Moderate -0.8 This is a straightforward 1-mark multiple choice question testing the basic property that complex roots of real polynomials come in conjugate pairs. Students simply need to recall that if 1-3i is a root, then 1+3i is also a root, and r equals their product: (1-3i)(1+3i) = 1+9 = 10. No problem-solving or extended reasoning required, just direct application of a standard result.
Given that \(z = 1 - 3\mathrm{i}\) is one root of the equation \(z^2 + pz + r = 0\), where \(p\) and \(r\) are real, find the value of \(r\).
Circle your answer.
[1 mark]
\(-8\) \quad \(-2\) \quad \(6\) \quad \(10\)
Given that $z = 1 - 3\mathrm{i}$ is one root of the equation $z^2 + pz + r = 0$, where $p$ and $r$ are real, find the value of $r$.
Circle your answer.
[1 mark]
$-8$ \quad $-2$ \quad $6$ \quad $10$
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q2 [1]}}