A-level · Complex numbers

Complex roots of polynomial equationsExam-style questions

Solving quadratics with a negative discriminant to get a conjugate pair of complex roots; the theorem that complex roots of a real-coefficient polynomial come in conjugate pairs; using one known complex root to find the others and factorise cubics and quartics; and finding the square roots of a + bi by equating real and imaginary parts.

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15 questions · A Level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 1 Level 2 Level 3 Level 4 Level 5

1
Solve the equation $z^2 - 4z + 13 = 0$ , giving your answers in the form $a + bi$ where $a$ and $b$ are real. (3 marks)
Level 1 · 3 marksTry it
2
Solve the equation $z^2 + 2z + 5 = 0$ , giving your answers in the form $a + bi$ . (3 marks)
Level 2 · 3 marksTry it
3
Solve the equation $z^{2} - 6z + 25 = 0$ , giving your answers in the form $a + bi$ . (3 marks)
Level 2 · 3 marksTry it
4
The complex number $1 + 2i$ is a root of a polynomial equation that has real coefficients.
(a) Write down another root of the equation. (1 mark)
(b) Hence find a quadratic factor of the polynomial that has real coefficients. (2 marks)
Level 3 · 3 marksTry it
5
(a) Solve the equation $z^{2} + 4z + 29 = 0$ , giving your answers in the form $a + bi$ . (3 marks)
(b) Verify that the product of your roots equals the constant term. (1 mark)
Level 3 · 4 marksTry it

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