Question 11
The vertices of a regular hexagon are positioned on the circumference of a unit circle as shown on the Argand plane.
Consider the complex number $w$, as shown on the plane.
Determine $w$, expressing your answer in the form $r\,\text{cis}(\theta)$.
[1 mark] b)Convert $w$ into Cartesian form.
[2 marks]Each vertex of the hexagon is a solution of an equation of the form $z^n=a$ where $z\in\mathbb{C}$.
State the value of $n$.
[1 mark] d)State the value of $a$.
[1 mark] e)Verify that $w$ satisfies the equation $z^n=a$ using the results from 11c) and 11d).
[2 marks]Six vertices, a sixth of a turn apart
Watch how the diagram gives you the modulus and argument of $w$, then what happens to the argument as you raise $w$ to higher powers.
De Moivre’s theorem multiplies the argument by the power. Six lots of $\frac{2\pi}{3}$ is $4\pi$, two full turns, so $w^6$ lands back on $1$.
Find the equation the hexagon solves
The black dots are the vertices of the hexagon and the orange dot is $w$. The blue rings are the solutions of $z^n=a$. Choose $n$ and $a$ so that each vertex is a solution and each solution is a vertex.
Each solution lands on a vertex and each vertex is a solution. This is the equation for parts c) and d), with $n=6$ and $a=1$. The last line of the grey box is the check in part e).
Every solution lands on a vertex, but there are only three solutions. The other three vertices are missed, so this can’t be the equation of the hexagon.
All six vertices are solutions, but so are the six points halfway between them. The hexagon’s equation has exactly one solution at each vertex and no others, so you need fewer solutions.
Six solutions spaced $\frac{\pi}{3}$ apart do make a regular hexagon, but this one is turned away from the hexagon in the question and none of its points is a vertex. The question’s hexagon has a vertex at $1$, and $1^6=1$.
This equation has solutions, equally spaced round the unit circle. A hexagon has six vertices, so this can’t be its equation.
None of these solutions lands on a vertex.
Only one of these solutions lands on a vertex.
Only two of these solutions land on a vertex.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide.
Every vertex is on the unit circle, so $|w|=1$. The six vertices split a full turn into six equal angles of $\frac{2\pi}{6}=\frac{\pi}{3}$. Start at the vertex on the positive real axis and count anticlockwise. $w$ is the second vertex, so its argument is two of those angles.
$$\arg(w)=2\times\frac{\pi}{3}=\frac{2\pi}{3}$$
You can leave the modulus out, as the marking guide does, or write it in as $1\,\text{cis}\left(\frac{2\pi}{3}\right)$. The QCAA also accept the argument in degrees, $w=\text{cis}(120^\circ)$.
Marker: correctly expresses $w$ in $r\,\text{cis}(\theta)$ form
Write the polar form out in full, using $r\,\text{cis}(\theta)=r\cos(\theta)+r\sin(\theta)i$. Here $r=1$.
$$\begin{aligned}w&=\cos\left(\frac{2\pi}{3}\right)\\[4pt]&\quad+\sin\left(\frac{2\pi}{3}\right)i\end{aligned}$$
If you made an error in part a), you can still earn both marks in part b) by converting your own answer correctly.
Marker: converts $w$ into $r\cos(\theta)+r\sin(\theta)i$ form
The marking guide prints this criterion as $r\cos(\theta)+r\cos(\theta)i$. The second cosine is a typo for $r\sin(\theta)i$, as the guide’s own sample response shows.
The angle $\frac{2\pi}{3}$ is in the second quadrant, with a reference angle of $\frac{\pi}{3}$. Cosine is negative there and sine is positive, so use the exact values for $\frac{\pi}{3}$ with those signs.
$$\begin{aligned}\cos\left(\frac{2\pi}{3}\right)&=-\frac12\\[4pt]\sin\left(\frac{2\pi}{3}\right)&=\frac{\sqrt3}{2}\end{aligned}$$
Check it against the diagram. $w$ is left of the imaginary axis and above the real axis, so its real part is negative and its imaginary part is positive.
Marker: uses exact values for $\text{Re}(w)$ and $\text{Im}(w)$
An equation of the form $z^n=a$ has $n$ solutions, equally spaced round a circle centred at the origin. The hexagon has six vertices, so the equation has six solutions.
Every vertex also solves equations such as $z^{12}=1$, but those have extra solutions between the vertices. The marking guide gives only $n=6$, the equation whose solutions are exactly the six vertices.
Marker: correctly states the value of $n$
Every vertex is a solution of $z^6=a$, so you can substitute any vertex to find $a$. The easiest one is the vertex on the positive real axis, $z=1$.
$$a=1^6=1$$
So the vertices are the sixth roots of unity, the solutions of $z^6=1$.
Marker: correctly states the value of $a$
Start by writing down what you need to show. With $n=6$ and $a=1$, you need to verify $w^6=1$. De Moivre’s theorem raises the modulus to the power and multiplies the argument by it.
$$\begin{aligned}w^6&=\left(\text{cis}\left(\frac{2\pi}{3}\right)\right)^6\\[4pt]&=\text{cis}\left(6\times\frac{2\pi}{3}\right)\\[4pt]&=\text{cis}(4\pi)\end{aligned}$$
The QCAA also accept other methods, such as a binomial expansion of the Cartesian form from part b). If your $n$ was wrong, you can still earn this mark by using De Moivre’s theorem correctly with your own $n$.
Marker: uses De Moivre’s theorem in the calculation of $w^n$ for the value of $n$
An argument of $4\pi$ is two full turns, so it points the same way as an argument of $0$.
$$\text{cis}(4\pi)=\text{cis}(0)=1$$
Finish by comparing the result with $a$ and saying that the result is verified.
If an earlier error means your $w^n$ doesn’t equal your $a$, you can still earn this mark for a correct calculation. Part e) must still state what you were trying to verify, such as $w^6=1$.
Marker: verifies the result by showing that the result of the calculation is the value of $a$
Each phrase of the question gave you something
Roots of unity on an Argand diagram are a routine you have practised, and the diagram hands you the modulus and the spacing of the vertices. The exact values of $\cos\frac{2\pi}{3}$ and $\sin\frac{2\pi}{3}$ are ones you know without a calculator.
If I had to guess where you might lose marks, it would be the argument in part a). Counting one step round instead of two gives $\frac{\pi}{3}$, which is the vertex in the first quadrant. In part e), write down what you are verifying before you start. A line that stops at $\text{cis}(4\pi)$ leaves the last mark behind, so simplify it to $1$ and compare it with $a$.
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 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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