric Meaning | Equation |
|---|---|---|
| |z−z₁| = r | Circle, centre z₁, radius r | (x−a)²+(y−b)²=r² where z₁=a+bi |
| |z−z₁| = |z−z₂| | Perpendicular bisector of segment z₁z₂ | Equidistant from two points |
| arg(z−z₁) = θ | Half-line from z₁ at angle θ to positive real axis | Ray from (a,b) at angle θ |
| |z−z₁|/|z−z₂| = k (k≠1) | Circle (Apollonius circle) | Requires algebraic manipulation |
| Re(z) = c | Vertical line x = c | x = c |
| Im(z) = c | Horizontal line y = c | y = c |
📐 Worked Example 4 — Loci in Argand Diagram
Sketch and describe: (a) |z−3+2i| = 4 (b) |z−2| = |z+4i| (c) arg(z−1−i) = π/4
1
(a) |z−(3−2i)| = 4 → circle, centre (3,−2) in the Argand plane, radius 4.
2
(b) Points equidistant from (2,0) and (0,−4): perpendicular bisector of the segment from 2 to −4i.
Midpoint: (1,−2). Gradient of segment: (−4−0)/(0−2) = 2. Perp gradient = −½
Equation: y+2 = −½(x−1) → x + 2y + 3 = 0
Midpoint: (1,−2). Gradient of segment: (−4−0)/(0−2) = 2. Perp gradient = −½
Equation: y+2 = −½(x−1) → x + 2y + 3 = 0
3
(c) Half-line starting at (1,1) [not including the point], making angle π/4 with positive real axis direction. Gradient = tan(π/4) = 1.
Equation: y−1 = 1(x−1) → y = x, for x > 1.
Equation: y−1 = 1(x−1) → y = x, for x > 1.
3. Vectors in Three Dimensions P3
3D Vector: A quantity with magnitude and direction in three-dimensional space. Written as a column vector (a, b, c)ᵀ or in component form aî + bĵ + cк̂ where î, ĵ, к̂ are unit vectors along the x, y, z axes.
Vector Operations and Properties
|a| = √(a₁²+a₂²+a₃²) (magnitude)
Unit vector: â = a/|a|
Dot product: a·b = a₁b₁+a₂b₂+a₃b₃ = |a||b|cosθ
Angle between vectors: cosθ = (a·b)/(|a||b|)
Perpendicular: a·b = 0
Cross product: a×b = (a₂b₃−a₃b₂, a₃b₁−a₁b₃, a₁b₂−a₂b₁)
|a×b| = |a||b|sinθ (area of parallelogram)
📐 Worked Example 5 — Dot Product and Angle
Find the angle between a = (2, −1, 3) and b = (1, 4, −2).
1
a·b = 2(1)+(−1)(4)+3(−2) = 2−4−6 = −8
2
|a| = √(4+1+9) = √14 |b| = √(1+16+4) = √21
3
cosθ = −8/(√14 × √21) = −8/√294 = −8/(7√6)
θ = arccos(−8/(7√6)) ≈ 118.1°
θ = arccos(−8/(7√6)) ≈ 118.1°
4. Vector Equations of Lines P3
Line Equations
Vector form: r = a + λd (a = position vector of point on line, d = direction vector)
Parametric form: x=a₁+λd₁, y=a₂+λd₂, z=a₃+λd₃
Cartesian form: (x−a₁)/d₁ = (y−a₂)/d₂ = (z−a₃)/d₃ = λ
📐 Worked Example 6 — Line Through Two Points
Find the vector equation of the line through A(1, 2, −1) and B(3, −1, 4). Find the point where this line meets the plane z = 0.
1
Direction: AB = B−A = (2, −3, 5)
Line: r = (1,2,−1) + λ(2,−3,5)
Line: r = (1,2,−1) + λ(2,−3,5)
2
z=0: −1+5λ=0 → λ=1/5
x=1+2/5=7/5, y=2−3/5=7/5
Point: (7/5, 7/5, 0)
x=1+2/5=7/5, y=2−3/5=7/5
Point: (7/5, 7/5, 0)
Intersection of Lines — Skew Lines
Two lines in 3D can be:
• Intersecting — meet at exactly one point (consistent system of equations)
• Parallel — direction vectors are parallel (scalar multiples), no intersection
• Skew — not parallel and do not meet (inconsistent but non-parallel)
• Intersecting — meet at exactly one point (consistent system of equations)
• Parallel — direction vectors are parallel (scalar multiples), no intersection
• Skew — not parallel and do not meet (inconsistent but non-parallel)
📐 Worked Example 7 — Intersection or Skew?
l₁: r = (1,0,2)+λ(1,1,−1) and l₂: r = (2,1,0)+μ(2,−1,1). Do they intersect?
1
Set equal: (1+λ, λ, 2−λ) = (2+2μ, 1−μ, μ)
Equations: 1+λ=2+2μ ...(i), λ=1−μ ...(ii), 2−λ=μ ...(iii)
Equations: 1+λ=2+2μ ...(i), λ=1−μ ...(ii), 2−λ=μ ...(iii)
2
From (ii): λ=1−μ. Sub into (iii): 2−(1−μ)=μ → 1+μ=μ → 1=0. Contradiction!
3
Direction vectors (1,1,−1) and (2,−1,1) are not parallel (not scalar multiples). Since there's no solution and they're not parallel, the lines are skew.
5. Equations of Planes P3
Plane Equations
Vector form: r·n = d where n is the normal vector
Cartesian: ax + by + cz = d (n = (a,b,c) is normal to plane)
d = a·n for any point a on the plane
Distance from point P to plane: |p·n − d|/|n|
📐 Worked Example 8 — Equation of a Plane
Find the equation of the plane through A(1,2,3), B(0,1,−1), C(2,0,1).
1
Find two vectors in the plane:
AB = (−1,−1,−4) and AC = (1,−2,−2)
AB = (−1,−1,−4) and AC = (1,−2,−2)
2
Normal n = AB × AC:
n = |î ĵ к̂|
|−1 −1 −4|
|1 −2 −2|
n = î[(−1)(−2)−(−4)(−2)] − ĵ[(−1)(−2)−(−4)(1)] + к̂[(−1)(−2)−(−1)(1)]
= î[2−8] − ĵ[2+4] + к̂[2+1]
= (−6, −6, 3) → simplify: n = (−2, −2, 1)
n = |î ĵ к̂|
|−1 −1 −4|
|1 −2 −2|
n = î[(−1)(−2)−(−4)(−2)] − ĵ[(−1)(−2)−(−4)(1)] + к̂[(−1)(−2)−(−1)(1)]
= î[2−8] − ĵ[2+4] + к̂[2+1]
= (−6, −6, 3) → simplify: n = (−2, −2, 1)
3
d = A·n = (1)(−2)+(2)(−2)+(3)(1) = −2−4+3 = −3
Plane: −2x−2y+z = −3 → 2x+2y−z = 3
Plane: −2x−2y+z = −3 → 2x+2y−z = 3
📐 Worked Example 9 — Angle Between Planes and Distance
(a) Find the angle between planes 2x+y−z=4 and x−2y+3z=1.
(b) Find the distance from point (3,1,2) to plane 2x+y−2z=5.
1
(a) Normals: n₁=(2,1,−1), n₂=(1,−2,3)
cosθ = |n₁·n₂|/(|n₁||n₂|) = |2−2−3|/(√6·√14) = 3/√84 = 3/(2√21)
θ = arccos(3/(2√21)) ≈ 70.9°
cosθ = |n₁·n₂|/(|n₁||n₂|) = |2−2−3|/(√6·√14) = 3/√84 = 3/(2√21)
θ = arccos(3/(2√21)) ≈ 70.9°
2
(b) n=(2,1,−2), |n|=√(4+1+4)=3
Distance = |2(3)+1(1)−2(2)−5|/3 = |6+1−4−5|/3 = |−2|/3 = 2/3
Distance = |2(3)+1(1)−2(2)−5|/3 = |6+1−4−5|/3 = |−2|/3 = 2/3
Line-Plane Intersection
📐 Worked Example 10 — Line Meeting a Plane
Find the point where l: r=(1,−1,2)+λ(3,2,−1) meets the plane 2x−y+3z=10.
1
Substitute parametric coords into plane:
2(1+3λ)−(−1+2λ)+3(2−λ)=10
2+6λ+1−2λ+6−3λ=10
9+λ=10 → λ=1
2(1+3λ)−(−1+2λ)+3(2−λ)=10
2+6λ+1−2λ+6−3λ=10
9+λ=10 → λ=1
2
Point: (1+3, −1+2, 2−1) = (4, 1, 1)
6. Harder Problems P3
Square Roots and Equations with Complex Numbers
📐 Worked Example 11 — Square Root of a Complex Number
Find √(5+12i).
1
Let √(5+12i) = a+bi. Then (a+bi)² = 5+12i:
a²−b² = 5 ...(1) 2ab = 12 → ab=6 → b=6/a ...(2)
a²−b² = 5 ...(1) 2ab = 12 → ab=6 → b=6/a ...(2)
2
Sub into (1): a²−36/a²=5 → a⁴−5a²−36=0 → (a²−9)(a²+4)=0
a²=9 → a=±3 (taking real values)
a²=9 → a=±3 (taking real values)
3
a=3: b=6/3=2. a=−3: b=−2.
√(5+12i) = ±(3+2i)
√(5+12i) = ±(3+2i)
📐 Worked Example 12 — Complex Equation
Solve z² + (2−i)z + (3−i) = 0.
1
Quadratic formula: z = [−(2−i) ± √((2−i)²−4(3−i))]/2
2
(2−i)² = 4−4i+i² = 3−4i
4(3−i) = 12−4i
Discriminant = (3−4i)−(12−4i) = −9 → √(−9) = 3i
4(3−i) = 12−4i
Discriminant = (3−4i)−(12−4i) = −9 → √(−9) = 3i
3
z = [−(2−i) ± 3i]/2 = (−2+i±3i)/2
z = (−2+4i)/2 = −1+2i or z = (−2−2i)/2 = −1−i
z = (−2+4i)/2 = −1+2i or z = (−2−2i)/2 = −1−i
Distance Between Skew Lines
Shortest Distance Between Skew Lines
d = |（a₂−a₁）·（d₁×d₂）| / |d₁×d₂|
where a₁, a₂ are points on the two lines and d₁, d₂ are their direction vectors.
📐 Worked Example 13 — Shortest Distance Between Skew Lines
Find the shortest distance between l₁: r=(1,0,0)+λ(1,1,0) and l₂: r=(0,1,0)+μ(0,1,1).
1
d₁=(1,1,0), d₂=(0,1,1). a₁=(1,0,0), a₂=(0,1,0).
d₁×d₂ = |î ĵ к̂| = î(1−0)−ĵ(1−0)+к̂(1−0) = (1,−1,1)
|1 1 0|
|0 1 1|
d₁×d₂ = |î ĵ к̂| = î(1−0)−ĵ(1−0)+к̂(1−0) = (1,−1,1)
|1 1 0|
|0 1 1|
2
a₂−a₁ = (−1,1,0)
(a₂−a₁)·(d₁×d₂) = (−1)(1)+(1)(−1)+(0)(1) = −2
|d₁×d₂| = √(1+1+1) = √3
(a₂−a₁)·(d₁×d₂) = (−1)(1)+(1)(−1)+(0)(1) = −2
|d₁×d₂| = √(1+1+1) = √3
3
d = |−2|/√3 = 2/√3 = 2√3/3
📝 Exam Practice Questions
Q1 [4 marks] — Given z₁ = 2+3i and z₂ = 1−i, find: (a) |z₁z₂| (b) arg(z₁/z₂) (c) z₁² in Cartesian form.
(a) |z₁z₂| = |z₁||z₂| = √13 × √2 = √26
(b) arg(z₁/z₂) = arg(z₁) − arg(z₂)
arg(z₁) = arctan(3/2) ≈ 0.9828, arg(z₂) = arctan(−1) = −π/4
arg(z₁/z₂) ≈ 0.9828+π/4 ≈ 1.768 rad
Or: z₁/z₂ = (2+3i)(1+i)/2 = (2+2i+3i−3)/2 = (−1+5i)/2. arg = arctan(−5) in Q2 ≈ π−arctan(5) ≈ 1.768 ✓
(c) z₁² = (2+3i)² = 4+12i−9 = −5+12i
(b) arg(z₁/z₂) = arg(z₁) − arg(z₂)
arg(z₁) = arctan(3/2) ≈ 0.9828, arg(z₂) = arctan(−1) = −π/4
arg(z₁/z₂) ≈ 0.9828+π/4 ≈ 1.768 rad
Or: z₁/z₂ = (2+3i)(1+i)/2 = (2+2i+3i−3)/2 = (−1+5i)/2. arg = arctan(−5) in Q2 ≈ π−arctan(5) ≈ 1.768 ✓
(c) z₁² = (2+3i)² = 4+12i−9 = −5+12i
Q2 [4 marks] — Show that z = 1+2i satisfies z²−2z+5=0. Write down the other root and verify. Hence factorise z²−2z+5 over the reals and over the complex numbers.
z²=(1+2i)²=1+4i−4=−3+4i. z²−2z+5=(−3+4i)−2(1+2i)+5=−3+4i−2−4i+5=0 ✓
Other root: z = 1−2i (conjugate)
Verify: (1−2i)²−2(1−2i)+5=−3−4i−2+4i+5=0 ✓
Over reals: z²−2z+5 (irreducible)
Over complex: (z−1−2i)(z−1+2i)
Other root: z = 1−2i (conjugate)
Verify: (1−2i)²−2(1−2i)+5=−3−4i−2+4i+5=0 ✓
Over reals: z²−2z+5 (irreducible)
Over complex: (z−1−2i)(z−1+2i)
Q3 [4 marks] — Sketch the locus |z−2i| = |z−3| and find its Cartesian equation. Hence find the complex number on this locus with the smallest modulus.
Perpendicular bisector of segment from (0,2) to (3,0).
Midpoint: (3/2, 1). Gradient of segment: (0−2)/(3−0)=−2/3. Perp gradient=3/2.
Equation: y−1=(3/2)(x−3/2) → y=3x/2−9/4+1=3x/2−5/4
6x−4y=5 (or 3x−2y=5/2)
Smallest modulus: foot of perpendicular from origin to line.
Perpendicular from O to 3x−2y=5/2: direction (3,−2). Parametric: (3t,−2t).
3(3t)−2(−2t)=5/2 → 13t=5/2 → t=5/26.
Point: (15/26,−10/26)=(15/26,−5/13). z=15/26−(5/13)i
|z|_min = √((15/26)²+(5/13)²) = 5√(9+4)/26 = 5√13/26
Midpoint: (3/2, 1). Gradient of segment: (0−2)/(3−0)=−2/3. Perp gradient=3/2.
Equation: y−1=(3/2)(x−3/2) → y=3x/2−9/4+1=3x/2−5/4
6x−4y=5 (or 3x−2y=5/2)
Smallest modulus: foot of perpendicular from origin to line.
Perpendicular from O to 3x−2y=5/2: direction (3,−2). Parametric: (3t,−2t).
3(3t)−2(−2t)=5/2 → 13t=5/2 → t=5/26.
Point: (15/26,−10/26)=(15/26,−5/13). z=15/26−(5/13)i
|z|_min = √((15/26)²+(5/13)²) = 5√(9+4)/26 = 5√13/26
Q4 [4 marks] — Lines l₁ and l₂ have equations r=(2,1,−1)+λ(1,−1,2) and r=(1,3,0)+μ(2,1,−1). (a) Show they intersect. (b) Find the intersection point. (c) Find the angle between them.
(a)&(b) 2+λ=1+2μ ...(i), 1−λ=3+μ ...(ii), −1+2λ=−μ ...(iii)
From (ii): λ=−2−μ. Sub into (i): 2+(−2−μ)=1+2μ → −μ=1+2μ → −3μ=1 → μ=−1/3
λ=−2+1/3=−5/3. Check (iii): −1+2(−5/3)=−1/3=−(−1/3)=1/3. Wait: −μ=1/3. ✓
Point: r=(2−5/3, 1+5/3, −1−10/3)=(1/3, 8/3, −13/3)
Intersection at (1/3, 8/3, −13/3)
(c) d₁=(1,−1,2), d₂=(2,1,−1). cosθ=|d₁·d₂|/(|d₁||d₂|)=|2−1−2|/(√6·√6)=1/6
θ = arccos(1/6) ≈ 80.4°
From (ii): λ=−2−μ. Sub into (i): 2+(−2−μ)=1+2μ → −μ=1+2μ → −3μ=1 → μ=−1/3
λ=−2+1/3=−5/3. Check (iii): −1+2(−5/3)=−1/3=−(−1/3)=1/3. Wait: −μ=1/3. ✓
Point: r=(2−5/3, 1+5/3, −1−10/3)=(1/3, 8/3, −13/3)
Intersection at (1/3, 8/3, −13/3)
(c) d₁=(1,−1,2), d₂=(2,1,−1). cosθ=|d₁·d₂|/(|d₁||d₂|)=|2−1−2|/(√6·√6)=1/6
θ = arccos(1/6) ≈ 80.4°
Exam Tip: When checking line intersection, solve any two equations for λ and μ, then verify in the third equation. If all three are consistent, the lines intersect; if not, they are skew.
Q5 [4 marks] — Find the equation of the plane through (2,1,0) perpendicular to both planes x+2y−z=3 and 2x−y+z=1.
Normal to required plane is perpendicular to normals of both given planes, so it equals n₁×n₂ where n₁=(1,2,−1), n₂=(2,−1,1).
n = (1,2,−1)×(2,−1,1) = î(2×1−(−1)(−1))−ĵ(1×1−(−1)×2)+к̂(1×(−1)−2×2)
= î(2−1)−ĵ(1+2)+к̂(−1−4) = (1,−3,−5)
d = (2,1,0)·(1,−3,−5) = 2−3+0=−1
x−3y−5z=−1
n = (1,2,−1)×(2,−1,1) = î(2×1−(−1)(−1))−ĵ(1×1−(−1)×2)+к̂(1×(−1)−2×2)
= î(2−1)−ĵ(1+2)+к̂(−1−4) = (1,−3,−5)
d = (2,1,0)·(1,−3,−5) = 2−3+0=−1
x−3y−5z=−1
Q6 [3 marks] — Use De Moivre's theorem to show that cos3θ = 4cos³θ − 3cosθ.
(cosθ+i sinθ)³ = cos3θ+i sin3θ (De Moivre)
Expand LHS: cos³θ+3cos²θ(i sinθ)+3cosθ(i sinθ)²+(i sinθ)³
= cos³θ+3i cos²θ sinθ−3cosθ sin²θ−i sin³θ
Real part: cos3θ = cos³θ−3cosθ sin²θ = cos³θ−3cosθ(1−cos²θ)
= 4cos³θ−3cosθ ✓
Expand LHS: cos³θ+3cos²θ(i sinθ)+3cosθ(i sinθ)²+(i sinθ)³
= cos³θ+3i cos²θ sinθ−3cosθ sin²θ−i sin³θ
Real part: cos3θ = cos³θ−3cosθ sin²θ = cos³θ−3cosθ(1−cos²θ)
= 4cos³θ−3cosθ ✓