2022 YIJC P1 Q1

2022 YIJC P1 Q1

6 marks
Prelims

Do not use a calculator in answering this question.

It is given that \(2-\mathbf{i}\) is a root of the equation \(2{{z}^{3}}+a{{z}^{2}}-2z+b=0\).

  1. Find the values of the real numbers \(a\) and \(b\) and the remaining roots of the equation.[4]
  2. Using these values of \(a\) and \(b\), deduce the roots of the equation \(b{{z}^{3}}-2{{z}^{2}}+az+2=0\).[2]

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Answer:(a) \(a=-5\), \(b=15\), remaining roots \(2+\mathrm{i}\) and \(-\dfrac32\) (b) \(z=-\dfrac23\), \(z=\dfrac25\pm\dfrac15\mathrm{i}\)

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