About Complex numbers in A Level Mathematics
Complex numbers is one of the 8 main topics in our A Level Mathematics (Cambridge (CAIE) 9709) question bank, and about 4% of its real questions come from it. Tayyari Karo has 17 real questions on this topic from 2021 to 2026, from Paper 3. By difficulty: 9 hard, 8 medium.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 7 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2026 · Paper 3 · Complex numbers
The square roots of −71 − (20√3)i can be expressed in the Cartesian form x + iy, where x and y are real and exact.
By first forming a quartic equation, find the square roots of −71 − (20√3)i. Give your answer in exact Cartesian form. [5]
Show model answer
(x + iy)² = −71 − 20√3 i gives x² − y² = −71 and 2xy = −20√3.
Eliminating y: y = −10√3/x, so x² − 300/x² = −71, i.e. x⁴ + 71x² − 300 = 0, which factorises as (x² + 75)(x² − 4) = 0. Hence x² = 4 (x² = −75 is impossible).
x = ±2, y = ∓5√3. The square roots are ±(2 − 5√3 i).
Square x + iy, equate real and imaginary parts, eliminate one variable to get a quartic in x (a quadratic in x²) and solve.
May/June 2025 · Paper 3 · Complex numbers
Find the complex numbers z for which (z + 5i)/(z − 5) is real and |z| = √17. Give your answers in the form z = x + iy, where x and y are real. [6]
Show model answer
Let z = x + iy. Multiplying numerator and denominator of (z+5i)/(z−5) by the conjugate (x−5)−iy, the numerator becomes x(x−5) + y(y+5) + i[(y+5)(x−5) − xy]. For the expression to be real, the imaginary part must be zero: (y+5)(x−5) − xy = 0, which simplifies to x − y = 5.
Using |z|² = 17: x² + y² = 17. Substituting y = x − 5: x² + (x−5)² = 17, so 2x² − 10x + 8 = 0, i.e. x² − 5x + 4 = 0, giving x = 1 or x = 4.
For x = 4, y = −1, so z = 4 − i. For x = 1, y = −4, so z = 1 − 4i.
Multiply by the conjugate to isolate the real condition, giving a linear relation between x and y, then combine with the modulus condition.
October/November 2025 · Paper 3 · Complex numbers
Solve the quadratic equation (2 + i)w² + 4w + 2 − i = 0. Give your answers in the form x + iy, where x and y are real. [5]
Show model answer
Using the quadratic formula: w = [−4 ± √(16 − 4(2+i)(2−i))] / [2(2+i)]. Since (2+i)(2−i)=5: w = [−4 ± √(16−20)]/(2(2+i)) = [−4 ± 2i]/(2(2+i)) = (−2±i)/(2+i).
Multiplying numerator and denominator by the conjugate (2−i):
w₁ = (−2+i)(2−i)/5 = (−3+4i)/5 = −3/5 + 4i/5.
w₂ = (−2−i)(2−i)/5 = −5/5 = −1.
Apply the quadratic formula treating w as the unknown, simplify using i²=−1, then rationalise each complex fraction by multiplying by the conjugate of the denominator.
May/June 2024 · Paper 3 · Complex numbers
The complex number u is given by u = −1 − i√3.
(a) Express u in the form r(cos θ + i sin θ), where r ≥ 0 and −π < θ ≤ π. Give the exact values of r and θ. [2]
The complex number v is given by v = 5(cos(⅙π) + i sin(⅙π)).
(b) Express the complex number v/u in the form re^(iθ) where r ≥ 0 and −π < θ ≤ π. [2]
Show model answer
(a) |u| = √((−1)² + (√3)²) = √4 = 2. u lies in the third quadrant with reference angle arctan(√3/1) = π/3, so θ = −(π − π/3) = −2π/3.
u = 2(cos(−2π/3) + i sin(−2π/3)), so r = 2, θ = −2π/3.
(b) v/u has modulus |v|/|u| = 5/2 and argument arg v − arg u = π/6 − (−2π/3) = π/6 + 4π/6 = 5π/6.
So v/u = (5/2)e^(i·5π/6).
Find the modulus and argument of u directly using its real/imaginary parts; for v/u divide the moduli and subtract the arguments.
May/June 2023 · Paper 3 · Complex numbers
The polynomial x³ + 5x² + 31x + 75 is denoted by p(x).
(a) Show that x + 3 is a factor of p(x). [2]
(b) Show that z = −1 + 2√6i is a root of p(z) = 0. [3]
(c) Hence find the complex numbers z which are roots of p(z²) = 0. [7]
Show model answer
(a) p(−3) = (−3)³+5(−3)²+31(−3)+75 = −27+45−93+75 = 0, so x+3 is a factor.
(b) z = −1+2√6i. z² = 1 − 4√6i + (2√6i)² = 1 − 4√6i − 24 = −23 − 4√6i.
z³ = z²·z = (−23−4√6i)(−1+2√6i) = 23 − 46√6i + 4√6i − 8(6)i² = 23+48 − 42√6i = 71 − 42√6i.
p(z) = z³+5z²+31z+75 = (71−42√6i) + 5(−23−4√6i) + 31(−1+2√6i) + 75
Real part: 71−115−31+75 = 0. Imaginary part (× √6): −42−20+62 = 0.
So p(z) = 0, as required.
(c) Since p has real coefficients, the third root is z̄ = −1−2√6i. So the roots of p(w)=0 are w=−3, w=−1+2√6i, w=−1−2√6i.
We need z² equal to each of these.
z² = −3: z = ±i√3.
z² = −1+2√6i: let z=x+iy, x²−y²=−1, 2xy=2√6 → xy=√6. Trying x=√2,y=√3: xy=√6 ✓, x²−y²=2−3=−1 ✓.
So z = √2+i√3 or z = −√2−i√3. …
The full model answer is in the app.
October/November 2023 · Paper 3 · Complex numbers
The complex number u is defined by u = (3 + 2i)/(a − 5i), where a is real.
(a) Express u in the Cartesian form x + iy, where x and y are in terms of a. [3]
(b) Given that arg u = (1/4)π, find the value of a. [2]
Show model answer
(a) Multiply numerator and denominator by (a + 5i): u = (3+2i)(a+5i)/[(a−5i)(a+5i)] = (3a + 15i + 2ai + 10i²)/(a² + 25) = (3a − 10 + (2a+15)i)/(a² + 25).
So x = (3a−10)/(a²+25), y = (2a+15)/(a²+25).
(b) arg u = π/4 means the real and imaginary parts of u are equal and positive, i.e. x = y (with x, y > 0): 3a − 10 = 2a + 15 → a = 25.
Rationalise the denominator by multiplying by its conjugate; for (b), use that arg u = π/4 means Re(u) = Im(u) > 0.
October/November 2022 · Paper 3 · Complex numbers
The complex numbers u and w are defined by u = 2e^(¼πi) and w = 3e^(⅓πi).
(a) Find u²/w, giving your answer in the form re^(iθ), where r > 0 and -π < θ ≤ π. Give the exact values of r and θ. [3]
(b) State the least positive integer n such that both Im(wⁿ) = 0 and Re(wⁿ) > 0. [1]
Show model answer
(a) u² = 4e^(iπ/2). So u²/w = 4e^(iπ/2)/(3e^(iπ/3)) = (4/3)e^(i(π/2 - π/3)) = (4/3)e^(iπ/6). So r = 4/3, θ = π/6.
(b) wⁿ has argument nπ/3. For Im(wⁿ) = 0 and Re(wⁿ) > 0, the argument must be a multiple of 2π. The least positive n with nπ/3 = 2πk is n = 6.
Use the laws of indices for the exponential (polar) form to combine u² and w, then find when wⁿ's argument is a multiple of 2π.
