Question 18
Consider the polynomial $P(z)=z^{3}+az^{2}+bz+c$, where $a,b,c\in\mathbb{R}$ and $z\in\mathbb{C}$.
Two of the roots of $P(z)$ are also roots of $z^{4}+z^{3}+z^{2}+z+1$. The remaining root of $P(z)$ is $z=2$.
Given $z^{5}-1=(z-1)\left(z^{4}+z^{3}+z^{2}+z+1\right)$, determine a possible expression for $P(z)$.
Leave your answer in expanded form.
Pick a conjugate pair from the roots of unity
Watch the quartic’s four roots appear on the unit circle. You will see why $P(z)$ can only take a matching pair of them.
There are two correct answers, one for each conjugate pair. The question asks for “a possible expression”, so either one gets full marks.
Choose two roots and check the coefficients
Pick two of the quartic’s roots. The page multiplies $(z-2)(z-w_{1})(z-w_{2})$ and shows you $a$, $b$ and $c$. They are only all real for a conjugate pair.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA’s first method with the pair $\operatorname{cis}\left(\pm\frac{2\pi}{5}\right)$.
You are given the factorisation, so the quartic is the other factor of $z^{5}-1$.
$$z^{5}-1=(z-1)\left(z^{4}+z^{3}+z^{2}+z+1\right)$$
So its roots are the 5th roots of unity, $z=\operatorname{cis}\left(\frac{2k\pi}{5}\right)$, without $z=1$. That leaves $\operatorname{cis}\left(\pm\frac{2\pi}{5}\right)$ and $\operatorname{cis}\left(\pm\frac{4\pi}{5}\right)$. Marking them on a unit circle also earns this mark.
Marker: correctly determines the roots of $z^{5}=1$
$P(z)$ has real coefficients, so by the conjugate root theorem its two complex roots have to be a conjugate pair.
$$z=\operatorname{cis}\left(\frac{2\pi}{5}\right)\quad\text{and}\quad z=\operatorname{cis}\left(-\frac{2\pi}{5}\right)$$
The other pair, $\operatorname{cis}\left(\pm\frac{4\pi}{5}\right)$, works too and gives a second answer. The interactive above shows what goes wrong if you pick two roots that are not conjugates.
Marker: correctly recognises one possible pair of roots
Each root gives a linear factor. Multiply the pair together.
$$\left(z-\operatorname{cis}\left(\frac{2\pi}{5}\right)\right)\left(z-\operatorname{cis}\left(-\frac{2\pi}{5}\right)\right)$$
You can also use decimals such as $\operatorname{cis}(1.257)$. This mark can be implied by your later working.
Marker: determines a quadratic factor of $P(z)$ in factorised form
Expand it. The imaginary parts cancel in the sum, and the product of a conjugate pair on the unit circle is 1.
$$\begin{aligned}&=z^{2}-\left(\operatorname{cis}\left(\frac{2\pi}{5}\right)+\operatorname{cis}\left(-\frac{2\pi}{5}\right)\right)z+\operatorname{cis}(0)\\&=z^{2}-2\cos\left(\tfrac{2\pi}{5}\right)z+1\end{aligned}$$
You keep follow-through marks from here if you slipped earlier.
Marker: expresses the determined quadratic factor of $P(z)$ in expanded form
The remaining root is $z=2$, so $z-2$ is the third factor.
$$P(z)=(z-2)\left(z^{2}-2\cos\left(\tfrac{2\pi}{5}\right)z+1\right)$$
Marker: uses the factor of $z=2$ to express $P(z)$ in factorised form
Expand the product. The question asks for expanded form, so you cannot stop at step 5.
$$P(z)=z^{3}-2\left(\cos\left(\tfrac{2\pi}{5}\right)+1\right)z^{2}+\left(4\cos\left(\tfrac{2\pi}{5}\right)+1\right)z-2$$
$P(z)\approx z^{3}-2.62z^{2}+2.24z-2$
With the other pair you get $P(z)\approx z^{3}-0.38z^{2}-2.24z-2$, which also earns full marks. The QCAA’s third method puts the roots straight into $P(z)$ and solves for $a$, $b$ and $c$ on the calculator.
Marker: determines $P(z)$ in expanded form
Each phrase of the question gave you something
You are not told the roots of $z^{4}+z^{3}+z^{2}+z+1$. You have to use the factorisation of $z^{5}-1$ to find them, and then use the conjugate root theorem to decide which two roots $P(z)$ can share.
I would check your choice of pair first. Picking $\operatorname{cis}\left(\frac{2\pi}{5}\right)$ and $\operatorname{cis}\left(\frac{4\pi}{5}\right)$ gives complex coefficients. The QCAA gave you the factorisation here, but in later years you had to spot it yourself. The factorisation is something that appears on our flash cards.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.