Further Complex Numbers Review — Grade 12 Specialist Maths — Practice Maths

Use De Moivre’s theorem to expand (cosθ + i sinθ)⁴ and hence express cos(4θ) in terms of cosθ only.

Expand using the binomial theorem (powers of i: i⁰=1, i¹=i, i²=−1, i³=−i, i⁴=1):

(c + is)⁴ = c⁴ + 4c³(is) + 6c²(is)² + 4c(is)³ + (is)⁴

= c⁴ + 4ic³s − 6c²s² − 4ics³ + s⁴

Real part: cos(4θ) = cos⁴θ − 6cos²θsin²θ + sin⁴θ

Using sin²θ = 1 − cos²θ:

= cos⁴θ − 6cos²θ(1−cos²θ) + (1−cos²θ)²

= cos⁴θ − 6cos²θ + 6cos⁴θ + 1 − 2cos²θ + cos⁴θ

cos(4θ) = 8cos⁴θ − 8cos²θ + 1

Compute (1 − i)¹⁰ using De Moivre’s theorem. Express your answer in the form a + bi.

|1 − i| = √2, arg(1 − i) = −π/4 (fourth quadrant, since Re > 0, Im < 0).

So 1 − i = √2(cos(−π/4) + i sin(−π/4)).

By De Moivre: (1 − i)¹⁰ = (√2)¹⁰(cos(−10π/4) + i sin(−10π/4))

= 32(cos(−5π/2) + i sin(−5π/2))

cos(−5π/2) = cos(−π/2) = 0 (since −5π/2 = −2π − π/2)

sin(−5π/2) = sin(−π/2) = −1

(1 − i)¹⁰ = 32(0 + i(−1)) = −32i

Find all cube roots of 8i. Express each root in polar form and in exact Cartesian form.

Write 8i in polar form: |8i| = 8, arg(8i) = π/2. So 8i = 8(cos(π/2) + i sin(π/2)).

The three cube roots have modulus 8^1/3 = 2 and arguments (π/2 + 2kπ)/3 for k = 0, 1, 2:

k=0: arg = π/6 ⇒ z₀ = 2(cos(π/6) + i sin(π/6)) = 2(√3/2 + i/2) = √3 + i

k=1: arg = π/6 + 2π/3 = 5π/6 ⇒ z₁ = 2(cos(5π/6) + i sin(5π/6)) = 2(−√3/2 + i/2) = −√3 + i

k=2: arg = π/6 + 4π/3 = 3π/2 ⇒ z₂ = 2(cos(3π/2) + i sin(3π/2)) = 2(0 − i) = −2i

Write z = 2e^i7π/6 in Cartesian form a + bi.

By Euler’s formula: e^i7π/6 = cos(7π/6) + i sin(7π/6).

cos(7π/6) = −√3/2 (third quadrant, reference angle π/6)

sin(7π/6) = −1/2

z = 2(−√3/2 + i(−1/2)) = −√3 − i

Use exponential form to simplify 3e^iπ/3 × 2e^iπ/4, and express the result in Cartesian form.

Using the multiplication rule for exponential form:

3e^iπ/3 × 2e^iπ/4 = 6e^{i(π/3 + π/4)} = 6e^{i(4π/12 + 3π/12)} = 6e^i7π/12

This is the exact exponential form. For Cartesian form:

7π/12 = 105° ⇒ cos(105°) = cos(60°+45°) = cos60cos45 − sin60sin45 = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4

sin(105°) = sin60cos45 + cos60sin45 = (√6 + √2)/4

z = 6[(√2 − √6)/4 + i(√6 + √2)/4] = (3√2 − 3√6)/2 + i(3√6 + 3√2)/2

Solve z⁴ = −16. Find all four roots in exact Cartesian form.

Write −16 = 16(cosπ + i sinπ) in polar form.

The four fourth roots have modulus 16^1/4 = 2 and arguments (π + 2kπ)/4 for k = 0, 1, 2, 3:

k=0: arg = π/4 ⇒ z₀ = 2(cosπ/4 + i sinπ/4) = √2 + i√2

k=1: arg = 3π/4 ⇒ z₁ = 2(cos3π/4 + i sin3π/4) = −√2 + i√2

k=2: arg = 5π/4 ⇒ z₂ = 2(cos5π/4 + i sin5π/4) = −√2 − i√2

k=3: arg = 7π/4 ⇒ z₃ = 2(cos7π/4 + i sin7π/4) = √2 − i√2

Describe the locus |z − 2 + i| = |z − 4 − 3i| geometrically and find its Cartesian equation.

Geometric description: The set of points equidistant from (2, −1) and (4, 3) — the perpendicular bisector of the segment joining these two points.

Write z = x + iy:

(x−2)² + (y+1)² = (x−4)² + (y−3)²

x²−4x+4+y²+2y+1 = x²−8x+16+y²−6y+9

−4x+4+2y+1 = −8x+16−6y+9

4x + 8y = 20

x + 2y = 5

Use exponential form to compute (√3 + i)⁶. Express as a real number.

|√3 + i| = √(3+1) = 2, arg(√3+i) = π/6.

Exponential form: √3 + i = 2e^iπ/6.

(√3+i)⁶ = 2⁶e^i(6π/6) = 64e^iπ = 64(cosπ + i sinπ) = 64(−1 + 0i) = −64

The five fifth roots of unity satisfy z⁵ = 1. Find all five roots, sketch their positions on the unit circle, and explain why their sum equals zero.

Write 1 = 1(cos 0 + i sin 0). The five roots have modulus 1 and arguments 2kπ/5 for k = 0, 1, 2, 3, 4:

z_k = cos(2kπ/5) + i sin(2kπ/5)

z₀=1, z₁=cos(2π/5)+i sin(2π/5), z₂=cos(4π/5)+i sin(4π/5), z₃=cos(6π/5)+i sin(6π/5), z₄=cos(8π/5)+i sin(8π/5).

These are equally spaced at 72° intervals on the unit circle, forming a regular pentagon.

Sum equals zero: The roots are the roots of z⁵−1=0, i.e. (z−1)(z⁴+z³+z²+z+1)=0. By Vieta’s formulas, the sum of all five roots equals the negative of the coefficient of z⁴ divided by the leading coefficient in z⁵−1, which is 0. Geometrically, the five equally spaced unit vectors cancel.

Sketch the region {z : 0 < arg(z − 1) ≤ π/2 and |z − 1| ≤ 3}.

Condition 1: 0 < arg(z − 1) ≤ π/2 — the argument of (z − 1) is strictly between 0 and π/2 (inclusive of π/2). This is a wedge/sector from point (1, 0), in the first quadrant direction.

Condition 2: |z − 1| ≤ 3 — closed disk centred at 1, radius 3.

Intersection: A quarter-disk (sector) centred at (1, 0) with radius 3, bounded by rays at angle 0 (excluded) and π/2 (included), and the arc of the circle at distance 3.

Boundary: the arc from (4, 0) to (1, 3) [quarter circle, included], the ray along the positive real direction from (1,0) [excluded, open endpoint], and the ray going straight up from (1,0) [included].

Use Euler’s formula to show that cos(2θ) = (e^2iθ + e^−2iθ)/2, and hence evaluate ∫₀^π/4 cos(2θ) dθ.

By Euler’s formula: e^2iθ = cos(2θ) + i sin(2θ) and e^−2iθ = cos(2θ) − i sin(2θ).

Adding: e^2iθ + e^−2iθ = 2cos(2θ), so cos(2θ) = (e^2iθ + e^−2iθ)/2. ✓

∫₀^π/4 cos(2θ) dθ = [sin(2θ)/2]₀^π/4 = sin(π/2)/2 − 0 = 1/2.

Answer: 1/2

Use De Moivre’s power reduction to express cos³θ in terms of cosθ and cos(3θ), then evaluate ∫₀^π/3 cos³θ dθ.

Let z = e^iθ. Then (z + z⁻¹)³ = (2cosθ)³ = 8cos³θ.

Expand: z³ + 3z + 3z⁻¹ + z⁻³ = (z³+z⁻³) + 3(z+z⁻¹) = 2cos(3θ) + 6cosθ.

So 8cos³θ = 2cos(3θ) + 6cosθ, giving cos³θ = ¼cos(3θ) + ¾cosθ.

∫₀^π/3 cos³θ dθ = ∫₀^π/3 [¼cos(3θ) + ¾cosθ] dθ

= [sin(3θ)/12 + ¾sinθ]₀^π/3

At θ=π/3: sin(π)/12 + ¾sin(π/3) = 0 + ¾ × √3/2 = 3√3/8

At θ=0: 0. Answer: 3√3/8

The locus of z satisfies |z − 2i|/|z − 2| = 1. Identify the locus and find its Cartesian equation.

|z − 2i| = |z − 2| means z is equidistant from 2i (the point (0, 2)) and 2 (the point (2, 0)).

Geometric description: Perpendicular bisector of the segment from (2, 0) to (0, 2).

Write z = x + iy:

x² + (y−2)² = (x−2)² + y²

x² + y² − 4y + 4 = x² − 4x + 4 + y²

−4y = −4x ⇒ y = x

The locus is the line y = x (the full line, not just a ray), passing through the midpoint (1, 1) of the segment from (2, 0) to (0, 2).

The polynomial z⁶ − 64 = 0 has six roots. Find all six roots in exponential form, and verify that they form a regular hexagon on the Argand diagram.

Write 64 = 64e^i·0 (since arg(64) = 0). The six sixth roots have modulus 64^1/6 = 2 and arguments 2kπ/6 = kπ/3 for k = 0, 1, 2, 3, 4, 5:

z₀ = 2e^i·0 = 2 z₁ = 2e^iπ/3 z₂ = 2e^i2π/3 z₃ = 2e^iπ = −2 z₄ = 2e^i4π/3 z₅ = 2e^i5π/3

Cartesian values: 2, 1+i√3, −1+i√3, −2, −1−i√3, 1−i√3.

Regular hexagon: All six roots lie on the circle of radius 2, equally spaced at 60° intervals. The distance between adjacent roots is constant (= 2, by the chord formula 2R sin(π/6) = 2×2×1/2 = 2), confirming they form a regular hexagon.

Prove Euler’s identity e^iπ + 1 = 0 using Euler’s formula, and explain why all five fundamental constants (e, i, π, 1, 0) appear.

Proof: Euler’s formula states e^iθ = cosθ + i sinθ for any real θ.

Substitute θ = π:

e^iπ = cosπ + i sinπ = −1 + i(0) = −1

Therefore e^iπ + 1 = −1 + 1 = 0. ✓

The five constants:

e — the base of natural logarithms (analysis/calculus)
i — the imaginary unit (√−1), from complex number theory
π — the ratio of a circle’s circumference to its diameter (geometry/trigonometry)
1 — the multiplicative identity (arithmetic)
0 — the additive identity (arithmetic)

The identity is remarkable because it connects these five fundamental constants from different areas of mathematics in a single equation.

Further Complex Numbers — Topic Review — Solutions

Use De Moivre’s theorem to expand (cosθ + i sinθ)4 and hence express cos(4θ) in terms of cosθ only.

Compute (1 − i)10 using De Moivre’s theorem. Express your answer in the form a + bi.

Find all cube roots of 8i. Express each root in polar form and in exact Cartesian form.

Write z = 2ei7π/6 in Cartesian form a + bi.

Use exponential form to simplify 3eiπ/3 × 2eiπ/4, and express the result in Cartesian form.

Solve z4 = −16. Find all four roots in exact Cartesian form.