Practice Maths

Unit 2 Review — Specialist Mathematics

← Back to Specialist Mathematics

This review spans all three topics in Unit 2 of Specialist Mathematics: Complex Numbers (Cartesian form, operations, Argand diagram and modulus, polar form, De Moivre’s theorem, and polynomial roots), Trigonometry and Functions (Pythagorean and reciprocal identities, compound and double angle formulas, trigonometric equations, and inverse trig functions), and Matrices and Systems (matrix operations, determinants, inverses, geometric transformations, and solving systems). Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.

Mixed Review Questions

  1. Fluency

    Q1 — Complex Number Operations

    Let z = 5 − 2i and w = 1 + 3i. Find: (a) z + w    (b) zw    (c) z/w in the form a + bi.

    (a) z + w = (5+1) + (−2+3)i = 6 + i

    (b) zw = (5−2i)(1+3i) = 5 + 15i − 2i − 6i² = 5 + 13i + 6 = 11 + 13i

    (c) z/w = (5−2i)(1−3i) / ((1+3i)(1−3i)) = (5 − 15i − 2i + 6i²) / (1+9)

    = (5 − 17i − 6) / 10 = (−1 − 17i)/10 = −1/10 − (17/10)i

  2. Fluency

    Q2 — Modulus and Argument

    For z = −√3 + i, find: (a) |z|    (b) arg(z) in exact radians    (c) Write z in polar form r cisθ.

    (a) |z| = √((−√3)² + 1²) = √(3+1) = 2

    (b) z is in quadrant II (x < 0, y > 0). Reference angle = arctan(1/√3) = π/6.

    arg(z) = π − π/6 = 5π/6

    (c) z = 2 cis(5π/6)

  3. Understanding

    Q3 — De Moivre’s Theorem

    Use De Moivre’s theorem to find (√3 + i)6. Express in Cartesian form.

    Convert to polar: |√3 + i| = √(3+1) = 2,   arg = arctan(1/√3) = π/6.

    √3 + i = 2 cis(π/6)

    Apply De Moivre’s: (√3 + i)6 = 26 cis(6 × π/6) = 64 cis(π) = 64(−1 + 0i) = −64

  4. Understanding

    Q4 — Polynomial Equations with Complex Roots

    Given that 3 + i is a root of P(x) = x³ − 7x² + 16x − 10, and P(x) has real coefficients, find all roots of P(x) and express P(x) as a product of linear and quadratic factors over ℝ.

    By the conjugate root theorem, 3 − i is also a root.

    Quadratic factor from the complex pair:

    (x − (3+i))(x − (3−i)) = x² − 6x + (9+1) = x² − 6x + 10

    Divide P(x) by x² − 6x + 10:

    x³ − 7x² + 16x − 10 = (x² − 6x + 10)(x − 1)

    Check: (x² − 6x + 10)(x − 1) = x³ − x² − 6x² + 6x + 10x − 10 = x³ − 7x² + 16x − 10 ✓

    Roots: x = 1, x = 3 + i, x = 3 − i

    P(x) = (x − 1)(x² − 6x + 10) over ℝ.

  5. Problem Solving

    Q5 — nth Roots of a Complex Number

    Find all fourth roots of −16 (i.e., solve z4 = −16). Express in exact polar and Cartesian form.

    Write −16 = 16 cis(π). The fourth roots are: z = 161/4 cis((π + 2kπ)/4) for k = 0, 1, 2, 3.

    161/4 = 2.

    k = 0: 2 cis(π/4) = 2(1/√2 + i/√2) = √2 + √2 i

    k = 1: 2 cis(3π/4) = 2(−1/√2 + i/√2) = −√2 + √2 i

    k = 2: 2 cis(5π/4) = 2(−1/√2 − i/√2) = −√2 − √2 i

    k = 3: 2 cis(7π/4) = 2(1/√2 − i/√2) = √2 − √2 i

    The four roots are equally spaced at 90° on a circle of radius 2, and come in two conjugate pairs.

  6. Fluency

    Q6 — Pythagorean and Reciprocal Identities

    Simplify: (a) sin²θ + cos²θ    (b) sec²θ − 1    (c) (1 − cos²θ)/sinθ (assuming sinθ ≠ 0).

    (a) sin²θ + cos²θ = 1 (Pythagorean identity)

    (b) sec²θ − 1 = tan²θ (from 1 + tan²θ = sec²θ)

    (c) (1 − cos²θ)/sinθ = sin²θ/sinθ = sinθ

  7. Fluency

    Q7 — Compound Angle Formulas

    Use compound angle formulas to find the exact value of: (a) sin(75°)    (b) cos(15°).

    (a) sin(75°) = sin(45° + 30°)

    = sin45°cos30° + cos45°sin30°

    = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2) = (√6 + √2)/4

    (b) cos(15°) = cos(45° − 30°)

    = cos45°cos30° + sin45°sin30°

    = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2) = (√6 + √2)/4

    Note: sin(75°) = cos(15°) since 75° + 15° = 90°. ✓

  8. Understanding

    Q8 — Double Angle Formula Application

    Given cosθ = −3/5 where π/2 < θ < π, find: (a) sin(2θ)    (b) cos(2θ).

    Since θ is in Q2: sinθ = +√(1 − 9/25) = √(16/25) = 4/5.

    (a) sin(2θ) = 2 sinθ cosθ = 2(4/5)(−3/5) = −24/25

    (b) cos(2θ) = cos²θ − sin²θ = 9/25 − 16/25 = −7/25

    Alternatively: cos(2θ) = 2cos²θ − 1 = 2(9/25) − 1 = 18/25 − 25/25 = −7/25 ✓

  9. Understanding

    Q9 — Solving a Trigonometric Equation

    Solve 2sin²θ − sinθ − 1 = 0 for θ ∈ [0, 2π].

    Let u = sinθ. The equation becomes 2u² − u − 1 = 0.

    Factor: (2u + 1)(u − 1) = 0

    u = −1/2   or   u = 1

    sinθ = 1: θ = π/2

    sinθ = −1/2: Reference angle = π/6. Solutions in Q3 and Q4:

    θ = π + π/6 = 7π/6   and   θ = 2π − π/6 = 11π/6

    Solutions: θ = π/2, 7π/6, 11π/6

  10. Understanding

    Q10 — Inverse Trigonometric Functions

    Find the exact value of: (a) arcsin(1/2)    (b) arccos(−√3/2)    (c) arctan(−1). State which quadrant each principal value lies in.

    (a) arcsin(1/2): The principal value range for arcsin is [−π/2, π/2] (Q4 to Q1).

    sin(π/6) = 1/2, so arcsin(1/2) = π/6 (Q1).

    (b) arccos(−√3/2): Range is [0, π] (Q1 to Q2).

    cos(5π/6) = −√3/2, so arccos(−√3/2) = 5π/6 (Q2).

    (c) arctan(−1): Range is (−π/2, π/2) (Q4 to Q1).

    tan(−π/4) = −1, so arctan(−1) = −π/4 (Q4).

  11. Problem Solving

    Q11 — Proving a Trigonometric Identity

    Prove that cos(3θ) = 4cos³θ − 3cosθ using only the compound and double angle formulas.

    cos(3θ) = cos(2θ + θ)

    = cos(2θ)cosθ − sin(2θ)sinθ

    = (2cos²θ − 1)cosθ − (2sinθcosθ)sinθ

    = 2cos³θ − cosθ − 2sin²θcosθ

    = 2cos³θ − cosθ − 2(1 − cos²θ)cosθ

    = 2cos³θ − cosθ − 2cosθ + 2cos³θ

    = 4cos³θ − 3cosθ ✓

  12. Fluency

    Q12 — Matrix Operations

    Find AB where A = [[2, 1], [−3, 0]] and B = [[1, −2], [4, 3]]. Also find det(A).

    AB:

    Row 1: [2(1)+1(4),   2(−2)+1(3)] = [6, −1]

    Row 2: [−3(1)+0(4),   −3(−2)+0(3)] = [−3, 6]

    AB = [[6, −1], [−3, 6]]

    det(A) = 2(0) − 1(−3) = 0 + 3 = 3

  13. Understanding

    Q13 — Inverse Matrix and Solving a System

    Solve the system using the inverse matrix method:
    3x + y = 11
    2x − 3y = −1

    A = [[3, 1], [2, −3]], B = [[11], [−1]].

    det(A) = 3(−3) − 1(2) = −9 − 2 = −11

    A−1 = (1/−11) [[−3, −1], [−2, 3]] = [[3/11, 1/11], [2/11, −3/11]]

    X = A−1B = [[3/11, 1/11], [2/11, −3/11]] × [[11], [−1]]

    x = (3/11)(11) + (1/11)(−1) = 3 − 1/11 = 32/11...

    Re-check: x = (3 × 11 + 1 × (−1))/11 = (33−1)/11 = 32/11. Let me verify with substitution instead:

    Multiply eq1 by 3: 9x + 3y = 33. Add to eq2: 11x = 32, x = 32/11.

    y = 11 − 3(32/11) = (121 − 96)/11 = 25/11.

    Verify: 3(32/11) + 25/11 = 96/11 + 25/11 = 121/11 = 11 ✓

    x = 32/11, y = 25/11

  14. Understanding

    Q14 — Geometric Transformation

    A reflection in the line y = x maps (x, y) to (y, x). (a) State the transformation matrix. (b) Apply it to the line y = 3x − 2 and state the equation of the image line.

    (a) Reflection in y = x: M = [[0, 1], [1, 0]].

    Check: M × [[x],[y]] = [[y],[x]] — swaps coordinates. ✓

    (b) If (x, y) maps to (x′, y′) = (y, x), then x = y′ and y = x′.

    Substitute into y = 3x − 2:

    x′ = 3y′ − 2

    Rearranging: y′ = (x′ + 2)/3 = (1/3)x′ + 2/3

    Image line: y = (1/3)x + 2/3 (the inverse function, as expected from reflection in y = x).

  15. Problem Solving

    Q15 — Trigonometric Equation with Identity

    Solve sin(2θ) + sinθ = 0 for θ ∈ [0, 2π].

    Apply the double angle formula: sin(2θ) = 2sinθcosθ.

    2sinθcosθ + sinθ = 0

    sinθ(2cosθ + 1) = 0

    sinθ = 0: θ = 0, π, 2π

    cosθ = −1/2: Reference angle = π/3. Solutions in Q2 and Q3:

    θ = π − π/3 = 2π/3   and   θ = π + π/3 = 4π/3

    Solutions: θ = 0, 2π/3, π, 4π/3, 2π

  16. Problem Solving

    Q16 — Systems with Parameter

    For the system kx + 2y = 4 and 3x + 6y = 12, find all values of k that give: (a) a unique solution    (b) infinitely many solutions    (c) no solution.

    Write the augmented matrix: [[k, 2, | 4], [3, 6, | 12]].

    Simplify row 2 by ÷3: [[k, 2, | 4], [1, 2, | 4]].

    Both equations are equivalent to “x + 2y = 4/k” and “x + 2y = 4” (if k ≠ 0).

    Note row 2 of the original is 3x + 6y = 12 → x + 2y = 4. The coefficient matrix has det = 6k − 6.

    (a) Unique solution: det ≠ 0 → 6k − 6 ≠ 0 → k ≠ 1.

    (b) Infinitely many solutions: det = 0 (k = 1) AND consistent.

    When k = 1: Row 1 becomes x + 2y = 4, same as Row 2. Both rows are the same line. → k = 1.

    (c) No solution: det = 0 AND inconsistent. Since when k = 1 the system is consistent, there is no value of k that gives no solution in this particular system.

    Note: If the RHS of Row 1 were different from 4, there would be a no-solution case. Here both rows reduce to the same equation when k = 1.

  17. Problem Solving

    Q17 — Argand Diagram Locus

    Find the Cartesian equation of the locus |z − 2| = |z + 2i| and describe the locus geometrically.

    Let z = x + iy.

    |z − 2| = |(x−2) + iy| = √((x−2)² + y²)

    |z + 2i| = |x + i(y+2)| = √(x² + (y+2)²)

    Set equal and square both sides:

    (x−2)² + y² = x² + (y+2)²

    x² − 4x + 4 + y² = x² + y² + 4y + 4

    −4x = 4y

    y = −x

    Geometrically: the locus is the straight line y = −x, which is the perpendicular bisector of the segment joining 2+0i and 0−2i (i.e., the points (2,0) and (0,−2) on the Argand diagram).

  18. Problem Solving

    Q18 — Composite Trigonometric Equation

    Solve cos(2θ) = 1 + cosθ for θ ∈ [0, 2π]. (Hint: use the identity cos(2θ) = 2cos²θ − 1.)

    Substitute cos(2θ) = 2cos²θ − 1:

    2cos²θ − 1 = 1 + cosθ

    2cos²θ − cosθ − 2 = 0

    Use the quadratic formula with u = cosθ:

    u = (1 ± √(1 + 16))/4 = (1 ± √17)/4

    u = (1 + √17)/4 ≈ 1.28   (rejected, since |cosθ| ≤ 1)

    u = (1 − √17)/4 ≈ −0.780

    cosθ ≈ −0.780  →  θ = arccos(−0.780) ≈ 141.2° or 218.8°

    θ ≈ 2.465 rad or 3.818 rad (to 3 decimal places)

  19. Understanding

    Q19 — Transforming a Unit Square

    Apply the matrix T = [[0, −1], [1, 0]] to the unit square with vertices (0,0), (1,0), (1,1), (0,1). Describe the transformation and state the area of the image.

    T = [[0,−1],[1,0]] is the 90° anticlockwise rotation matrix.

    Apply to each vertex:

    (0,0) → (0,0)

    (1,0) → [[0,−1],[1,0]][[1],[0]] = [[0],[1]] = (0,1)

    (1,1) → [[0,−1],[1,0]][[1],[1]] = [[−1],[1]] = (−1,1)

    (0,1) → [[0,−1],[1,0]][[0],[1]] = [[−1],[0]] = (−1,0)

    The image square has vertices (0,0), (0,1), (−1,1), (−1,0). This is a rotation of 90° anticlockwise.

    Area of image = |det(T)| × area of original = |0(0) − (−1)(1)| × 1 = 1 × 1 = 1 square unit (unchanged, as expected for a rotation).

  20. Problem Solving

    Q20 — Complex Number and Trigonometry Connection

    Using the expansion of (cosθ + i sinθ)4 via De Moivre’s theorem and the binomial theorem, derive an expression for cos(4θ) in terms of cosθ only.

    By De Moivre’s: (cosθ + i sinθ)4 = cos(4θ) + i sin(4θ)   … (1)

    By the binomial theorem (let c = cosθ, s = sinθ):

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

    = c4 + 4ic³s − 6c²s² − 4ics³ + s4

    = (c4 − 6c²s² + s4) + i(4c³s − 4cs³)   … (2)

    Equating real parts from (1) and (2):

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

    Replace sin²θ = 1 − cos²θ:

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

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

    cos(4θ) = 8cos4θ − 8cos²θ + 1