Question 18
Consider the complex solutions to the following equation, where $0<\arg(z)<\pi$.
$$(z+1)\left(z^{14}-z^{13}+z^{12}-z^{11}+\cdots+z^{4}-z^{3}+z^{2}-z\right)=1-z$$
Let $w_{1}$ be the solution with the maximum possible real part and $w_{2}$ be the solution with the maximum possible imaginary part.
Show that $\dfrac{w_{1}^{4}}{w_{2}}\in\mathbb{Z}$.
Four turns of w₁ land exactly on w₂
Watch the equation turn into fifteen points on a circle. You will see which two points are $w_{1}$ and $w_{2}$, and why dividing them gives you 1.
The roots are 24° apart, so each extra power of $w_{1}$ moves you forward by one root.
Pick w₂ carefully, then raise w₁ to a power
Choose which root you think is $w_{2}$, then move the slider to change the power of $w_{1}$. You only get an integer when the power lands you exactly on $w_{2}$.
Work through the proof one mark at a time
Try each step yourself before you reveal it. You can check decimals on your calculator, but you should keep the roots in polar form.
Expand the left-hand side. The signs alternate, so when you multiply by $z+1$ almost every term cancels with its neighbour.
$$\begin{aligned}(z+1)\left(z^{14}-\cdots-z\right)&=\left(z^{15}-z^{14}+\cdots-z^{2}\right)\\&\quad+\left(z^{14}-z^{13}+\cdots-z\right)\\&=z^{15}-z\end{aligned}$$
$$z^{15}-z=1-z\quad\Rightarrow\quad z^{15}=1$$
Only the first and last terms survive. Adding $z$ to both sides turns the whole equation into $z^{15}=1$, so the solutions are the 15th roots of unity.
Marker: correctly simplifies the original equation
The solutions of $z^{15}=1$ are the 15th roots of unity, spaced $\frac{2\pi}{15}$ apart around the unit circle.
$$z=\operatorname{cis}\left(\frac{2\pi n}{15}\right),\quad n\in\mathbb{Z}$$
The condition $0<\arg(z)<\pi$ keeps you in the top half, so you only need $n=1$ to $n=7$. A sketch of equally spaced roots on an Argand diagram also earns you this mark.
Marker: describes the location of the solutions
On the unit circle the real part is $\cos(\arg z)$, so you want the argument closest to 0.
$$w_{1}=\operatorname{cis}\left(\frac{2\pi}{15}\right)$$
You can also give it as $\operatorname{cis}(24^{\circ})$ or about $0.91+0.41i$. This mark can be implied by your later working.
Marker: determines $w_{1}$
The imaginary part is $\sin(\arg z)$, so you want the argument closest to $\frac{\pi}{2}$. Be careful here, because two roots sit either side of it.
$$\frac{6\pi}{15}=72^{\circ}\qquad\frac{8\pi}{15}=96^{\circ}$$
96° is only 6° from 90°, but 72° is 18° away, so you take the larger argument.
$$w_{2}=\operatorname{cis}\left(\frac{8\pi}{15}\right)$$
You can also give it as $\operatorname{cis}(96^{\circ})$ or about $-0.10+0.99i$.
Marker: determines $w_{2}$
Use De Moivre's theorem to raise $w_{1}$ to the fourth power, then divide.
$$\frac{w_{1}^{4}}{w_{2}}=\frac{\operatorname{cis}\left(\frac{8\pi}{15}\right)}{\operatorname{cis}\left(\frac{8\pi}{15}\right)}=1$$
$\dfrac{w_{1}^{4}}{w_{2}}=1\in\mathbb{Z}$
You only get this mark if you actually reach an integer, so you need to finish with the 1 and say that it is in $\mathbb{Z}$.
Marker: shows that $\frac{w_{1}^{4}}{w_{2}}$ is an integer
Each phrase of the question gave you something
The equation looks like a degree 15 polynomial that you could never solve by hand. You have to expand it and see that almost everything cancels, which turns it into a question about 15th roots of unity.
I would expect some of you to choose $\operatorname{cis}\left(\frac{6\pi}{15}\right)$ for $w_{2}$, because 72° looks close to 90°. The QCAA like to make you compare two neighbouring roots, so you need to check which argument is actually closer to $\frac{\pi}{2}$.
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 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.