nth roots via factorization

Questions that require factorizing a polynomial equation (often z^n - a)(z^m - b) = 0 or similar forms) before finding roots using De Moivre's theorem, rather than directly solving a single equation z^n = w.

3 questions · Standard +0.9

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CAIE Further Paper 2 2021 June Q1
7 marks Standard +0.8
1
  1. Find \(a\) and \(b\) such that $$z ^ { 8 } - i z ^ { 5 } - z ^ { 3 } + i = \left( z ^ { 5 } - a \right) \left( z ^ { 3 } - b \right) .$$
  2. Hence find the roots of $$z ^ { 8 } - i z ^ { 5 } - z ^ { 3 } + i = 0$$ giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(0 \leqslant \theta < 2 \pi\).
AQA FP2 2009 January Q8
11 marks Standard +0.8
8
  1. Show that $$\left( z ^ { 4 } - \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z ^ { 4 } - \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 8 } - 2 z ^ { 4 } \cos \theta + 1$$ (2 marks)
  2. Hence solve the equation $$z ^ { 8 } - z ^ { 4 } + 1 = 0$$ giving your answers in the form \(\mathrm { e } ^ { \mathrm { i } \phi }\), where \(- \pi < \phi \leqslant \pi\).
  3. Indicate the roots on an Argand diagram.
AQA FP2 2007 June Q8
13 marks Challenging +1.2
8
    1. Given that \(z ^ { 6 } - 4 z ^ { 3 } + 8 = 0\), show that \(z ^ { 3 } = 2 \pm 2 \mathrm { i }\).
    2. Hence solve the equation $$z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$$ giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(- \pi < \theta \leqslant \pi\).
  2. Show that, for any real values of \(k\) and \(\theta\), $$\left( z - k \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z - k \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 2 } - 2 k z \cos \theta + k ^ { 2 }$$
  3. Express \(z ^ { 6 } - 4 z ^ { 3 } + 8\) as the product of three quadratic factors with real coefficients.