2022HSCShort answer

Q13 (c) (ii)

2 marks

Question

Show that if $z$ is a solution of $z^5+1=0$ and $z\ne-1$ , then $u=z+\frac1z$ is a solution of $u^2-u-1=0$ .

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Solution

Criteria	Marks
Provides correct solution	2
Uses the given substitution and attempts to simplify, or equivalent merit	1

Sample answer:
Let $z$ be a solution to $z^5+1=0$ with $z\ne-1$ .
$u=z+\frac1z$

\begin{aligned} u^2-u-1&=\left(z+\frac1z\right)^2-\left(z+\frac1z\right)-1\\ &=z^2+2+\frac1{z^2}-z-\frac1z-1\\ &=z^2-z+1-\frac1z+\frac1{z^2}\\ &=\frac{z^4-z^3+z^2-z+1}{z^2}\\ &=\frac1{z^2}\left(\frac{1-(-z)^5}{1-(-z)}\right)\quad\text{using Geometric series, as }z\ne-1\\ &=\frac{1+z^5}{z^2(1+z)}=0\quad\text{since }z^5+1=0 \end{aligned}

$\therefore\ u=z+\frac1z\text{ is a solution of }u^2-u-1=0$

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Syllabus alignment

ProofComplex Numbers

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.
Solving equations with complex numbersUse De Moivre's theorem and solve equations and polynomial problems involving complex and conjugate roots.

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