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
It is given that $3 + 2i$ is a root of the equation $z^3 - 8z^2 + 25z - 26 = 0$ , whose coefficients are all real.
(a) Write down a second root of the equation. (1 mark)
(b) Hence find the real root of the equation and factorise $z^3 - 8z^2 + 25z - 26$ completely. (4 marks)
Level 4 · 5 marksTry it
2
Find the two square roots of $3 + 4i$ , giving each in the form $a + bi$ where $a$ and $b$ are real. (5 marks)
Level 4 · 5 marksTry it
3
It is given that $1 + 2i$ is a root of the equation

$z^{3} - 3z^{2} + 7z - 5 = 0$
(a) Explain why $1 - 2i$ must also be a root. (1 mark)
(b) Find all three roots of the equation. (4 marks)
Level 4 · 5 marksTry it
4
Find the two square roots of $-8 + 6i$ , giving your answers in the form $a + bi$ where $a$ and $b$ are real. (5 marks)
Level 4 · 5 marksTry it
5
(a) Factorise $z^{4} + 3z^{2} - 28$ into two quadratic factors with real coefficients. (2 marks)
(b) Hence solve the equation $z^{4} + 3z^{2} - 28 = 0$ . (3 marks)
Level 4 · 5 marksTry it