Question 19
Consider complex numbers of the form $w=x+i$, where $x$ is a positive real number.
If $\operatorname{Re}\left(w^{7}\right)=0$, determine all possible values of $x$.
Three points on the line give a w⁷ that points straight up or down
Watch $w$ slide along the line while the dial on the right shows the direction of $w^{7}$. You will see it hit the imaginary axis exactly three times.
As $w$ moves left, $\theta$ grows towards $\frac{\pi}{2}$, and the arrow for $w^{7}$ spins seven times as fast.
Slide w until the arrow for w⁷ is vertical
Move the slider to change $x$. You can see $\arg(w)$ on the left and the direction of $w^{7}$ on the right, and you need that arrow on the imaginary axis.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA's first method, which uses De Moivre's theorem and the argument of $w$.
Write $w$ in polar form so that you can raise it to the 7th power with De Moivre's theorem.
$$w=x+i=r\operatorname{cis}(\theta),\qquad r=|w|=\sqrt{x^{2}+1}$$
De Moivre's theorem says that when you raise $w$ to a power, you raise the modulus to that power and multiply the argument by it.
$$w^{7}=\left(r\operatorname{cis}(\theta)\right)^{7}=r^{7}\operatorname{cis}(7\theta)$$
Writing $\operatorname{cis}$ out in full shows you the real part straight away.
$$w^{7}=r^{7}\cos(7\theta)+i\,r^{7}\sin(7\theta)\quad\Rightarrow\quad\operatorname{Re}\left(w^{7}\right)=r^{7}\cos(7\theta)$$
This mark can be implied by your later working. You could also expand $(x+i)^{7}$ with the binomial theorem, but that leaves you with a degree 7 polynomial to solve.
Marker: correctly uses De Moivre's theorem
$w=x+i$ has real part $x$ and imaginary part 1, and $x>0$ puts it in the first quadrant.
$$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\operatorname{Im}(w)}{\operatorname{Re}(w)}=\frac{1}{x}$$
$$\arg(w)=\theta=\tan^{-1}\left(\frac{1}{x}\right),\quad x\in\mathbb{R}^{+}$$
Because $x$ and 1 are both positive, $w$ is in the first quadrant, so $0<\theta<\frac{\pi}{2}$. You will need this in step 6.
A right-angled triangle with sides $x$ and 1 marked also earns you this mark.
Marker: correctly determines an expression representing $\arg(w)$ in terms of $x$
The 7 comes from step 1. De Moivre's theorem multiplied the argument by 7, so $\arg(w^{7})=7\theta$. Now you use the condition you were given.
$$\operatorname{Re}\left(w^{7}\right)=r^{7}\cos(7\theta)=0$$
$r=\sqrt{x^{2}+1}$ can never be zero, so you need the cosine to be zero. You get that at every odd multiple of $\frac{\pi}{2}$, which is where $w^{7}$ points straight up or straight down.
$$\cos(7\theta)=0\quad\Rightarrow\quad7\theta=\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ \ldots=\frac{(2n+1)\pi}{2}$$
Put your expression for $\theta$ from step 2 into this.
$$\arg(w^{7})=7\tan^{-1}\left(\frac{1}{x}\right)=\frac{(2n+1)\pi}{2},\quad n\in\mathbb{Z}$$
Follow-through marks are allowed here if you made a slip earlier.
Marker: determines a relationship involving $\arg(w^{7})$ using the condition $\operatorname{Re}(w^{7})=0$
You divide both sides by 7 to get $\theta$ on its own. Then you take the tangent of both sides, which undoes the $\tan^{-1}$, and flip the fraction to get $x$.
$$\begin{aligned}\tan^{-1}\left(\frac{1}{x}\right)&=\frac{(2n+1)\pi}{14}\\x&=\frac{1}{\tan\left(\frac{(2n+1)\pi}{14}\right)}=\cot\left(\frac{(2n+1)\pi}{14}\right)\end{aligned}$$
This mark can be implied by your later working, and listing the values directly is also accepted.
Marker: determines a general expression representing possible values of $x$
Because $x>0$, $\theta$ has to be in the first quadrant, so you start with $n=0$.
$$n=0:\quad\theta=\frac{\pi}{14},\qquad x=\cot\left(\frac{\pi}{14}\right)$$
It is worth checking that it works. You get $7\theta=\frac{7\pi}{14}=\frac{\pi}{2}$, so $w^{7}$ points straight up and its real part is 0.
Writing it as $\frac{1}{\tan\left(\frac{\pi}{14}\right)}$ is accepted, and so is any one of the three correct values.
Marker: determines one value of $x$
Use the range of $\theta$ from step 2 to decide which values of $n$ you are allowed.
$$0<\frac{(2n+1)\pi}{14}<\frac{\pi}{2}\quad\Rightarrow\quad0<2n+1<7\quad\Rightarrow\quad n=0,\ 1,\ 2$$
So you get two more values after $n=0$.
$$n=1:\ x=\cot\left(\tfrac{3\pi}{14}\right)\qquad n=2:\ x=\cot\left(\tfrac{5\pi}{14}\right)$$
$x=\cot\left(\tfrac{\pi}{14}\right),\ \cot\left(\tfrac{3\pi}{14}\right),\ \cot\left(\tfrac{5\pi}{14}\right)$
You have to stop here, and you should say why. $n=3$ gives $\theta=\frac{7\pi}{14}=\frac{\pi}{2}$, so $x=0$. Any other value of $n$ puts $\theta$ outside $0<\theta<\frac{\pi}{2}$, and you showed in step 2 that $\theta$ has to be inside that range. If you prefer a picture, the QCAA's other methods place the seven roots of $w^{7}=\pm ai$ on an Argand diagram and keep the three in the first quadrant.
Marker: evaluates the reasonableness of solution by determining the remaining two values of $x$
Each phrase of the question gave you something
Nothing in the question tells you to use polar form. You have to see that a condition on $w^{7}$ is much easier to handle through its argument than by expanding $(x+i)^{7}$.
There are two slips I would watch for here. The first is stopping after $\cot\left(\frac{\pi}{14}\right)$, and the second is including $n=3$, which gives you $x=0$. The QCAA love to say "all possible values", because it makes you decide exactly where your list ends.
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 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.